## Section2.7Subspace Basis and Dimension (VS7)

### Subsection2.7.1Class Activities

#### Observation2.7.1.

Recall from section Section 2.4 that a subspace of a vector space is a subset that is itself a vector space.

One easy way to construct a subspace is to take the span of set, but a linearly dependent set contains “redundant” vectors. For example, only two of the three vectors in the following image are needed to span the planar subspace.

#### Activity2.7.2.

Consider the subspace of $$\IR^4$$ given by $$W=\vspan\left\{ \left[\begin{array}{c}2\\3\\0\\1\end{array}\right], \left[\begin{array}{c}2\\0\\1\\-1\end{array}\right], \left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right], \left[\begin{array}{c}1\\5\\-1\\0\end{array}\right] \right\} \text{.}$$

##### (a)

Mark the part of $$\RREF\left[\begin{array}{cccc} 2&2&2&1\\ 3&0&-3&5\\ 0&1&2&-1\\ 1&-1&-3&0 \end{array}\right]$$ that shows that $$W$$'s spanning set is linearly dependent.

##### (b)

Find a basis for $$W$$ by removing a vector from its spanning set to make it linearly independent.

rref([2,2,2,1; 3,0,-3,5; 0,1,2,-1; 1,-1,-3,0])


#### Activity2.7.4.

Let $$W$$ be the subspace of $$\IR^4$$ given by

\begin{equation*} W = \vspan \left\{ \left[\begin{array}{c} 1 \\ 3 \\ 1 \\ -1 \end{array}\right], \left[\begin{array}{c} 2 \\ -1 \\ 1 \\ 2 \end{array}\right], \left[\begin{array}{c} 4 \\ 5 \\ 3 \\ 0 \end{array}\right], \left[\begin{array}{c} 3 \\ 2 \\ 2 \\ 1 \end{array}\right] \right\} \text{.} \end{equation*}

Find a basis for $$W\text{.}$$

#### Activity2.7.5.

Let $$W$$ be the subspace of $$\P_3$$ given by

\begin{equation*} W = \vspan \left\{x^3+3x^2+x-1, 2x^3-x^2+x+2, 4x^3+5x^2+3x, 3x^3+2x^2+2x+1 \right\} \end{equation*}

Find a basis for $$W\text{.}$$

#### Activity2.7.6.

Let $$W$$ be the subspace of $$M_{2,2}$$ given by

\begin{equation*} W = \vspan \left\{ \left[\begin{array}{cc} 1 & 3 \\ 1 & -1 \end{array}\right], \left[\begin{array}{cc} 2 & -1 \\ 1 & 2 \end{array}\right], \left[\begin{array}{cc} 4 & 5 \\ 3 & 0 \end{array}\right], \left[\begin{array}{cc} 3 & 2 \\ 2 & 1 \end{array}\right] \right\}. \end{equation*}

Find a basis for $$W\text{.}$$

#### Activity2.7.7.

Let

\begin{equation*} S=\left\{ \left[\begin{array}{c}2\\3\\0\\1\end{array}\right], \left[\begin{array}{c}2\\0\\1\\-1\end{array}\right], \left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right], \left[\begin{array}{c}1\\5\\-1\\0\end{array}\right] \right\} \end{equation*}

and

\begin{equation*} T=\left\{ \left[\begin{array}{c}2\\0\\1\\-1\end{array}\right], \left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right], \left[\begin{array}{c}1\\5\\-1\\0\end{array}\right], \left[\begin{array}{c}2\\3\\0\\1\end{array}\right] \right\}\text{.} \end{equation*}
##### (a)

Find a basis for $$\vspan S\text{.}$$

##### (b)

Find a basis for $$\vspan T\text{.}$$

#### Observation2.7.8.

Even though we found different bases for them, $$\vspan S$$ and $$\vspan T$$ are exactly the same subspace of $$\IR^4\text{,}$$ since

\begin{equation*} S=\left\{ \left[\begin{array}{c}2\\3\\0\\1\end{array}\right], \left[\begin{array}{c}2\\0\\1\\-1\end{array}\right], \left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right], \left[\begin{array}{c}1\\5\\-1\\0\end{array}\right] \right\} = \left\{ \left[\begin{array}{c}2\\0\\1\\-1\end{array}\right], \left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right], \left[\begin{array}{c}1\\5\\-1\\0\end{array}\right], \left[\begin{array}{c}2\\3\\0\\1\end{array}\right] \right\}=T\text{.} \end{equation*}

#### Definition2.7.10.

The dimension of a vector space is equal to the size of any basis for the vector space.

