## SectionLISSLinear Independence and Spanning Sets

A vector space is defined as a set with two operations, meeting ten properties (Definition VS). Just as the definition of span of a set of vectors only required knowing how to add vectors and how to multiply vectors by scalars, so it is with linear independence. A definition of a linearly independent set of vectors in an arbitrary vector space only requires knowing how to form linear combinations and equating these with the zero vector. Since every vector space must have a zero vector (Property Z), we always have a zero vector at our disposal.

In this section we will also put a twist on the notion of the span of a set of vectors. Rather than beginning with a set of vectors and creating a subspace that is the span, we will instead begin with a subspace and look for a set of vectors whose span equals the subspace.

The combination of linear independence and spanning will be very important going forward.

### SubsectionLILinear Independence

Our previous definition of linear independence (Definition LICV) employed a relation of linear dependence that was a linear combination on one side of an equality and a zero vector on the other side. As a linear combination in a vector space (Definition LC) depends only on vector addition and scalar multiplication, and every vector space must have a zero vector (Property Z), we can extend our definition of linear independence from the setting of $$\complex{m}$$ to the setting of a general vector space $$V$$ with almost no changes. Compare these next two definitions with Definition RLDCV and Definition LICV.

#### DefinitionRLD.Relation of Linear Dependence.

Suppose that $$V$$ is a vector space. Given a set of vectors $$S=\set{\vectorlist{u}{n}}\text{,}$$ an equation of the form

\begin{equation*} \lincombo{\alpha}{u}{n}=\zerovector \end{equation*}

is a relation of linear dependence on $$S\text{.}$$ If this equation is formed in a trivial fashion, i.e. $$\alpha_i=0\text{,}$$ $$1\leq i\leq n\text{,}$$ then we say it is a trivial relation of linear dependence on $$S\text{.}$$

#### DefinitionLI.Linear Independence.

Suppose that $$V$$ is a vector space. The set of vectors $$S=\set{\vectorlist{u}{n}}$$ from $$V$$ is linearly dependent if there is a relation of linear dependence on $$S$$ that is not trivial. In the case where the only relation of linear dependence on $$S$$ is the trivial one, then $$S$$ is a linearly independent set of vectors.

Notice the emphasis on the word “only.” This might remind you of the definition of a nonsingular matrix, where if the matrix is employed as the coefficient matrix of a homogeneous system then the only solution is the trivial one.

In the vector space of polynomials with degree 4 or less, $$P_4$$ (Example VSP) consider the set $$S$$ below.

\begin{gather*} \set{ 2x^4+3x^3+2x^2-x+10,\, -x^4-2x^3+x^2+5x-8,\, 2x^4+x^3+10x^2+17x-2 }\text{.} \end{gather*}

Is this set of vectors linearly independent or dependent? Consider that

\begin{align*} &3\left(2x^4+3x^3+2x^2-x+10\right) +4\left(-x^4-2x^3+x^2+5x-8\right)\\ &\quad +(-1)\left(2x^4+x^3+10x^2+17x-2\right) =0x^4+0x^3+0x^2+0x+0=\zerovector\text{.} \end{align*}

This is a nontrivial relation of linear dependence (Definition RLD) on the set $$S$$ and so convinces us that $$S$$ is linearly dependent (Definition LI).

Now, I hear you say, “Where did those scalars come from?” Do not worry about that right now, just be sure you understand why the above explanation is sufficient to prove that $$S$$ is linearly dependent. The remainder of the example will demonstrate how we might find these scalars if they had not been provided so readily.

Let us look at another set of vectors (polynomials) from $$P_4\text{.}$$ Let

\begin{align*} T&=\left\{ 3x^4-2x^3+4x^2+6x-1,\, -3x^4+1x^3+0x^2+4x+2,\right.\\ &\quad \left.4x^4+5x^3-2x^2+3x+1,\, 2x^4-7x^3+4x^2+2x+1\right\}\text{.} \end{align*}

Suppose we have a relation of linear dependence on this set,

\begin{align*} \zerovector&=0x^4+0x^3+0x^2+0x+0\\ &=\alpha_1\left(3x^4-2x^3+4x^2+6x-1\right)+\alpha_2\left(-3x^4+1x^3+0x^2+4x+2\right)\\ &\quad +\alpha_3\left(4x^4+5x^3-2x^2+3x+1\right)+\alpha_4\left(2x^4-7x^3+4x^2+2x+1\right)\text{.} \end{align*}

Using our definitions of vector addition and scalar multiplication in $$P_4$$ (Example VSP), we arrive at,

Equating coefficients, we arrive at the homogeneous system of equations,

\begin{align*} 3\alpha_1-3\alpha_2+4\alpha_3+2\alpha_4&=0\\ -2\alpha_1+\alpha_2+5\alpha_3-7\alpha_4&=0\\ 4\alpha_1-2\alpha_3+4\alpha_4&=0\\ 6\alpha_1+4\alpha_2+3\alpha_3+2\alpha_4&=0\\ -\alpha_1+2\alpha_2+\alpha_3+\alpha_4&=0\text{.} \end{align*}

We form the coefficient matrix of this homogeneous system of equations and row-reduce to find

\begin{equation*} \begin{bmatrix} \leading{1} & 0 & 0 & 0\\ 0 & \leading{1} & 0 & 0\\ 0 & 0 & \leading{1} & 0\\ 0 & 0 & 0 & \leading{1}\\ 0 & 0 & 0 & 0 \end{bmatrix}\text{.} \end{equation*}

We expected the system to be consistent (Theorem HSC) and so can compute $$n-r=4-4=0$$ and Theorem CSRN tells us that the solution is unique. Since this is a homogeneous system, this unique solution is the trivial solution (Definition TSHSE), $$\alpha_1=0\text{,}$$ $$\alpha_2=0\text{,}$$ $$\alpha_3=0\text{,}$$ $$\alpha_4=0\text{.}$$ So by Definition LI the set $$T$$ is linearly independent.

A few observations. If we had discovered infinitely many solutions, then we could have used one of the nontrivial solutions to provide a linear combination in the manner we used to show that $$S$$ was linearly dependent. It is important to realize that it is not interesting that we can create a relation of linear dependence with zero scalars — we can always do that — but for $$T\text{,}$$ this is the only way to create a relation of linear dependence. It was no accident that we arrived at a homogeneous system of equations in this example, it is related to our use of the zero vector in defining a relation of linear dependence. It is easy to present a convincing statement that a set is linearly dependent (just exhibit a nontrivial relation of linear dependence) but a convincing statement of linear independence requires demonstrating that there is no relation of linear dependence other than the trivial one. Notice how we relied on theorems from Chapter SLE to provide this demonstration. Whew! There is a lot going on in this example. Spend some time with it, we will be waiting patiently right here when you get back.

Consider the two sets of vectors $$R$$ and $$S$$ from the vector space of all $$3\times 2$$ matrices, $$M_{32}$$ (Example VSM).

\begin{align*} R&=\set{ \begin{bmatrix} 3 & -1\\1 & 4\\6 & -6 \end{bmatrix},\, \begin{bmatrix} -2 & 3\\1 & -3\\-2 & -6 \end{bmatrix},\, \begin{bmatrix} 6 & -6\\-1 & 0\\7 & -9 \end{bmatrix},\, \begin{bmatrix} 7 & 9\\-4 & -5\\2 & 5 \end{bmatrix} }\\ S&=\set{ \begin{bmatrix} 2 & 0\\ 1 & -1\\ 1 & 3 \end{bmatrix},\, \begin{bmatrix} -4 & 0\\ -2 & 2\\ -2 & -6 \end{bmatrix},\, \begin{bmatrix} 1 & 1\\ -2 & 1\\ 2 & 4 \end{bmatrix},\, \begin{bmatrix} -5 & 3\\ -10 & 7\\ 2 & 0 \end{bmatrix} }\text{.} \end{align*}

