SectionVRVector Representations

You may have noticed that many questions about elements of abstract vector spaces eventually become questions about column vectors or systems of equations. Example SM32 would be an example of this. We will make this vague idea more precise in this section.

SubsectionVRVector Representation

We begin by establishing an invertible linear transformation between any vector space $$V$$ of dimension $$n$$ and $$\complex{n}\text{.}$$ This will allow us to “go back and forth” between the two vector spaces, no matter how abstract the definition of $$V$$ might be.

DefinitionVR.Vector Representation.

Suppose that $$V$$ is a vector space with a basis $$B=\set{\vectorlist{v}{n}}\text{.}$$ Define a function $$\ltdefn{\vectrepname{B}}{V}{\complex{n}}$$ as follows. For $$\vect{w}\in V$$ define the column vector $$\vectrep{B}{\vect{w}}\in\complex{n}$$ by

\begin{align*} \vect{w} &= \vectorentry{\vectrep{B}{\vect{w}}}{1}\vect{v}_1+ \vectorentry{\vectrep{B}{\vect{w}}}{2}\vect{v}_2+ \vectorentry{\vectrep{B}{\vect{w}}}{3}\vect{v}_3+ \cdots+ \vectorentry{\vectrep{B}{\vect{w}}}{n}\vect{v}_n\text{.} \end{align*}

This definition looks more complicated than it really is, though the form above will be useful in proofs. Simply stated, given $$\vect{w}\in V\text{,}$$ we write $$\vect{w}$$ as a linear combination of the basis elements of $$B\text{.}$$ It is key to realize that Theorem VRRB guarantees that we can do this for every $$\vect{w}\text{,}$$ and furthermore this expression as a linear combination is unique. The resulting scalars are just the entries of the vector $$\vectrep{B}{\vect{w}}\text{.}$$ This discussion should convince you that $$\vectrepname{B}$$ is “well-defined” as a function. We can determine a precise output for any input. Now we want to establish that $$\vectrepname{B}$$ is a function with additional properties — it is a linear transformation.

We will take a novel approach in this proof. We will construct another function, which we will easily determine is a linear transformation, and then show that this second function is really $$\vectrepname{B}$$ in disguise. Here we go.

Since $$B$$ is a basis, we can define $$\ltdefn{T}{V}{\complex{n}}$$ to be the unique linear transformation such that $$\lteval{T}{\vect{v}_i}=\vect{e}_i\text{,}$$ $$1\leq i\leq n\text{,}$$ as guaranteed by Theorem LTDB, and where the $$\vect{e}_i$$ are the standard unit vectors (Definition SUV). Then suppose for an arbitrary $$\vect{w}\in V$$ we have,

\begin{align*} \vectorentry{\lteval{T}{\vect{w}}}{i} &= \vectorentry{\lteval{T}{\sum_{j=1}^{n}\vectorentry{\vectrep{B}{\vect{w}}}{j}\vect{v}_j}}{i}&& \knowl{./knowl/definition-VR.html}{\text{Definition VR}}\\ &= \vectorentry{\sum_{j=1}^{n}\vectorentry{\vectrep{B}{\vect{w}}}{j}\lteval{T}{\vect{v}_j}}{i}&& \knowl{./knowl/theorem-LTLC.html}{\text{Theorem LTLC}}\\ &= \vectorentry{\sum_{j=1}^{n}\vectorentry{\vectrep{B}{\vect{w}}}{j}\vect{e}_j}{i}\\ &= \sum_{j=1}^{n}\vectorentry{\vectorentry{\vectrep{B}{\vect{w}}}{j}\vect{e}_j}{i}&& \knowl{./knowl/definition-CVA.html}{\text{Definition CVA}}\\ &= \sum_{j=1}^{n}\vectorentry{\vectrep{B}{\vect{w}}}{j}\vectorentry{\vect{e}_j}{i}&& \knowl{./knowl/definition-CVSM.html}{\text{Definition CVSM}}\\ &= \vectorentry{\vectrep{B}{\vect{w}}}{i}\vectorentry{\vect{e}_i}{i} + \sum_{\substack{j=1\\j\neq i}}^{n}\vectorentry{\vectrep{B}{\vect{w}}}{j}\vectorentry{\vect{e}_j}{i}&& \knowl{./knowl/property-CC.html}{\text{Property CC}}\\ &= \vectorentry{\vectrep{B}{\vect{w}}}{i}\left(1\right) + \sum_{\substack{j=1\\j\neq i}}^{n}\vectorentry{\vectrep{B}{\vect{w}}}{j}\left(0\right)&& \knowl{./knowl/definition-SUV.html}{\text{Definition SUV}}\\ &= \vectorentry{\vectrep{B}{\vect{w}}}{i}\text{.} \end{align*}

As column vectors, Definition CVE implies that $$\lteval{T}{\vect{w}}=\vectrep{B}{\vect{w}}\text{.}$$ Since $$\vect{w}$$ was an arbitrary element of $$V\text{,}$$ as functions $$T=\vectrepname{B}\text{.}$$ Now, since $$T$$ is known to be a linear transformation, it must follow that $$\vectrepname{B}$$ is also a linear transformation.