One set is linearly independent, the other is not. Which is which? Let us examine $$R$$ first. Build a generic relation of linear dependence (Definition RLD),

\begin{equation*} \alpha_1\begin{bmatrix} 3 & -1\\1 & 4\\6 & -6 \end{bmatrix}+ \alpha_2\begin{bmatrix} -2 & 3\\1 & -3\\-2 & -6 \end{bmatrix}+ \alpha_3\begin{bmatrix} 6 & -6\\-1 & 0\\7 & -9 \end{bmatrix}+ \alpha_4\begin{bmatrix} 7 & 9\\-4 & -5\\2 & 5 \end{bmatrix}= \zerovector\text{.} \end{equation*}

Massaging the left-hand side with our definitions of vector addition and scalar multiplication in $$M_{32}$$ (Example VSM) we obtain,

\begin{equation*} \begin{bmatrix} 3\alpha_1-2\alpha_2+6\alpha_3+7\alpha_4 & -\alpha_1+3\alpha_2-6\alpha_3+9\alpha_4 \\ \alpha_1+\alpha_2-\alpha_3-4\alpha_4 & 4\alpha_1-3\alpha_2+ -5\alpha_4 \\ 6\alpha_1-2\alpha_2+7\alpha_3+2\alpha_4 & -6\alpha_1-6\alpha_2-9\alpha_3+5\alpha_4 \end{bmatrix} =\begin{bmatrix} 0&0\\0&0\\0&0 \end{bmatrix}\text{.} \end{equation*}

Using our definition of matrix equality (Definition ME) to equate entries we get the homogeneous system of six equations in four variables,

\begin{align*} 3\alpha_1-2\alpha_2+6\alpha_3+7\alpha_4&=0\\ -\alpha_1+3\alpha_2-6\alpha_3+9\alpha_4&=0\\ \alpha_1+\alpha_2-\alpha_3-4\alpha_4&=0\\ 4\alpha_1-3\alpha_2+ -5\alpha_4&=0\\ 6\alpha_1-2\alpha_2+7\alpha_3+2\alpha_4&=0\\ -6\alpha_1-6\alpha_2-9\alpha_3+5\alpha_4&=0\text{.} \end{align*}

Form the coefficient matrix of this homogeneous system and row-reduce to obtain

\begin{equation*} \begin{bmatrix} \leading{1} & 0 & 0 & 0\\ 0 & \leading{1} & 0 & 0\\ 0 & 0 & \leading{1} & 0\\ 0 & 0 & 0 & \leading{1}\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}\text{.} \end{equation*}

Analyzing this matrix we are led to conclude that $$\alpha_1=0\text{,}$$ $$\alpha_2=0\text{,}$$ $$\alpha_3=0\text{,}$$ $$\alpha_4=0\text{.}$$ This means there is only a trivial relation of linear dependence on the vectors of $$R$$ and so we call $$R$$ a linearly independent set (Definition LI).

So it must be that $$S$$ is linearly dependent. Let us see if we can find a nontrivial relation of linear dependence on $$S\text{.}$$ We will begin as with $$R\text{,}$$ by constructing a relation of linear dependence (Definition RLD) with unknown scalars,

\begin{equation*} \alpha_1\begin{bmatrix} 2 & 0\\ 1 & -1\\ 1 & 3 \end{bmatrix}+ \alpha_2\begin{bmatrix} -4 & 0\\ -2 & 2\\ -2 & -6 \end{bmatrix}+ \alpha_3\begin{bmatrix} 1 & 1\\ -2 & 1\\ 2 & 4 \end{bmatrix}+ \alpha_4\begin{bmatrix} -5 & 3\\ -10 & 7\\ 2 & 0 \end{bmatrix}= \zerovector\text{.} \end{equation*}

Massaging the left-hand side with our definitions of vector addition and scalar multiplication in $$M_{32}$$ (Example VSM) we obtain,

\begin{equation*} \begin{bmatrix} 2\alpha_1-4\alpha_2+\alpha_3-5\alpha_4& \alpha_3+3\alpha_4\\ \alpha_1-2\alpha_2-2\alpha_3-10\alpha_4& -\alpha_1+2\alpha_2+\alpha_3+7\alpha_4\\ \alpha_1-2\alpha_2+2\alpha_3+2\alpha_4& 3\alpha_1-6\alpha_2+4\alpha_3 \end{bmatrix} =\begin{bmatrix} 0&0\\0&0\\0&0 \end{bmatrix}\text{.} \end{equation*}

Using our definition of matrix equality (Definition ME) to equate entries we get the homogeneous system of six equations in four variables,

\begin{align*} 2\alpha_1-4\alpha_2+\alpha_3-5\alpha_4&=0\\ \alpha_3+3\alpha_4&=0\\ \alpha_1-2\alpha_2-2\alpha_3-10\alpha_4&=0\\ -\alpha_1+2\alpha_2+\alpha_3+7\alpha_4&=0\\ \alpha_1-2\alpha_2+2\alpha_3+2\alpha_4&=0\\ 3\alpha_1-6\alpha_2+4\alpha_3 &=0\text{.} \end{align*}

Form the coefficient matrix of this homogeneous system and row-reduce to obtain

\begin{equation*} \begin{bmatrix} \leading{1} & -2 & 0 & -4\\ 0 & 0 & \leading{1} & 3\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}\text{.} \end{equation*}

Analyzing this we see that the system is consistent (we expected this since the system is homogeneous, Theorem HSC) and has $$n-r=4-2=2$$ free variables, namely $$\alpha_2$$ and $$\alpha_4\text{.}$$ This means there are infinitely many solutions, and in particular, we can find a nontrivial solution, so long as we do not pick all of our free variables to be zero. The mere presence of a nontrivial solution for these scalars is enough to conclude that $$S$$ is a linearly dependent set (Definition LI). But let us go ahead and explicitly construct a nontrivial relation of linear dependence.

Choose $$\alpha_2=1$$ and $$\alpha_4=-1\text{.}$$ There is nothing special about this choice, there are infinitely many possibilities, some “easier” than this one, just avoid picking both variables to be zero. (Why not?) Then we find the dependent variables to have values $$\alpha_1=-2$$ and $$\alpha_3=3\text{.}$$ So the relation of linear dependence,

\begin{equation*} (-2)\begin{bmatrix} 2 & 0\\ 1 & -1\\ 1 & 3 \end{bmatrix}+ (1)\begin{bmatrix} -4 & 0\\ -2 & 2\\ -2 & -6 \end{bmatrix}+ (3)\begin{bmatrix} 1 & 1\\ -2 & 1\\ 2 & 4 \end{bmatrix}+ (-1)\begin{bmatrix} -5 & 3\\ -10 & 7\\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 0&0\\0&0\\0&0 \end{bmatrix} \end{equation*}

is an iron-clad demonstration that $$S$$ is linearly dependent. Can you construct another such demonstration?

Is the set $$R=\set{(1,\,0),\,(6,\,3)}$$ linearly independent in the crazy vector space $$C$$ (Example CVS)?

We begin with an arbitrary relation of linear dependence on $$R$$

\begin{align*} \zerovector &= a_1(1,\,0) + a_2(6,\,3)&& \knowl{./knowl/definition-RLD.html}{\text{Definition RLD}} \end{align*}

and then massage it to a point where we can apply the definition of equality in $$C\text{.}$$ Recall the definitions of vector addition and scalar multiplication in $$C$$ are not what you would expect.

\begin{align*} (&-1,\,-1)\\ &=\zerovector&& \knowl{./knowl/example-CVS.html}{\text{Example CVS}}\\ &=a_1(1,\,0) + a_2(6,\,3)&& \knowl{./knowl/definition-RLD.html}{\text{Definition RLD}}\\ &=(1a_1+a_1-1,\,0a_1+a_1-1) + (6a_2+a_2-1,\,3a_2+a_2-1)&& \knowl{./knowl/example-CVS.html}{\text{Example CVS}}\\ &=(2a_1-1,\,a_1-1) + (7a_2-1,\,4a_2-1)\\ &=(2a_1-1+7a_2-1+1,\,a_1-1+4a_2-1+1)&& \knowl{./knowl/example-CVS.html}{\text{Example CVS}}\\ &=(2a_1+7a_2-1,\,a_1+4a_2-1) \end{align*}

Equality in $$C$$ (Example CVS) then yields the two equations,

\begin{align*} 2a_1+7a_2-1&=-1\\ a_1+4a_2-1&=-1 \end{align*}

which becomes the homogeneous system

\begin{align*} 2a_1+7a_2&=0\\ a_1+4a_2&=0\text{.} \end{align*}

Since the coefficient matrix of this system is nonsingular (check this!) the system has only the trivial solution $$a_1=a_2=0\text{.}$$ By Definition LI the set $$R$$ is linearly independent. Notice that even though the zero vector of $$C$$ is not what we might have first suspected, a question about linear independence still concludes with a question about a homogeneous system of equations. Hmmm.

### SubsectionSSSpanning Sets

In a vector space $$V\text{,}$$ suppose we are given a set of vectors $$S\subseteq V\text{.}$$ Then we can immediately construct a subspace, $$\spn{S}\text{,}$$ using Definition SS and then be assured by Theorem SSS that the construction does provide a subspace. We now turn the situation upside-down. Suppose we are first given a subspace $$W\subseteq V\text{.}$$ Can we find a set $$S$$ so that $$\spn{S}=W\text{?}$$ Typically $$W$$ is infinite and we are searching for a finite set of vectors $$S$$ that we can combine in linear combinations and “build” all of $$W\text{.}$$

I like to think of $$S$$ as the raw materials that are sufficient for the construction of $$W\text{.}$$ If you have nails, lumber, wire, copper pipe, drywall, plywood, carpet, shingles, paint (and a few other things), then you can combine them in many different ways to create a house (or infinitely many different houses for that matter). A fast-food restaurant may have beef, chicken, beans, cheese, tortillas, taco shells and hot sauce and from this small list of ingredients build a wide variety of items for sale. Or maybe a better analogy comes from Ben Cordes — the additive primary colors (red, green and blue) can be combined to create many different colors by varying the intensity of each. The intensity is like a scalar multiple, and the combination of the three intensities is like vector addition. The three individual colors, red, green and blue, are the elements of the spanning set.

Because we will use terms like “spanned by” and “spanning set,” there is the potential for confusion with “the span.” Come back and reread the first paragraph of this subsection whenever you are uncertain about the difference. Here is the working definition.

#### DefinitionSSVS.Spanning Set of a Vector Space.

Suppose $$V$$ is a vector space. A subset $$S$$ of $$V$$ is a spanning set of $$V$$ if $$\spn{S}=V\text{.}$$ In this case, we also frequently say $$S$$ spans $$V\text{.}$$

The definition of a spanning set requires that two sets (subspaces actually) be equal. If $$S$$ is a subset of $$V\text{,}$$ then $$\spn{S}\subseteq V\text{,}$$ always. Thus it is usually only necessary to prove that $$V\subseteq\spn{S}\text{.}$$ Now would be a good time to review Definition SE.

In Example SP4 we showed that

\begin{equation*} W=\setparts{p(x)}{p\in P_4,\ p(2)=0} \end{equation*}

is a subspace of $$P_4\text{,}$$ the vector space of polynomials with degree at most $$4$$ (Example VSP). In this example, we will show that the set

\begin{equation*} S=\set{x-2,\,x^2-4x+4,\,x^3-6x^2+12x-8,\,x^4-8x^3+24x^2-32x+16} \end{equation*}

is a spanning set for $$W\text{.}$$ To do this, we require that $$W=\spn{S}\text{.}$$ This is an equality of sets. We can check that every polynomial in $$S$$ has $$x=2$$ as a root and therefore $$S\subseteq W\text{.}$$ Since $$W$$ is closed under addition and scalar multiplication, $$\spn{S}\subseteq W$$ also.

So it remains to show that $$W\subseteq \spn{S}$$ (Definition SE). To do this, begin by choosing an arbitrary polynomial in $$W\text{,}$$ say $$r(x)=ax^4+bx^3+cx^2+dx+e\in W\text{.}$$ This polynomial is not as arbitrary as it would appear, since we also know it must have $$x=2$$ as a root. This translates to

\begin{equation*} 0=a(2)^4+b(2)^3+c(2)^2+d(2)+e=16a+8b+4c+2d+e \end{equation*}

as a condition on $$r\text{.}$$

We wish to show that $$r$$ is a polynomial in $$\spn{S}\text{,}$$ that is, we want to show that $$r$$ can be written as a linear combination of the vectors (polynomials) in $$S\text{.}$$ So let us try.

\begin{align*} r(x)&=ax^4+bx^3+cx^2+dx+e\\ &=\alpha_1\left(x-2\right)+\alpha_2\left(x^2-4x+4\right)+\alpha_3\left(x^3-6x^2+12x-8\right)\\ &\quad +\alpha_4\left(x^4-8x^3+24x^2-32x+16\right)\\ &=\alpha_4x^4+ \left(\alpha_3-8\alpha_4\right)x^3+ \left(\alpha_2-6\alpha_3+24\alpha_4\right)x^2\\ &\quad + \left(\alpha_1-4\alpha_2+12\alpha_3-32\alpha_4\right)x+ \left(-2\alpha_1+4\alpha_2-8\alpha_3+16\alpha_4\right) \end{align*}

Equating coefficients (vector equality in $$P_4$$) gives the system of five equations in four variables,

\begin{align*} \alpha_4&=a\\ \alpha_3-8\alpha_4&=b\\ \alpha_2-6\alpha_3+24\alpha_4&=c\\ \alpha_1-4\alpha_2+12\alpha_3-32\alpha_4&=d\\ -2\alpha_1+4\alpha_2-8\alpha_3+16\alpha_4&=e\text{.} \end{align*}

Any solution to this system of equations will provide the linear combination we need to determine if $$r\in\spn{S}\text{,}$$ but we need to be convinced there is a solution for any values of $$a,\,b,\,c,\,d,\,e$$ that qualify $$r$$ to be a member of $$W\text{.}$$ So the question is: is this system of equations consistent? We will form the augmented matrix, and row-reduce. (We probably need to do this by hand, since the matrix is symbolic — reversing the order of the first four rows is the best way to start). We obtain a matrix in reduced row-echelon form

For your results to match our first matrix, you may find it necessary to multiply the final row of your row-reduced matrix by the appropriate scalar, and/or add multiples of this row to some of the other rows. To obtain the second version of the matrix, the last entry of the last column has been simplified to zero according to the one condition we were able to impose on an arbitrary polynomial from $$W\text{.}$$ Since the last column is not a pivot column, Theorem RCLS tells us this system is consistent. Therefore, any polynomial from $$W$$ can be written as a linear combination of the polynomials in $$S\text{,}$$ so $$W\subseteq\spn{S}\text{.}$$ Therefore, $$W=\spn{S}$$ and $$S$$ is a spanning set for $$W$$ by Definition SSVS.

Notice that an alternative to row-reducing the augmented matrix by hand would be to appeal to Theorem FS by expressing the column space of the coefficient matrix as a null space, and then verifying that the condition on $$r$$ guarantees that $$r$$ is in the column space, thus implying that the system is always consistent. Give it a try, we will wait. This has been a complicated example, but worth studying carefully.

Given a subspace and a set of vectors, as in Example SSP4 it can take some work to determine that the set actually is a spanning set. An even harder problem is to be confronted with a subspace and required to construct a spanning set with no guidance. We will now work an example of this flavor, but some of the steps will be unmotivated. Fortunately, we will have some better tools for this type of problem later on.