The proof of Theorem VRLT provides an alternate definition of vector representation relative to a basis $$B$$ that we could state as a corollary (Proof Technique LC): $$\vectrepname{B}$$ is the unique linear transformation that takes $$B$$ to the standard unit basis.

Consider the vector $$\vect{y}\in\complex{4}$$

\begin{equation*} \vect{y}=\colvector{6\\14\\6\\7}\text{.} \end{equation*}

We will find several vector representations of $$\vect{y}$$ in this example. Notice that $$\vect{y}$$ never changes, but the representations of $$\vect{y}$$ do change. One basis for $$\complex{4}$$ is

\begin{equation*} B=\set{\vect{u}_1,\,\vect{u}_2,\,\vect{u}_3,\,\vect{u}_4}= \set{ \colvector{-2\\1\\2\\-3},\, \colvector{3\\-6\\2\\-4},\, \colvector{1\\2\\0\\5},\, \colvector{4\\3\\1\\6} } \end{equation*}

as can be seen by making these vectors the columns of a matrix, checking that the matrix is nonsingular and applying Theorem CNMB. To find $$\vectrep{B}{\vect{y}}\text{,}$$ we need to find scalars, $$a_1,\,a_2,\,a_3,\,a_4$$ such that

\begin{equation*} \vect{y}=a_1\vect{u}_1+a_2\vect{u}_2+a_3\vect{u}_3+a_4\vect{u}_4\text{.} \end{equation*}

By Theorem SLSLC the desired scalars are a solution to the linear system of equations with a coefficient matrix whose columns are the vectors in $$B$$ and with a vector of constants $$\vect{y}\text{.}$$ With a nonsingular coefficient matrix, the solution is unique, but this is no surprise as this is the content of Theorem VRRB. This unique solution is

\begin{align*} a_1&=2&a_2&=-1&a_3&=-3&a_4&=4\text{.} \end{align*}

Then by Definition VR, we have

\begin{equation*} \vectrep{B}{\vect{y}}=\colvector{2\\-1\\-3\\4}\text{.} \end{equation*}

Suppose now that we construct a representation of $$\vect{y}$$ relative to another basis of $$\complex{4}\text{,}$$

\begin{equation*} C=\set{ \colvector{-15\\9\\-4\\-2},\, \colvector{16\\-14\\5\\2},\, \colvector{-26\\14\\-6\\-3},\, \colvector{14\\-13\\4\\6} }\text{.} \end{equation*}

As with $$B\text{,}$$ it is easy to check that $$C$$ is a basis. Writing $$\vect{y}$$ as a linear combination of the vectors in $$C$$ leads to solving a system of four equations in the four unknown scalars with a nonsingular coefficient matrix. The unique solution can be expressed as

\begin{equation*} \vect{y}=\colvector{6\\14\\6\\7}= (-28)\colvector{-15\\9\\-4\\-2}+ (-8)\colvector{16\\-14\\5\\2}+ 11\colvector{-26\\14\\-6\\-3}+ 0\colvector{14\\-13\\4\\6} \end{equation*}

so that Definition VR gives

\begin{equation*} \vectrep{C}{\vect{y}}=\colvector{-28\\-8\\11\\0}\text{.} \end{equation*}

We often perform representations relative to standard bases, but for vectors in $$\complex{m}$$ this is a little silly. Let us find the vector representation of $$\vect{y}$$ relative to the standard basis (Theorem SUVB),

\begin{equation*} D=\set{\vect{e}_1,\,\vect{e}_2,\,\vect{e}_3,\,\vect{e}_4}\text{.} \end{equation*}

Then, without any computation, we can check that

\begin{equation*} \vect{y}=\colvector{6\\14\\6\\7}=6\vect{e}_1+14\vect{e}_2+6\vect{e}_3+7\vect{e}_4 \end{equation*}

so by Definition VR,

\begin{equation*} \vectrep{D}{\vect{y}}=\colvector{6\\14\\6\\7} \end{equation*}

which is not very exciting. Notice however that the order in which we place the vectors in the basis is critical to the representation. Let us keep the standard unit vectors as our basis, but rearrange the order we place them in the basis. So a fourth basis is

\begin{equation*} E=\set{\vect{e}_3,\,\vect{e}_4,\,\vect{e}_2,\,\vect{e}_1}\text{.} \end{equation*}

Then,

\begin{equation*} \vect{y}=\colvector{6\\14\\6\\7}=6\vect{e}_3+7\vect{e}_4+14\vect{e}_2+6\vect{e}_1 \end{equation*}

so by Definition VR,

\begin{equation*} \vectrep{E}{\vect{y}}=\colvector{6\\7\\14\\6}\text{.} \end{equation*}

So for every possible basis of $$\complex{4}$$ we could construct a different representation of $$\vect{y}\text{.}$$

Vector representations are most interesting for vector spaces that are not $$\complex{m}\text{.}$$

Consider the vector $$\vect{u}=15+10x-6x^2\in P_2$$ from the vector space of polynomials with degree at most 2 (Example VSP). A nice basis for $$P_2$$ is

\begin{equation*} B=\set{1,\,x,\,x^2} \end{equation*}

so that

\begin{equation*} \vect{u}=15+10x-6x^2=15(1)+10(x)+(-6)(x^2) \end{equation*}

so by Definition VR

\begin{equation*} \vectrep{B}{\vect{u}}=\colvector{15\\10\\-6}\text{.} \end{equation*}

Another nice basis for $$P_2$$ is

\begin{equation*} C=\set{1,\,1+x,\,1+x+x^2} \end{equation*}

so that now it takes a bit of computation to determine the scalars for the representation. We want $$a_1,\,a_2,\,a_3$$ so that

\begin{equation*} 15+10x-6x^2=a_1(1)+a_2(1+x)+a_3(1+x+x^2)\text{.} \end{equation*}

Performing the operations in $$P_2$$ on the right-hand side, and equating coefficients, gives the three equations in the three unknown scalars,

\begin{align*} 15&=a_1+a_2+a_3\\ 10&=a_2+a_3\\ -6&=a_3\text{.} \end{align*}

The coefficient matrix of this sytem is nonsingular, leading to a unique solution (no surprise there, see Theorem VRRB),

\begin{align*} a_1&=5&a_2&=16&a_3&=-6 \end{align*}

so by Definition VR

\begin{equation*} \vectrep{C}{\vect{u}}=\colvector{5\\16\\-6}\text{.} \end{equation*}

While we often form vector representations relative to “nice” bases, nothing prevents us from forming representations relative to “nasty” bases. For example, the set

\begin{equation*} D=\set{ -2-x+3x^2,\, 1-2x^2,\, 5+4x+x^2 } \end{equation*}

can be verified as a basis of $$P_2$$ by checking linear independence with Definition LI and then arguing that 3 vectors from $$P_2\text{,}$$ a vector space of dimension 3 (Theorem DP), must also be a spanning set (Theorem G).

Now we desire scalars $$a_1,\,a_2,\,a_3$$ so that

\begin{equation*} 15+10x-6x^2=a_1(-2-x+3x^2)+a_2(1-2x^2)+a_3(5+4x+x^2)\text{.} \end{equation*}

Performing the operations in $$P_2$$ on the right-hand side, and equating coefficients, gives the three equations in the three unknown scalars,

\begin{align*} 15&=-2a_1+a_2+5a_3\\ 10&=-a_1+4a_3\\ -6&=3a_1-2a_2+a_3\text{.} \end{align*}

The coefficient matrix of this sytem is nonsingular, leading to a unique solution (no surprise there, see Theorem VRRB),

\begin{align*} a_1&=-2&a_2&=1&a_3&=2 \end{align*}

so by Definition VR

\begin{equation*} \vectrep{D}{\vect{u}}=\colvector{-2\\1\\2}\text{.} \end{equation*}

We will appeal to Theorem KILT. Suppose $$U$$ is a vector space of dimension $$n\text{,}$$ so vector representation is $$\ltdefn{\vectrepname{B}}{U}{\complex{n}}\text{.}$$ Let $$B=\set{\vectorlist{u}{n}}$$ be the basis of $$U$$ used in the definition of $$\vectrepname{B}\text{.}$$ Suppose $$\vect{u}\in\krn{\vectrepname{B}}\text{.}$$ We write $$\vect{u}$$ as a linear combination of the vectors in the basis $$B$$ where the scalars are the components of the vector representation, $$\lteval{\vectrepname{B}}{\vect{u}}\text{.}$$ We have

\begin{align*} \vect{u} &= \vectorentry{\lteval{\vectrepname{B}}{\vect{u}}}{1}\vect{u}_1+ \vectorentry{\lteval{\vectrepname{B}}{\vect{u}}}{2}\vect{u}_2+ \cdots+ \vectorentry{\lteval{\vectrepname{B}}{\vect{u}}}{n}\vect{u}_n&& \knowl{./knowl/definition-VR.html}{\text{Definition VR}}\\ &= \vectorentry{\zerovector}{1}\vect{u}_1+ \vectorentry{\zerovector}{2}\vect{u}_2+ \cdots+ \vectorentry{\zerovector}{n}\vect{u}_n&& \knowl{./knowl/definition-KLT.html}{\text{Definition KLT}}\\ &= 0\vect{u}_1+ 0\vect{u}_2+ \cdots+ 0\vect{u}_n&& \knowl{./knowl/definition-ZCV.html}{\text{Definition ZCV}}\\ &=\zerovector+\zerovector+\cdots+\zerovector&& \knowl{./knowl/theorem-ZSSM.html}{\text{Theorem ZSSM}}\\ &=\zerovector&& \knowl{./knowl/property-Z.html}{\text{Property Z}}\text{.} \end{align*}

Thus an arbitrary vector, $$\vect{u}\text{,}$$ from the kernel ,$$\krn{\vectrepname{B}}\text{,}$$ must equal the zero vector of $$U\text{.}$$ So $$\krn{\vectrepname{B}}=\set{\zerovector}$$ and by Theorem KILT, $$\vectrepname{B}$$ is injective.