In the space of all $$2\times 2$$ matrices, $$M_{22}$$ consider the subspace

\begin{equation*} Z=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{a+3b-c-5d=0,\ -2a-6b+3c+14d=0} \end{equation*}

and find a spanning set for $$Z\text{.}$$

We need to construct a limited number of matrices in $$Z$$ so that every matrix in $$Z$$ can be expressed as a linear combination of this limited number of matrices. Suppose that $$B=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ is a matrix in $$Z\text{.}$$ Then we can form a column vector with the entries of $$B$$ and write

\begin{equation*} \colvector{a\\b\\c\\d}\in\nsp{\begin{bmatrix}1 & 3 & -1 & -5\\-2 & -6 & 3 & 14\end{bmatrix}}\text{.} \end{equation*}

Row-reducing this matrix and applying Theorem REMES we obtain the equivalent statement,

\begin{equation*} \colvector{a\\b\\c\\d}\in\nsp{\begin{bmatrix}\leading{1} & 3 & 0 & -1\\0 & 0 & \leading{1} & 4\end{bmatrix}}\text{.} \end{equation*}

We can then express the subspace $$Z$$ in the following equal forms,

\begin{align*} Z&=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{a+3b-c-5d=0,\ -2a-6b+3c+14d=0}\\ &=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{a+3b-d=0,\ c+4d=0}\\ &=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{a=-3b+d,\ c=-4d}\\ &=\setparts{\begin{bmatrix}-3b+d&b\\-4d&d\end{bmatrix}}{b,\,d\in\complexes}\\ &=\setparts{ \begin{bmatrix}-3b&b\\0&0\end{bmatrix}+ \begin{bmatrix}d&0\\-4d&d\end{bmatrix} }{b,\,d\in\complexes}\\ &=\setparts{ b\begin{bmatrix}-3&1\\0&0\end{bmatrix}+ d\begin{bmatrix}1&0\\-4&1\end{bmatrix} }{b,\,d\in\complexes}\\ &=\spn{\set{ \begin{bmatrix}-3&1\\0&0\end{bmatrix},\, \begin{bmatrix}1&0\\-4&1\end{bmatrix} }}\text{.} \end{align*}

So the set

\begin{equation*} Q=\set{ \begin{bmatrix}-3&1\\0&0\end{bmatrix},\, \begin{bmatrix}1&0\\-4&1\end{bmatrix} } \end{equation*}

spans $$Z$$ by Definition SSVS.

In Example LIC we determined that the set $$R=\set{(1,\,0),\,(6,\,3)}$$ is linearly independent in the crazy vector space $$C$$ (Example CVS). We now show that $$R$$ is a spanning set for $$C\text{.}$$

Given an arbitrary vector $$(x,\,y)\in C$$ we desire to show that it can be written as a linear combination of the elements of $$R\text{.}$$ In other words, are there scalars $$a_1$$ and $$a_2$$ so that

\begin{equation*} (x,\,y)=a_1(1,\,0) + a_2(6,\,3)\text{.} \end{equation*}

We will act as if this equation is true and try to determine just what $$a_1$$ and $$a_2$$ would be (as functions of $$x$$ and $$y$$). Recall that our vector space operations are unconventional and are defined in Example CVS.

\begin{align*} (x,\,y)&=a_1(1,\,0) + a_2(6,\,3)\\ &= (1a_1+a_1-1,\,0a_1+a_1-1) + (6a_2+a_2-1,\,3a_2+a_2-1)\\ &= (2a_1-1,\,a_1-1) + (7a_2-1,\,4a_2-1)\\ &= (2a_1-1+7a_2-1+1,\,a_1-1+4a_2-1+1)\\ &= (2a_1+7a_2-1,\,a_1+4a_2-1) \end{align*}

Equality in $$C$$ then yields the two equations,

\begin{align*} 2a_1+7a_2-1&=x\\ a_1+4a_2-1&=y \end{align*}

which becomes the linear system with a matrix representation

\begin{equation*} \begin{bmatrix} 2 & 7 \\ 1 & 4 \end{bmatrix} \colvector{a_1\\a_2} = \colvector{x+1\\y+1}\text{.} \end{equation*}

The coefficient matrix of this system is nonsingular, hence invertible (Theorem NI), and we can employ its inverse to find a solution (Theorem TTMI, Theorem SNCM),

\begin{equation*} \colvector{a_1\\a_2}= \inverse{\begin{bmatrix} 2 & 7 \\ 1 & 4 \end{bmatrix}}\colvector{x+1\\y+1}= \begin{bmatrix} 4 & -7 \\ -1 & 2 \end{bmatrix}\colvector{x+1\\y+1}= \colvector{4x-7y-3\\-x+2y+1}\text{.} \end{equation*}

We could chase through the above implications backwards and take the existence of these solutions as sufficient evidence for $$R$$ being a spanning set for $$C\text{.}$$ Instead, let us view the above as simply scratchwork and now get serious with a simple direct proof that $$R$$ is a spanning set. Ready? Suppose $$(x,\,y)$$ is any vector from $$C\text{,}$$ then compute the following linear combination using the definitions of the operations in $$C\text{,}$$

\begin{align*} (4x&-7y-3)(1,\,0)+(-x+2y+1)(6,\,3)\\ &=\left(1(4x-7y-3)+(4x-7y-3)-1,\,0(4x-7y-3)+(4x-7y-3)-1\right)+\\ &\quad\left(6(-x+2y+1)+(-x+2y+1)-1,\,3(-x+2y+1)+(-x+2y+1)-1\right)\\ &=(8x-14y-7,\,4x-7y-4)+(-7x+14y+6,\,-4x+8y+3)\\ &=((8x-14y-7)+(-7x+14y+6)+1,\,(4x-7y-4)+(-4x+8y+3)+1)\\ &=(x,\,y)\text{.} \end{align*}

This final sequence of computations in $$C$$ is sufficient to demonstrate that any element of $$C$$ can be written (or expressed) as a linear combination of the two vectors in $$R\text{,}$$ so $$C\subseteq\spn{R}\text{.}$$ Since the reverse inclusion $$\spn{R}\subseteq C$$ is trivially true, $$C=\spn{R}$$ and we say $$R$$ spans $$C$$ (Definition SSVS). Notice that this demonstration is no more or less valid if we hide from the reader our scratchwork that suggested $$a_1=4x-7y-3$$ and $$a_2=-x+2y+1\text{.}$$

### SubsectionVRVector Representation

In Chapter R we will take up the matter of representations fully, where Theorem VRRB will be critical for Definition VR. We will now motivate and prove a critical theorem that tells us how to “represent” a vector. This theorem could wait, but working with it now will provide some extra insight into the nature of linearly independent spanning sets. First an example, then the theorem.

Consider the set

\begin{equation*} S=\set{\colvector{-7\\5\\1},\,\colvector{-6\\5\\0},\,\colvector{-12\\7\\4}} \end{equation*}

from the vector space $$\complex{3}\text{.}$$ Let $$A$$ be the matrix whose columns are the set $$S\text{,}$$ and verify that $$A$$ is nonsingular. By Theorem NMLIC the elements of $$S$$ form a linearly independent set. Suppose that $$\vect{b}\in\complex{3}\text{.}$$ Then $$\linearsystem{A}{\vect{b}}$$ has a (unique) solution (Theorem NMUS) and hence is consistent. By Theorem SLSLC, $$\vect{b}\in\spn{S}\text{.}$$ Since $$\vect{b}$$ is arbitrary, this is enough to show that $$\spn{S}=\complex{3}\text{,}$$ and therefore $$S$$ is a spanning set for $$\complex{3}$$ (Definition SSVS). (This set comes from the columns of the coefficient matrix of Archetype B.)