We will appeal to Theorem RSLT. Suppose $$U$$ is a vector space of dimension $$n\text{,}$$ so vector representation is $$\ltdefn{\vectrepname{B}}{U}{\complex{n}}\text{.}$$ Let $$B=\set{\vectorlist{u}{n}}$$ be the basis of $$U$$ used in the definition of $$\vectrepname{B}\text{.}$$ Suppose $$\vect{v}\in\complex{n}\text{.}$$ Define the vector $$\vect{u}$$ by

\begin{equation*} \vect{u} = \vectorentry{\vect{v}}{1}\vect{u}_1+ \vectorentry{\vect{v}}{2}\vect{u}_2+ \vectorentry{\vect{v}}{3}\vect{u}_3+ \cdots+ \vectorentry{\vect{v}}{n}\vect{u}_n\text{.} \end{equation*}

Then for $$1\leq i\leq n\text{,}$$ by Definition VR,

\begin{align*} \vectorentry{\vectrep{B}{\vect{u}}}{i} &=\vectorentry{\vectrep{B}{ \vectorentry{\vect{v}}{1}\vect{u}_1+ \vectorentry{\vect{v}}{2}\vect{u}_2+ \vectorentry{\vect{v}}{3}\vect{u}_3+ \cdots+ \vectorentry{\vect{v}}{n}\vect{u}_n }}{i} =\vectorentry{\vect{v}}{i}&&\text{} \end{align*}

so the entries of vectors $$\vectrep{B}{\vect{u}}$$ and $$\vect{v}$$ are equal and Definition CVE yields the vector equality $$\vectrep{B}{\vect{u}}=\vect{v}\text{.}$$ This demonstrates that $$\vect{v}\in\rng{\vectrepname{B}}\text{,}$$ so $$\complex{n}\subseteq\rng{\vectrepname{B}}\text{.}$$ Since $$\rng{\vectrepname{B}}\subseteq\complex{n}$$ by Definition RLT, we have $$\rng{\vectrepname{B}}=\complex{n}$$ and Theorem RSLT says $$\vectrepname{B}$$ is surjective.

We will have many occasions later to employ the inverse of vector representation, so we will record the fact that vector representation is an invertible linear transformation.

Informally, we will refer to the application of $$\vectrepname{B}$$ as coordinatizing a vector, while the application of $$\ltinverse{\vectrepname{B}}$$ will be referred to as un-coordinatizing a vector.

SageVR.Vector Representations.

Vector representation is described in the text in a fairly abstract fashion. Sage will support this view (which will be useful in the next section), as well as provide a more practical approach. We will explain both approaches. We begin with an arbitrarily chosen basis. We then create an alternate version of QQ^4with this basis as a“user basis”, namely V.

v0 = vector(QQ, [ 1, 1, 1, 0])
v1 = vector(QQ, [ 1, 2, 3, 2])
v2 = vector(QQ, [ 2, 2, 3, 2])
v3 = vector(QQ, [-1, 3, 5, 5])
B = [v0, v1, v2, v3]
V = (QQ^4).subspace_with_basis(B)
V

V.echelonized_basis_matrix()


Now, the construction of a linear transformation will use the basis provided for V. In the proof of Theorem VRLT we defined a linear transformation $$T$$ that equaled $$\vectrepname{B}\text{.}$$ $$T$$ was defined by taking the basis vectors of $$B$$ to the basis composed of standard unit vectors (Definition SUV). This is exactly what we will accomplish in the following construction. Note how the basis associated with the domain is automatically paired with the elements of the basis for the codomain.

rho = linear_transformation(V, QQ^4, (QQ^4).basis())
rho


First, we verify Theorem VRILT:

rho.is_invertible()


Notice that the matrix of the linear transformation is the identity matrix. This might look odd now, but we will have a full explanation soon. Let us see if this linear transformation behaves as it should. We will “coordinatize” an arbitrary vector, w.

w = vector(QQ, [-13, 28, 45, 43])
rho(w)

lincombo = 3*v0 + 5*v1 + (-6)*v2 + 9*v3
lincombo

lincombo == w


Notice how the expression for lincombois exactly the messy expression displayed in Definition VR. More precisely, we could even write this as:

w == sum([rho(w)[i]*B[i] for i in range(4)])


Or we can test this equality repeatedly with random vectors.

u = random_vector(QQ, 4)
u == sum([rho(u)[i]*B[i] for i in range(4)])


Finding a vector representation is such a fundamental operation that Sage has an easier command, bypassing the need to create a linear transformation. It does still require constructing a vector space with the alternate basis. Here goes, repeating the prior example.

w = vector(QQ, [-13, 28, 45, 43])
V.coordinate_vector(w)


Boom!

SubsectionCVSCharacterization of Vector Spaces

Limiting our attention to vector spaces with finite dimension, we now describe every possible vector space. All of them. Really.

Since $$V$$ has dimension $$n$$ we can find a basis of $$V$$ of size $$n$$ (Definition D) which we will call $$B\text{.}$$ The linear transformation $$\vectrepname{B}$$ is an invertible linear transformation from $$V$$ to $$\complex{n}\text{,}$$ so by Definition IVS, we have that $$V$$ and $$\complex{n}$$ are isomorphic.

Theorem CFDVS is the first of several surprises in this chapter, though it might be a bit demoralizing too. It says that there really are not all that many different (finite dimensional) vector spaces, and none are really any more complicated than $$\complex{n}\text{.}$$ Hmmm. The following examples should make this point.

The vector space of polynomials with degree 8 or less, $$P_8\text{,}$$ has dimension 9 (Theorem DP). By Theorem CFDVS, $$P_8$$ is isomorphic to $$\complex{9}\text{.}$$

In Example DSP4 we determined that a certain subspace $$W$$ of $$P_4$$ has dimension $$4\text{.}$$ By Theorem CFDVS, $$W$$ is isomorphic to $$\complex{4}\text{.}$$

(⇐)

This is the advertised converse of Theorem IVSED. We will assume $$U$$ and $$V$$ have equal dimension and discover that they are isomorphic vector spaces. Let $$n$$ be the common dimension of $$U$$ and $$V\text{.}$$ Then by Theorem CFDVS there are isomorphisms $$\ltdefn{T}{U}{\complex{n}}$$ and $$\ltdefn{S}{V}{\complex{n}}\text{.}$$

$$T$$ is therefore an invertible linear transformation by Definition IVS. Similarly, $$S$$ is an invertible linear transformation, and so $$\ltinverse{S}$$ is an invertible linear transformation (Theorem IILT). The composition of invertible linear transformations is again invertible (Theorem CIVLT) so the composition of $$\ltinverse{S}$$ with $$T$$ is invertible. Then $$\ltdefn{\left(\compose{\ltinverse{S}}{T}\right)}{U}{V}$$ is an invertible linear transformation from $$U$$ to $$V$$ and Definition IVS says $$U$$ and $$V$$ are isomorphic.

$$\complex{10}\text{,}$$ $$P_{9}\text{,}$$ $$M_{25}$$ and $$M_{52}$$ are all vector spaces and each has dimension 10. By Theorem IFDVS each is isomorphic to any other.

The subspace of $$M_{44}$$ that contains all the symmetric matrices (Definition SYM) has dimension $$10\text{,}$$ so this subspace is also isomorphic to each of the four vector spaces above.

SubsectionCPCoordinatization Principle

With $$\vectrepname{B}$$ available as an invertible linear transformation, we can translate between vectors in a vector space $$U$$ of dimension $$m$$ and $$\complex{m}\text{.}$$ Furthermore, as a linear transformation, $$\vectrepname{B}$$ respects the addition and scalar multiplication in $$U\text{,}$$ while $$\vectrepinvname{B}$$ respects the addition and scalar multiplication in $$\complex{m}\text{.}$$ Since our definitions of linear independence, spans, bases and dimension are all built up from linear combinations, we will finally be able to translate fundamental properties between abstract vector spaces ($$U$$) and concrete vector spaces ($$\complex{m}$$).

The linear transformation $$\vectrepname{B}$$ is an isomorphism between $$U$$ and $$\complex{n}$$ (Theorem VRILT). As an invertible linear transformation, $$\vectrepname{B}$$ is an injective linear transformation (Theorem ILTIS), and $$\ltinverse{\vectrepname{B}}$$ is also an injective linear transformation (Theorem IILT, Theorem ILTIS).

(⇒)

Since $$\vectrepname{B}$$ is an injective linear transformation and $$S$$ is linearly independent, Theorem ILTLI says that $$R$$ is linearly independent.

(⇐)

If we apply $$\ltinverse{\vectrepname{B}}$$ to each element of $$R\text{,}$$ we will create the set $$S\text{.}$$ Since we are assuming $$R$$ is linearly independent and $$\ltinverse{\vectrepname{B}}$$ is injective, Theorem ILTLI says that $$S$$ is linearly independent.

(⇒)

Suppose $$\vect{u}\in\spn{\set{\vectorlist{u}{k}}}\text{.}$$ Then we know there are scalars, $$\scalarlist{a}{k}\text{,}$$ such that

\begin{equation*} \vect{u}=\lincombo{a}{u}{k}\text{.} \end{equation*}

Then, by Theorem LTLC,

\begin{align*} \vectrep{B}{\vect{u}}&=\vectrep{B}{\lincombo{a}{u}{k}}\\ &=a_1\vectrep{B}{\vect{u}_1}+a_2\vectrep{B}{\vect{u}_2}+a_3\vectrep{B}{\vect{u}_3}+\cdots+a_k\vectrep{B}{\vect{u}_k} \end{align*}

which says that $$\vectrep{B}{\vect{u}}\in\spn{\set{\vectrep{B}{\vect{u}_1},\,\vectrep{B}{\vect{u}_2},\,\vectrep{B}{\vect{u}_3},\,\ldots,\,\vectrep{B}{\vect{u}_k}}}\text{.}$$

(⇐)