Now examine the situation for a particular choice of $$\vect{b}\text{,}$$ say $$\vect{b}=\colvector{-33\\24\\5}\text{.}$$ Because $$S$$ is a spanning set for $$\complex{3}\text{,}$$ we know we can write $$\vect{b}$$ as a linear combination of the vectors in $$S\text{,}$$

\begin{equation*} \colvector{-33\\24\\5}=(-3)\colvector{-7\\5\\1}+(5)\colvector{-6\\5\\0}+(2)\colvector{-12\\7\\4}\text{.} \end{equation*}

The nonsingularity of the matrix $$A$$ tells that the scalars in this linear combination are unique. More precisely, it is the linear independence of $$S$$ that provides the uniqueness. We will refer to the scalars $$a_1=-3\text{,}$$ $$a_2=5\text{,}$$ $$a_3=2$$ as a “representation of $$\vect{b}$$ relative to $$S\text{.}$$” In other words, once we settle on $$S$$ as a linearly independent set that spans $$\complex{3}\text{,}$$ the vector $$\vect{b}$$ is recoverable just by knowing the scalars $$a_1=-3\text{,}$$ $$a_2=5\text{,}$$ $$a_3=2$$ (use these scalars in a linear combination of the vectors in $$S$$). This is all an illustration of the following important theorem, which we prove in the setting of a general vector space.

That $$\vect{w}$$ can be written as a linear combination of the vectors in $$B$$ follows from the spanning property of the set (Definition SSVS). This is good, but not the meat of this theorem. We now know that for any choice of the vector $$\vect{w}$$ there exist some scalars that will create $$\vect{w}$$ as a linear combination of the basis vectors. The real question is: Is there more than one way to write $$\vect{w}$$ as a linear combination of $$\{\vectorlist{v}{m}\}\text{?}$$ Are the scalars $$a_1,\,a_2,\,a_3,\,\ldots,\,a_m$$ unique? (Proof Technique U)

Assume there are two different linear combinations of $$\{\vectorlist{v}{m}\}$$ that equal the vector $$\vect{w}\text{.}$$ In other words there exist scalars $$a_1,\,a_2,\,a_3,\,\ldots,\,a_m$$ and $$b_1,\,b_2,\,b_3,\,\ldots,\,b_m$$ so that

\begin{align*} \vect{w}&=\lincombo{a}{v}{m}\\ \vect{w}&=\lincombo{b}{v}{m}\text{.} \end{align*}

Then notice that

But this is a relation of linear dependence on a linearly independent set of vectors (Definition RLD)! Now we are using the other assumption about $$B\text{,}$$ that $$\{\vectorlist{v}{m}\}$$ is a linearly independent set. So by Definition LI it must happen that the scalars are all zero. That is,

\begin{align*} (a_1-b_1)&=0&(a_2-b_2)&=0&(a_3-b_3)&=0&\ldots&&(a_m-b_m)&=0\\ a_1&=b_1&a_2&=b_2&a_3&=b_3&\ldots&&a_m&=b_m\text{.} \end{align*}

And so we find that the scalars are unique.

The converse of Theorem VRRB is true as well, but is not important enough to rise beyond an exercise (see Exercise LISS.T51).

This is a very typical use of the hypothesis that a set is linearly independent — obtain a relation of linear dependence and then conclude that the scalars must all be zero. The result of this theorem tells us that we can write any vector in a vector space as a linear combination of the vectors in a linearly independent spanning set, but only just. There is only enough raw material in the spanning set to write each vector one way as a linear combination. So in this sense, we could call a linearly independent spanning set a “minimal spanning set.” These sets are so important that we will give them a simpler name (“basis”) and explore their properties further in the next section.

#### 1.

Is the set of matrices below linearly independent or linearly dependent in the vector space $$M_{22}\text{?}$$ Why or why not?

\begin{equation*} \set{ \begin{bmatrix} 1&3\\-2&4 \end{bmatrix},\, \begin{bmatrix} -2&3\\3&-5 \end{bmatrix},\, \begin{bmatrix} 0&9\\-1&3 \end{bmatrix} } \end{equation*}

#### 2.

Explain the difference between the following two uses of the term “span”:

1. $$S$$ is a subset of the vector space $$V$$ and the span of $$S$$ is a subspace of $$V\text{.}$$

2. $$W$$ is a subspace of the vector space $$Y$$ and $$T$$ spans $$W\text{.}$$

#### 3.

The set

\begin{equation*} S=\set{ \colvector{6\\2\\1},\, \colvector{4\\-3\\1},\, \colvector{5\\8\\2} } \end{equation*}

is linearly independent and spans $$\complex{3}\text{.}$$ Write the vector $$\vect{x}=\colvector{-6\\2\\2}$$ as a linear combination of the elements of $$S\text{.}$$ How many ways are there to answer this question, and which theorem allows you to say so?

### ExercisesLISSExercises

#### C20.

In the vector space of $$2\times 2$$ matrices, $$M_{22}\text{,}$$ determine if the set $$S$$ below is linearly independent.

\begin{equation*} S=\set{ \begin{bmatrix} 2&-1\\1&3 \end{bmatrix},\, \begin{bmatrix} 0&4\\-1&2 \end{bmatrix},\, \begin{bmatrix} 4&2\\1&3 \end{bmatrix} } \end{equation*}
Solution.

Begin with a relation of linear dependence on the vectors in $$S$$ and massage it according to the definitions of vector addition and scalar multiplication in $$M_{22}\text{,}$$

\begin{align*} \zeromatrix&= a_1 \begin{bmatrix} 2&-1\\1&3 \end{bmatrix}+ a_2 \begin{bmatrix} 0&4\\-1&2 \end{bmatrix}+ a_3 \begin{bmatrix} 4&2\\1&3 \end{bmatrix}\\ \begin{bmatrix} 0&0\\0&0 \end{bmatrix} &= \begin{bmatrix} 2a_1+4a_3& -a_1+4a_2+2a_3\\ a_1-a_2+a_3& 3a_1+2a_2+3a_3 \end{bmatrix}\text{.} \end{align*}

By our definition of matrix equality (Definition ME) we arrive at a homogeneous system of linear equations,

\begin{align*} 2a_1+4a_3&=0\\ -a_1+4a_2+2a_3&=0\\ a_1-a_2+a_3&=0\\ 3a_1+2a_2+3a_3&=0\text{.} \end{align*}

The coefficient matrix of this system row-reduces to the matrix,

\begin{equation*} \begin{bmatrix} \leading{1} & 0 & 0\\ 0 & \leading{1} & 0\\ 0 & 0 & \leading{1}\\ 0 & 0 & 0 \end{bmatrix} \end{equation*}

and from this we conclude that the only solution is $$a_1=a_2=a_3=0\text{.}$$ Since the relation of linear dependence (Definition RLD) is trivial, the set $$S$$ is linearly independent (Definition LI).

#### C21.

In the crazy vector space $$C$$ (Example CVS), is the set $$S=\set{(0,\,2),\ (2,\,8)}$$ linearly independent?

Solution.

We begin with a relation of linear dependence using unknown scalars $$a$$ and $$b\text{.}$$ We wish to know if these scalars must both be zero. Recall that the zero vector in $$C$$ is $$(-1,\,-1)$$ and that the definitions of vector addition and scalar multiplication in Example CVS are not what we might expect.

\begin{align*} \zerovector &=(-1,\,-1)\\ &=a(0,\,2) +b(2,\,8)&& \knowl{./knowl/definition-RLD.html}{\text{Definition RLD}}\\ &=(0a+a-1,\,2a+a-1) + (2b+b-1,\,8b+b-1)&& \knowl{./knowl/example-CVS.html}{\text{Example CVS}}\\ &=(a-1,\,3a-1) + (3b-1,\,9b-1)\\ &=(a-1+3b-1+1,\,3a-1+9b-1+1)&& \knowl{./knowl/example-CVS.html}{\text{Example CVS}}\\ &=(a+3b-1,\,3a+9b-1)\text{.} \end{align*}

From this we obtain two equalities, which can be converted to a homogeneous system of equations,

\begin{align*} -1&=a+3b-1&a+3b&=0\\ -1&=3a+9b-1&3a+9b&=0\text{.} \end{align*}

This homogeneous system has a singular coefficient matrix, and so has more than just the trivial solution (Definition NM). Any nontrivial solution will give us a nontrivial relation of linear dependence on $$S\text{.}$$ So $$S$$ is linearly dependent (Definition LI).