Suppose that $$\vectrep{B}{\vect{u}}\in\spn{\set{\vectrep{B}{\vect{u}_1},\,\vectrep{B}{\vect{u}_2},\,\vectrep{B}{\vect{u}_3},\,\ldots,\,\vectrep{B}{\vect{u}_k}}}\text{.}$$ Then there are scalars $$\scalarlist{b}{k}$$ such that

\begin{equation*} \vectrep{B}{\vect{u}}=b_1\vectrep{B}{\vect{u}_1}+b_2\vectrep{B}{\vect{u}_2}+b_3\vectrep{B}{\vect{u}_3}+\cdots+b_k\vectrep{B}{\vect{u}_k}\text{.} \end{equation*}

Recall that $$\vectrepname{B}$$ is invertible (Theorem VRILT), so

\begin{align*} \vect{u}&=\lteval{I_U}{\vect{u}}&& \knowl{./knowl/definition-IDLT.html}{\text{Definition IDLT}}\\ &=\lteval{\left(\compose{\ltinverse{\vectrepname{B}}}{\vectrepname{B}}\right)}{\vect{u}}&& \knowl{./knowl/definition-IVLT.html}{\text{Definition IVLT}}\\ &=\lteval{\ltinverse{\vectrepname{B}}}{\lteval{\vectrepname{B}}{\vect{u}}}&& \knowl{./knowl/definition-LTC.html}{\text{Definition LTC}}\\ &=\lteval{\ltinverse{\vectrepname{B}}}{b_1\vectrep{B}{\vect{u}_1}+b_2\vectrep{B}{\vect{u}_2}+\cdots+b_k\vectrep{B}{\vect{u}_k}}\\ &=b_1\lteval{\ltinverse{\vectrepname{B}}}{\vectrep{B}{\vect{u}_1}}+b_2\lteval{\ltinverse{\vectrepname{B}}}{\vectrep{B}{\vect{u}_2}} +\cdots+b_k\lteval{\ltinverse{\vectrepname{B}}}{\vectrep{B}{\vect{u}_k}}&& \knowl{./knowl/theorem-LTLC.html}{\text{Theorem LTLC}}\\ &=b_1\lteval{I_U}{\vect{u}_1}+b_2\lteval{I_U}{\vect{u}_2}+\cdots+b_k\lteval{I_U}{\vect{u}_k}&& \knowl{./knowl/definition-IVLT.html}{\text{Definition IVLT}}\\ &=\lincombo{b}{u}{k}&& \knowl{./knowl/definition-IDLT.html}{\text{Definition IDLT}} \end{align*}

which says that $$\vect{u}\in\spn{\set{\vectorlist{u}{k}}}\text{.}$$

Here is a fairly simple example that illustrates a very, very important idea.

In Example VRP2 we needed to know that

\begin{equation*} D=\set{ -2-x+3x^2,\, 1-2x^2,\, 5+4x+x^2 } \end{equation*}

is a basis for $$P_2\text{.}$$ With Theorem CLI and Theorem CSS this task is much easier.

First, choose a known basis for $$P_2\text{,}$$ a basis that forms vector representations easily. We will choose

\begin{equation*} B=\set{1,\,x,\,x^2}\text{.} \end{equation*}

Now, form the subset of $$\complex{3}$$ that is the result of applying $$\vectrepname{B}$$ to each element of $$D\text{,}$$

\begin{align*} F&=\set{\vectrep{B}{-2-x+3x^2},\,\vectrep{B}{1-2x^2},\,\vectrep{B}{5+4x+x^2}}\\ &= \set{ \colvector{-2\\-1\\3},\, \colvector{1\\0\\-2},\, \colvector{5\\4\\1} } \end{align*}

and ask if $$F$$ is a linearly independent spanning set for $$\complex{3}\text{.}$$ This is easily seen to be the case by forming a matrix $$A$$ whose columns are the vectors of $$F\text{,}$$ row-reducing $$A$$ to the identity matrix $$I_3\text{,}$$ and then using the nonsingularity of $$A$$ to assert that $$F$$ is a basis for $$\complex{3}$$ (Theorem CNMB). Now, since $$F$$ is a basis for $$\complex{3}\text{,}$$ Theorem CLI and Theorem CSS tell us that $$D$$ is also a basis for $$P_2\text{.}$$

Example CP2 illustrates the broad notion that computations in abstract vector spaces can be reduced to computations in $$\complex{m}\text{.}$$ You may have noticed this phenomenon as you worked through examples in Chapter VS or Chapter LT employing vector spaces of matrices or polynomials. These computations seemed to invariably result in systems of equations or the like from Chapter SLE, Chapter V and Chapter M. It is vector representation, $$\vectrepname{B}\text{,}$$ that allows us to make this connection formal and precise.

Knowing that vector representation allows us to translate questions about linear combinations, linear independence and spans from general vector spaces to $$\complex{m}$$ allows us to prove a great many theorems about how to translate other properties. Rather than prove these theorems, each of the same style as the other, we will offer some general guidance about how to best employ Theorem VRLT, Theorem CLI and Theorem CSS. This comes in the form of a “principle”: a basic truth, but most definitely not a theorem (hence, no proof).

This is a simple example of the The Coordinatization Principle, depending only on the fact that coordinatizing is an invertible linear transformation (Theorem VRILT). Suppose we have a linear combination to perform in $$M_{32}\text{,}$$ the vector space of $$3\times 2$$ matrices, but we are adverse to doing the operations of $$M_{32}$$ (Definition MA, Definition MSM). More specifically, suppose we are faced with the computation

\begin{equation*} 6 \begin{bmatrix} 3 & 7\\ -2 & 4\\ 0 & -3 \end{bmatrix} +2 \begin{bmatrix} -1 & 3\\ 4 & 8\\ -2 & 5 \end{bmatrix}\text{.} \end{equation*}

We choose a nice basis for $$M_{32}$$ (or a nasty basis if we are so inclined),

\begin{equation*} B=\set{ \begin{bmatrix}1&0\\0&0\\0&0\end{bmatrix},\, \begin{bmatrix}0&0\\1&0\\0&0\end{bmatrix},\, \begin{bmatrix}0&0\\0&0\\1&0\end{bmatrix},\, \begin{bmatrix}0&1\\0&0\\0&0\end{bmatrix},\, \begin{bmatrix}0&0\\0&1\\0&0\end{bmatrix},\, \begin{bmatrix}0&0\\0&0\\0&1\end{bmatrix} } \end{equation*}

and apply $$\vectrepname{B}$$ to each vector in the linear combination. This gives us a new computation, now in the vector space $$\complex{6}\text{,}$$ which we can compute with operations in $$\complex{6}$$ (Definition CVA, Definition CVSM),

\begin{gather*} 6\colvector{3\\-2\\0\\7\\4\\-3}+2\colvector{-1\\4\\-2\\3\\8\\5}=\colvector{16\\-4\\-4\\48\\40\\-8}\text{.} \end{gather*}

We are after the result of a computation in $$M_{32}\text{,}$$ so we now can apply $$\ltinverse{\vectrepname{B}}$$ to obtain a $$3\times 2$$ matrix,

which is exactly the matrix we would have computed had we just performed the matrix operations in the first place. So this was not meant to be an easier way to compute a linear combination of two matrices, just a different way.

SageSUTH2.Sage Under The Hood, Round 2.

You will have noticed that we have never constructed examples involving our favorite abstract vector spaces, such as the vector space of polynomials with fixed maximum degree, the vector space of matrices of a fixed size, or even the crazy vector space. There is nothing to stop us (or you) from implementing these examples in Sage as vector spaces. Maybe someday it will happen. But since Sage is built to be a tool for serious mathematical research, the designers recognize that this is not necessary.

Theorem CFDVS tells us that every finite-dimensional vector space can be described (loosely speaking) by just a field of scalars (for us, $$\complexes$$ in the text, QQin Sage) and the dimension. You can study whatever whacky vector space you might dream up, or whatever very complicated vector space that is important for particle physics, and through vector representation (“coordinatization”), you can convert your calculations to and from Sage.

1.

The vector space of $$3\times 5$$ matrices, $$M_{35}$$ is isomorphic to what fundamental vector space?

2.

A basis for $$\complex{3}$$ is

\begin{equation*} B=\set{ \colvector{1\\2\\-1},\, \colvector{3\\-1\\2},\, \colvector{1\\1\\1} }\text{.} \end{equation*}

Compute $$\vectrep{B}{\colvector{5\\8\\-1}}\text{.}$$

3.

What is the first “surprise,” and why is it surprising?

ExercisesVRExercises

C10.

In the vector space $$\complex{3}\text{,}$$ compute the vector representation $$\vectrep{B}{\vect{v}}$$ for the basis $$B$$ and vector $$\vect{v}\text{,}$$

\begin{align*} B&=\set{ \colvector{2\\-2\\2},\, \colvector{1\\3\\1},\, \colvector{3\\5\\2} } & \vect{v}&=\colvector{11\\5\\8}\text{.} \end{align*}
Solution.

We need to express the vector $$\vect{v}$$ as a linear combination of the vectors in $$B\text{.}$$ Theorem VRRB tells us we will be able to do this, and do it uniquely. The vector equation

\begin{equation*} a_1\colvector{2\\-2\\2}+ a_2\colvector{1\\3\\1}+ a_3\colvector{3\\5\\2} = \colvector{11\\5\\8} \end{equation*}

becomes (via Theorem SLSLC) a system of linear equations with augmented matrix,

\begin{equation*} \begin{bmatrix} 2 & 1 & 3 & 11\\ -2 & 3 & 5 & 5\\ 2 & 1 & 2 & 8 \end{bmatrix}\text{.} \end{equation*}

This system has the unique solution $$a_1=2\text{,}$$ $$a_2=-2\text{,}$$ $$a_3=3\text{.}$$ So by Definition VR,

\begin{equation*} \vectrep{B}{\vect{v}}=\vectrep{B}{\colvector{11\\5\\8}} = \vectrep{B}{ 2\colvector{2\\-2\\2}+ (-2)\colvector{1\\3\\1}+ 3\colvector{3\\5\\2} } =\colvector{2\\-2\\3}\text{.} \end{equation*}

C20.