#### C22.

In the vector space of polynomials $$P_3\text{,}$$ determine if the set $$S$$ is linearly independent or linearly dependent.

\begin{equation*} S=\set{2 +x -3x^2 -8x^3,\, 1+ x + x^2 +5x^3,\, 3 -4x^2 -7x^3} \end{equation*}
Solution.

Begin with a relation of linear dependence (Definition RLD),

\begin{equation*} a_1\left(2 +x -3x^2 -8x^3\right)+a_2\left(1+ x + x^2 +5x^3\right)+a_3\left(3 -4x^2 -7x^3\right)=\zerovector\text{.} \end{equation*}

Massage according to the definitions of scalar multiplication and vector addition in the definition of $$P_3$$ (Example VSP) and use the zero vector for this vector space,

\begin{equation*} \left(2a_1+a_2+3a_3\right)+ \left(a_1+a_2\right)x+ \left(-3a_1+a_2-4a_3\right)x^2+ \left(-8a_1+5a_2-7a_3\right)x^3 =0+0x+0x^2+0x^3\text{.} \end{equation*}

The definition of the equality of polynomials allows us to deduce the following four equations,

\begin{align*} 2a_1+a_2+3a_3&=0\\ a_1+a_2&=0\\ -3a_1+a_2-4a_3&=0\\ -8a_1+5a_2-7a_3&=0\text{.} \end{align*}

Row-reducing the coefficient matrix of this homogeneous system leads to the unique solution $$a_1=a_2=a_3=0\text{.}$$ So the only relation of linear dependence on $$S$$ is the trivial one, and this is linear independence for $$S$$ (Definition LI).

#### C23.

Determine if the set $$S=\set{(3,\,1),\,(7,\,3)}$$ is linearly independent in the crazy vector space $$C$$ (Example CVS).

Solution.

Notice, or discover, that the following gives a nontrivial relation of linear dependence on $$S$$ in $$C\text{,}$$ so by Definition LI, the set $$S$$ is linearly dependent.

\begin{equation*} 2(3,\,1)+(-1)(7,\,3)=(7,\,3)+(-9,\,-5)=(-1,\,-1)=\zerovector \end{equation*}

#### C24.

In the vector space of real-valued functions $$F = \setparts{f}{f:\mathbb{R}\rightarrow\mathbb{R}}\text{,}$$ determine if the following set $$S$$ is linearly independent.

\begin{equation*} S = \set{\sin^2{x}, \cos^2{x}, 2} \end{equation*}
Solution.

One of the fundamental identities of trigonometry is $$\sin^2(x) + \cos^2(x) = 1\text{.}$$ Thus, we have a dependence relation $$2(\sin^2{x}) + 2(\cos^2{x}) + (-1)(2) = 0\text{,}$$ and the set is linearly dependent.

#### C25.

Let

\begin{equation*} S = \set{ \begin{bmatrix} 1& 2\\2 & 1 \end{bmatrix}, \begin{bmatrix} 2 & 1\\ -1 & 2\end{bmatrix}, \begin{bmatrix} 0 & 1\\ 1 & 2\end{bmatrix} }\text{.} \end{equation*}
1. Determine if $$S$$ spans $$M_{22}\text{.}$$

2. Determine if $$S$$ is linearly independent.

Solution.
1. If $$S$$ spans $$M_{22}\text{,}$$ then for every $$2 \times 2$$ matrix $$B = \begin{bmatrix}x & y \\ z & w\end{bmatrix}\text{,}$$ there exist constants $$\alpha, \beta, \gamma$$ so that

\begin{align*} \begin{bmatrix}x & y \\ z & w\end{bmatrix} &= \alpha\begin{bmatrix} 1& 2\\2 & 1 \end{bmatrix} + \beta\begin{bmatrix} 2 & 1\\ -1 & 2\end{bmatrix} + \gamma\begin{bmatrix} 0 & 1\\ 1 & 2\end{bmatrix}\text{.} \end{align*}

Applying Definition ME, this leads to the linear system

\begin{align*} \alpha + 2\beta &= x\\ 2\alpha + \beta + \gamma &= y\\ 2\alpha - \beta + \gamma &= z\\ \alpha + 2\beta + 2\gamma &= w\text{.} \end{align*}

We need to row-reduce the augmented matrix of this system by hand due to the symbols $$x\text{,}$$ $$y\text{,}$$ $$z\text{,}$$ and $$w$$ in the vector of constants.

\begin{align*} \begin{bmatrix} 1 & 2 & 0 & x\\ 2 & 1 & 1 & y\\ 2 & -1 & 1 & z\\ 1 & 2 & 2 & w \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & x - y + z\\ 0 & \leading{1} & 0 & \frac{1}{2}(y - z)\\ 0 & 0 & \leading{1} & \frac{1}{2}(w - x)\\ 0 & 0 & 0 & \frac{1}{2}(5y - 3x - 3z - w) \end{bmatrix} \end{align*}

It is easy to find values of $$x, y, z$$ and $$w$$ so that the final column will become a pivot column after a few more row operations, namely any choice with $$5y - 3x - 3z - w \ne 0\text{.}$$ Thus, by Theorem RCLS the system would be inconsistent and there is a matrix in $$M_{22}$$ that is not in the span of $$S\text{.}$$ One such matrix is $$B = \begin{bmatrix} 3 & 3 \\ 3 & 2\end{bmatrix}\text{.}$$) So $$S$$ does not span $$M_{22}\text{.}$$

2. To check for linear independence, we need to see if there are nontrivial coefficients $$\alpha, \beta, \gamma$$ that solve

\begin{align*} \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} &= \alpha\begin{bmatrix} 1& 2\\2 & 1 \end{bmatrix} + \beta\begin{bmatrix} 2 & 1\\ -1 & 2\end{bmatrix}+ \gamma\begin{bmatrix} 0 & 1\\ 1 & 2\end{bmatrix}\text{.} \end{align*}

This requires the same work that was done in part (a), with $$x = y = z = w = 0\text{.}$$In that case, the coefficient matrix row-reduces so the first three columns are pivot columns and a row of zeros on the bottom, so we know that the only solution to the matrix equation is $$\alpha = \beta = \gamma = 0\text{.}$$ So the set $$S$$ is linearly independent.

#### C26.

Let

\begin{equation*} S = \set{ \begin{bmatrix} 1& 2\\2 & 1 \end{bmatrix}, \begin{bmatrix} 2 & 1\\ -1 & 2\end{bmatrix}, \begin{bmatrix} 0 & 1\\ 1 & 2\end{bmatrix}, \begin{bmatrix} 1 & 0\\1 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 4\\0 & 3\end{bmatrix} }\text{.} \end{equation*}
1. Determine if $$S$$ spans $$M_{22}\text{.}$$

2. Determine if $$S$$ is linearly independent.

Solution.
1. The matrices in $$S$$ will span $$M_{22}$$ if for any $$\begin{bmatrix} x& y\\z& w\end{bmatrix}\text{,}$$ there are coefficients $$a, b, c, d, e$$ so that

\begin{align*} a\begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix} + b\begin{bmatrix} 2 & 1\\-1 & 2 \end{bmatrix} + c\begin{bmatrix} 0 & 1 \\ 1 & 2\end{bmatrix} + d\begin{bmatrix} 1 & 0 \\ 1 & 1\end{bmatrix} + e\begin{bmatrix} 1 & 4 \\ 0 & 3 \end{bmatrix} &= \begin{bmatrix} x & y \\ z & w \end{bmatrix}\text{.} \end{align*}

Thus, we have

\begin{align*} \begin{bmatrix} a + 2b + d + e & 2a + b + c + 4e\\ 2a - b + c + d& a + 2b + 2c + d + 3e \end{bmatrix} &= \begin{bmatrix} x & y \\ z & w \end{bmatrix}\\ \end{align*}

so we have the matrix equation

\begin{align*} \begin{bmatrix} 1 & 2 & 0 & 1 & 1 \\ 2 & 1 & 1 & 0 & 4\\ 2 & -1 & 1 & 1 & 0\\ 1 & 2 & 2 & 1 & 3 \end{bmatrix} \colvector{a\\b\\c\\d\\e} &= \colvector{x\\y\\z\\w}\text{.} \end{align*}