Rework Example CM32 replacing the basis $$B$$ by the basis

\begin{equation*} C= \set{ \begin{bmatrix} -14 & -9 \\ 10 & 10 \\ -6 & -2 \end{bmatrix},\, \begin{bmatrix} -7 & -4 \\ 5 & 5 \\ -3 & -1 \end{bmatrix},\, \begin{bmatrix} -3 & -1 \\ 0 & -2 \\ 1 & 1 \end{bmatrix},\, \begin{bmatrix} -7 & -4 \\ 3 & 2 \\ -1 & 0 \end{bmatrix},\, \begin{bmatrix} 4 & 2 \\ -3 & -3 \\ 2 & 1 \end{bmatrix},\, \begin{bmatrix} 0 & 0 \\ -1 & -2 \\ 1 & 1 \end{bmatrix} }\text{.} \end{equation*}
Solution.

The following computations replicate the computations given in Example CM32, only using the basis $$C\text{,}$$

\begin{align*} \vectrep{C}{\begin{bmatrix}3 & 7 \\ -2 & 4 \\ 0 & -3\end{bmatrix}}&= \colvector{-9\\12\\-6\\7\\-2\\-1} & \vectrep{C}{\begin{bmatrix}-1 & 3 \\ 4 & 8 \\ -2 & 5\end{bmatrix}}&= \colvector{-11\\34\\-4\\-1\\16\\5}\\ 6\colvector{-9\\12\\-6\\7\\-2\\-1} + 2\colvector{-11\\34\\-4\\-1\\16\\5} &=\colvector{-76\\140\\-44\\40\\20\\4} & \vectrepinv{C}{\colvector{-76\\140\\-44\\40\\20\\4}}&= \begin{bmatrix}16 & 48 \\ -4 & 40 \\ -4 & -8\end{bmatrix}\text{.} \end{align*}

M10.

Prove that the set $$S$$ below is a basis for the vector space of $$2\times 2$$ matrices, $$M_{22}\text{.}$$ Do this by choosing a natural basis for $$M_{22}$$ and coordinatizing the elements of $$S$$ with respect to this basis. Examine the resulting set of column vectors from $$\complex{4}$$ and apply the The Coordinatization Principle,

\begin{equation*} S=\set{ \begin{bmatrix} 33 &99\\ 78 & -9 \end{bmatrix},\, \begin{bmatrix} -16 & -47\\ -36 & 2 \end{bmatrix},\, \begin{bmatrix} 10 & 27\\ 17 & 3 \end{bmatrix},\, \begin{bmatrix} -2 & -7\\ -6 & 4 \end{bmatrix} }\text{.} \end{equation*}

M20.

The set $$B=\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4}$$ is a basis of the vector space $$P_3\text{,}$$ polynomials with degree 3 or less. Therefore $$\vectrepname{B}$$ is a linear transformation, according to Theorem VRLT. Find a “formula” for $$\vectrepname{B}\text{.}$$ In other words, find an expression for $$\vectrep{B}{a+bx+cx^2+dx^3}\text{.}$$

\begin{align*} \vect{v}_1&=1 - 5x - 22x^2 + 3x^3 & \vect{v}_2&=-2 + 11x + 49x^2 - 8x^3\\ \vect{v}_3&=-1 + 7x + 33x^2 - 8x^3 & \vect{v}_4&=-1 + 4x + 16x^2 + x^3 \end{align*}
Solution.

Our strategy is to determine the values of the linear transformation on a “nice” basis for the domain, and then apply the ideas of Theorem LTDB to obtain our formula. $$\vectrepname{B}$$ is a linear transformation of the form $$\ltdefn{\vectrepname{B}}{P_3}{\complex{4}}\text{,}$$ so for a basis of the domain we choose a very simple one: $$C=\set{1,\,x,\,x^2,\,x^3}\text{.}$$ We now give the vector representations of the elements of $$C\text{,}$$ which are obtained by solving the relevant systems of equations obtained from linear combinations of the elements of $$B\text{.}$$

\begin{align*} \vectrep{B}{1} &=\colvector{20 \\ 7 \\ 1 \\ 4} & \vectrep{B}{x} &=\colvector{17 \\ 14 \\ -8 \\ -3} & \vectrep{B}{x^2}&=\colvector{-3 \\ -3 \\ 2 \\ 1} & \vectrep{B}{x^3}&=\colvector{0 \\ -1 \\ 1 \\ 1} &\text{.} \end{align*}

This is enough information to determine the linear transformation uniquely, and in particular, to allow us to use Theorem LTLC to construct a formula. We have

\begin{align*} \vectrep{B}{a+bx+cx^2+dx^3} &= a\vectrep{B}{1} + b\vectrep{B}{x} + c\vectrep{B}{x^2} + d\vectrep{B}{x^3}\\ &= a\colvector{20 \\ 7 \\ 1 \\ 4} + b\colvector{17 \\ 14 \\ -8 \\ -3} + c\colvector{-3 \\ -3 \\ 2 \\ 1} + d\colvector{0 \\ -1 \\ 1 \\ 1}\\ &= \colvector{ 20a + 17b - 3c\\ 7a + 14b - 3c - d\\ a - 8b + 2c + d\\ 4a - 3b + c + d }\text{.} \end{align*}