This system will have a solution for every vector on the right side if the row-reduced coefficient matrix has a leading one in every row, since then it is never possible to have a pivot column appear in the final column of a row-reduced augmented matrix.

\begin{align*} \begin{bmatrix} 1 & 2 & 0 & 1 & 1 \\ 2 & 1 & 1 & 0 & 4\\ 2 & -1 & 1 & 1 & 0\\ 1 & 2 & 2 & 1 & 3 \end{bmatrix} \rref \begin{bmatrix} \leading{1}& 0 & 0 & 0 & 1\\ 0 & \leading{1} & 0 & 0 & 1\\ 0 & 0 & \leading{1} & 0 & 1\\ 0 & 0 & 0 & \leading{1} & -2 \end{bmatrix} \end{align*}

Since there is a leading one in each row of the row-reduced coefficient matrix, there is a solution for every vector $$\colvector{x\\y\\z\\w}\text{,}$$ which means that there is a solution to the original equation for every matrix $$\begin{bmatrix} x & y\\ z & w \end{bmatrix}\text{.}$$ Thus, the original five matrices span $$M_{22}\text{.}$$

2. The matrices in $$S$$ are linearly independent if the only solution to

\begin{align*} a\begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix} + b\begin{bmatrix} 2 & 1\\-1 & 2 \end{bmatrix} + c\begin{bmatrix} 0 & 1 \\ 1 & 2\end{bmatrix} + d\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} + e\begin{bmatrix} 1 & 4 \\ 0 & 3 \end{bmatrix} &= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \end{align*}

is $$a = b = c = d = e = 0\text{.}$$

We have

\begin{align*} \begin{bmatrix} a + 2b + d + e & 2a + b + c + 4e\\ 2a - b + c + d & a + 2b + 2c + d + 3e \end{bmatrix} &= \begin{bmatrix} 1 & 2 & 0 & 1 & 1 \\ 2 & 1 & 1 & 0 & 4\\ 2 & -1 & 1 & 1 & 0\\ 1 & 2 & 2 & 1 & 3 \end{bmatrix} \colvector{a\\b\\c\\d\\e} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \end{align*}

so we need to find the nullspace of the matrix

\begin{align*} \begin{bmatrix} 1 & 2 & 0 & 1 & 1 \\ 2 & 1 & 1 & 0 & 4\\ 2 & -1 & 1 & 1 & 0\\ 1 & 2 & 2 & 1 & 3 \end{bmatrix}\text{.} \end{align*}

We row-reduced this matrix in part (a), and found a pivot column, which provides a free variable in a description of the solution set to the homogeneous system, so the nullspace is nontrivial and there are an infinite number of solutions to

\begin{align*} a\begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix} + b\begin{bmatrix} 2 & 1\\-1 & 2 \end{bmatrix} + c \begin{bmatrix} 0 & 1 \\ 1 & 2\end{bmatrix} + d \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} + e \begin{bmatrix} 1 & 4 \\ 0 & 3 \end{bmatrix} &= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\text{.} \end{align*}

Thus, this set of matrices is not linearly independent.

#### C30.

In Example LIM32, find another nontrivial relation of linear dependence on the linearly dependent set of $$3\times 2$$ matrices, $$S\text{.}$$

#### C40.

Determine if the set $$T=\set{x^2-x+5,\,4x^3-x^2+5x,\,3x+2}$$ spans the vector space of polynomials with degree 4 or less, $$P_4\text{.}$$

Solution.

The polynomial $$x^4$$ is an element of $$P_4\text{.}$$ Can we write this element as a linear combination of the elements of $$T\text{?}$$ To wit, are there scalars $$a_1\text{,}$$ $$a_2\text{,}$$ $$a_3$$ such that

\begin{align*} x^4&=a_1\left(x^2-x+5\right)+a_2\left(4x^3-x^2+5x\right)+a_3\left(3x+2\right)\text{.} \end{align*}

Massaging the right side of this equation, according to the definitions of Example VSP, and then equating coefficients, leads to an inconsistent system of equations (check this!). As such, $$T$$ is not a spanning set for $$P_4\text{.}$$

#### C41.

The set $$W$$ is a subspace of $$M_{22}\text{,}$$ the vector space of all $$2\times 2$$ matrices. Prove that $$S$$ is a spanning set for $$W\text{.}$$

\begin{align*} W&=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{2a-3b+4c-d=0} & S=&\set{ \begin{bmatrix}1&0\\0&2\end{bmatrix},\, \begin{bmatrix}0&1\\0&-3\end{bmatrix},\, \begin{bmatrix}0&0\\1&4\end{bmatrix} } \end{align*}
Solution.

We want to show that $$W=\spn{S}$$ (Definition SSVS), which is an equality of sets (Definition SE).

First, show that $$\spn{S}\subseteq W\text{.}$$ Begin by checking that each of the three matrices in $$S$$ is a member of the set $$W\text{.}$$ Then, since $$W$$ is a vector space, the closure properties (Property AC, Property SC) guarantee that every linear combination of elements of $$S$$ remains in $$W\text{.}$$

Second, show that $$W\subseteq\spn{S}\text{.}$$ We want to convince ourselves that an arbitrary element of $$W$$ is a linear combination of elements of $$S\text{.}$$ Choose

\begin{equation*} \vect{x}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\in W \end{equation*}

The values of $$a,\,b,\,c,\,d$$ are not totally arbitrary, since membership in $$W$$ requires that $$2a-3b+4c-d=0\text{.}$$ Now, rewrite as follows,

\begin{align*} \vect{x} &=\begin{bmatrix}a&b\\c&d\end{bmatrix}\\ &=\begin{bmatrix}a&b\\c&2a-3b+4c\end{bmatrix}&&2a-3b+4c-d=0\\ &= \begin{bmatrix}a&0\\0&2a\end{bmatrix}+ \begin{bmatrix}0&b\\0&-3b\end{bmatrix}+ \begin{bmatrix}0&0\\c&4c\end{bmatrix}&& \knowl{./knowl/definition-MA.html}{\text{Definition MA}}\\ &= a\begin{bmatrix}1&0\\0&2\end{bmatrix}+ b\begin{bmatrix}0&1\\0&-3\end{bmatrix}+ c\begin{bmatrix}0&0\\1&4\end{bmatrix}&& \knowl{./knowl/definition-MSM.html}{\text{Definition MSM}}\\ &\in\spn{S}&& \knowl{./knowl/definition-SS.html}{\text{Definition SS}}\text{.} \end{align*}

#### C42.

Determine if the set $$S=\set{(3,\,1),\,(7,\,3)}$$ spans the crazy vector space $$C$$ (Example CVS).

Solution.

We will try to show that $$S$$ spans $$C\text{.}$$ Let $$(x,\,y)$$ be an arbitrary element of $$C$$ and search for scalars $$a_1$$ and $$a_2$$ such that

\begin{align*} (x,\,y)&=a_1(3,\,1) + a_2(7,\,3)\\ &=(4a_1-1,\,2a_1-1)+(8a_2-1,\,4a_2-1)\\ &=(4a_1+8a_2-1,2a_1+4a_2-1)\text{.} \end{align*}

Equality in $$C$$ leads to the system

\begin{align*} 4a_1+8a_2&=x+1\\ 2a_1+4a_2&=y+1\text{.} \end{align*}

This system has a singular coefficient matrix whose column space is simply $$\spn{\set{\colvector{2\\1}}}\text{.}$$ So any choice of $$x$$ and $$y$$ that causes the column vector $$\colvector{x+1\\y+1}$$ to lie outside the column space will lead to an inconsistent system, and hence create an element $$(x,\,y)$$ that is not in the span of $$S\text{.}$$ So $$S$$ does not span $$C\text{.}$$

For example, choose $$x=0$$ and $$y=5\text{,}$$ and then we can see that $$\colvector{1\\6}\not\in\spn{\set{\colvector{2\\1}}}$$ and we know that $$(0,\,5)$$ cannot be written as a linear combination of the vectors in $$S\text{.}$$ A shorter solution might begin by asserting that $$(0,\,5)$$ is not in $$\spn{S}$$ and then establishing this claim alone.

#### M10.

Halfway through Example SSP4, we need to show that the system of equations

\begin{equation*} \linearsystem{ \begin{bmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & -8\\ 0 & 1 & -6 & 24\\ 1 & -4 & 12 & -32\\ -2 & 4 & -8 & 16\\ \end{bmatrix} } {\colvector{a\\b\\c\\d\\e}} \end{equation*}

is consistent for every choice of the vector of constants satisfying $$16a+8b+4c+2d+e=0\text{.}$$

Express the column space of the coefficient matrix of this system as a null space, using Theorem FS. From this use Theorem CSCS to establish that the system is always consistent. Notice that this approach removes from Example SSP4 the need to row-reduce a symbolic matrix.

Solution.

Theorem FS provides the matrix

\begin{equation*} L= \begin{bmatrix} \leading{1} & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} \end{bmatrix} \end{equation*}

and so if $$A$$ denotes the coefficient matrix of the system, then $$\csp{A}=\nsp{L}\text{.}$$ The single homogeneous equation in $$\homosystem{L}$$ is equivalent to the condition on the vector of constants (use $$a,\,b,\,c,\,d,\,e$$ as variables and then multiply by 16).

#### T20.

Suppose that $$S$$ is a finite linearly independent set of vectors from the vector space $$V\text{.}$$ Let $$T$$ be any subset of $$S\text{.}$$ Prove that $$T$$ is linearly independent.

Solution.

We will prove the contrapositive (Proof Technique CP): If $$T$$ is linearly dependent, then $$S$$ is linearly dependent. This might be an interesting statement in its own right.

Write $$S=\set{\vectorlist{v}{m}}$$ and without loss of generality we can assume that the subset $$T$$ is the first $$t$$ vectors of $$S\text{,}$$ $$t\leq m\text{,}$$ so $$T=\set{\vectorlist{v}{t}}\text{.}$$ Since $$T$$ is linearly dependent, by Definition LI there are scalars, not all zero, $$\scalarlist{a}{t}\text{,}$$ so that

\begin{align*} \zerovector &= \lincombo{a}{v}{t}\\ &= \lincombo{a}{v}{t}+\zerovector+\zerovector+\dots+\zerovector\\ &= \lincombo{a}{v}{t}+0\vect{v}_{t+1}+0\vect{v}_{t+2}+\dots+0\vect{v}_{m} \end{align*}

which is a nontrivial relation of linear dependence (Definition RLD) on the set $$S\text{,}$$ so we can say $$S$$ is linearly dependent.

#### T40.

Prove the following variant of Theorem EMMVP that has a weaker hypothesis: Suppose that $$C=\set{\vectorlist{u}{p}}$$ is a linearly independent spanning set for $$\complex{n}\text{.}$$ Suppose also that $$A$$ and $$B$$ are $$m\times n$$ matrices such that $$A\vect{u}_i=B\vect{u}_i$$ for every $$1\leq i\leq n\text{.}$$ Then $$A=B\text{.}$$

Can you weaken the hypothesis even further while still preserving the conclusion?

#### T50.

Suppose that $$V$$ is a vector space and $$\vect{u},\,\vect{v}\in V$$ are two vectors in $$V\text{.}$$ Use the definition of linear independence to prove that $$S=\set{\vect{u},\,\vect{v}}$$ is a linearly dependent set if and only if one of the two vectors is a scalar multiple of the other. Prove this directly in the context of an abstract vector space ($$V$$), without simply giving an upgraded version of Theorem DLDS for the special case of just two vectors.

Solution.

If $$S$$ is linearly dependent, then there are scalars $$\alpha$$ and $$\beta\text{,}$$ not both zero, such that $$\alpha\vect{u}+\beta\vect{v}=\zerovector\text{.}$$ Suppose that $$\alpha\neq 0\text{,}$$ the proof proceeds similarly if $$\beta\neq 0\text{.}$$ Now,

\begin{align*} \vect{u} &=1\vect{u}&& \knowl{./knowl/property-O.html}{\text{Property O}}\\ &=\left(\frac{1}{\alpha}\alpha\right)\vect{u}&& \knowl{./knowl/property-MICN.html}{\text{Property MICN}}\\ &=\frac{1}{\alpha}\left(\alpha\vect{u}\right)&& \knowl{./knowl/property-SMA.html}{\text{Property SMA}}\\ &=\frac{1}{\alpha}\left(\alpha\vect{u}+\zerovector\right)&& \knowl{./knowl/property-Z.html}{\text{Property Z}}\\ &=\frac{1}{\alpha}\left(\alpha\vect{u}+\beta\vect{v}-\beta\vect{v}\right)&& \knowl{./knowl/property-AI.html}{\text{Property AI}}\\ &=\frac{1}{\alpha}\left(\zerovector-\beta\vect{v}\right)&& \knowl{./knowl/definition-LI.html}{\text{Definition LI}}\\ &=\frac{1}{\alpha}\left(-\beta\vect{v}\right)&& \knowl{./knowl/property-Z.html}{\text{Property Z}}\\ &=\frac{-\beta}{\alpha}\vect{v}&& \knowl{./knowl/property-SMA.html}{\text{Property SMA}} \end{align*}

which shows that $$\vect{u}$$ is a scalar multiple of $$\vect{v}\text{.}$$

⇐ Suppose now that $$\vect{u}$$ is a scalar multiple of $$\vect{v}\text{.}$$ More precisely, suppose there is a scalar $$\gamma$$ such that $$\vect{u}=\gamma\vect{v}\text{.}$$ Then

\begin{align*} (-1)\vect{u}+\gamma\vect{v} &=(-1)\vect{u}+\vect{u}\\ &=(-1)\vect{u}+(1)\vect{u}&& \knowl{./knowl/property-O.html}{\text{Property O}}\\ &=\left((-1)+1\right)\vect{u}&& \knowl{./knowl/property-DSA.html}{\text{Property DSA}}\\ &=0\vect{u}&& \knowl{./knowl/property-AICN.html}{\text{Property AICN}}\\ &=\zerovector&& \knowl{./knowl/theorem-ZSSM.html}{\text{Theorem ZSSM}}\text{.} \end{align*}

This is a relation of linear of linear dependence on $$S$$ (Definition RLD), which is nontrivial since one of the scalars is $$-1\text{.}$$ Therefore $$S$$ is linearly dependent by Definition LI.

Be careful using this theorem. It is only applicable to sets of two vectors. In particular, linear dependence in a set of three or more vectors can be more complicated than just one vector being a scalar multiple of another.

#### T51.

Carefully formulate the converse of Theorem VRRB and provide a proof.

Solution.

The converse could read: “Suppose that $$V$$ is a vector space and $$S=\set{\vectorlist{v}{m}}$$ is a set of vectors in $$V\text{.}$$ If, for each $$\vect{w}\in V\text{,}$$ there are unique scalars $$a_1,\,a_2,\,a_3,\,\ldots,\,a_m$$ such that

\begin{equation*} \vect{w}=\lincombo{a}{v}{m} \end{equation*}
then $$S$$ is a linearly independent set that spans $$V\text{.}$$

Since every vector $$\vect{w}\in V$$ is assumed to be a linear combination of the elements of $$S\text{,}$$ it is easy to see that $$S$$ is a spanning set for $$V$$ (Definition SSVS).

To establish linear independence, begin with an arbitrary relation of linear dependence on the vectors in $$S$$ (Definition RLD). One way to form such a relation is the trivial way, where each scalar is zero. But our hypothesis of uniqueness then implies that the only way to form this relation of linear dependence is the trivial way. But this establishes the linear independence of $$S$$ (Definition LI).