As you'd expect, $$\IR^n$$ has dimension $$n\text{.}$$ For example, $$\IR^3$$ has dimension $$3$$ because any basis for $$\IR^3$$ such as

\begin{equation*} \setList{\vec e_1,\vec e_2,\vec e_3} \text{ and } \setList{ \left[\begin{array}{c}1\\0\\0\end{array}\right], \left[\begin{array}{c}0\\1\\0\end{array}\right], \left[\begin{array}{c}1\\1\\1\end{array}\right] } \text{ and } \setList{ \left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\-2\\1\end{array}\right], \left[\begin{array}{c}3\\-2\\5\end{array}\right] } \end{equation*}

contains exactly three vectors.

#### Activity2.7.11.

Find the dimension of each subspace of $$\IR^4$$ by finding $$\RREF$$ for each corresponding matrix.

##### (a)
\begin{equation*} \vspan\left\{ \left[\begin{array}{c}2\\3\\0\\-1\end{array}\right], \left[\begin{array}{c}2\\0\\0\\3\end{array}\right], \left[\begin{array}{c}4\\3\\0\\2\end{array}\right], \left[\begin{array}{c}-3\\0\\1\\3\end{array}\right] \right\} \end{equation*}
##### (b)
\begin{equation*} \vspan\left\{ \left[\begin{array}{c}2\\3\\0\\-1\end{array}\right], \left[\begin{array}{c}2\\0\\0\\3\end{array}\right], \left[\begin{array}{c}3\\13\\7\\16\end{array}\right], \left[\begin{array}{c}-1\\10\\7\\14\end{array}\right], \left[\begin{array}{c}4\\3\\0\\2\end{array}\right] \right\} \end{equation*}
##### (c)
\begin{equation*} \vspan\left\{ \left[\begin{array}{c}2\\3\\0\\-1\end{array}\right], \left[\begin{array}{c}4\\3\\0\\2\end{array}\right], \left[\begin{array}{c}-3\\0\\1\\3\end{array}\right], \left[\begin{array}{c}3\\6\\1\\5\end{array}\right] \right\} \end{equation*}
##### (d)
\begin{equation*} \vspan\left\{ \left[\begin{array}{c}5\\3\\0\\-1\end{array}\right], \left[\begin{array}{c}-2\\1\\0\\3\end{array}\right], \left[\begin{array}{c}4\\5\\1\\3\end{array}\right] \right\} \end{equation*}

### Subsection2.7.3Slideshow

Slideshow of activities available at https://teambasedinquirylearning.github.io/linear-algebra/2022/VS7.slides.html.

### Subsection2.7.5Mathematical Writing Explorations

#### Exploration2.7.12.

Prove each of the following statements is true.
• If $$\{\vec{b}_1, \vec{b}_2,\ldots, \vec{b}_m\}$$ and $$\{\vec{c}_1,\vec{c}_2,\ldots,\vec{c}_n\}$$ are each a basis for a vector space $$V\text{,}$$ then $$m=n.$$

• If $$\{\vec{v}_1,\vec{v}_2\ldots, \vec{v}_n\}$$ is linearly independent, then so is $$\{\vec{v}_1,\vec{v}_1 + \vec{v}_2, \ldots, \vec{v}_1 + \vec{v}_2 + \cdots + \vec{v}_n\}\text{.}$$

• Let $$V$$ be a vector space of dimension $$n\text{,}$$ and $$\vec{v} \in V\text{.}$$ Then there exists a basis for $$V$$ which contains $$\vec{v}\text{.}$$

#### Exploration2.7.13.

Suppose we have the set of all function $$f:S \rightarrow \mathbb{R}\text{.}$$ We claim that this is a vector space under the usual operation of function addition and scalar multiplication. What is the dimension of this space for each choice of $$S$$ below:
• $$\displaystyle S = \{1\}$$

• $$\displaystyle S = \{1,2\}$$

• $$\displaystyle S = \{1,2,\ldots ,n\}$$

• $$\displaystyle S = \mathbb{R}$$

#### Exploration2.7.14.

Suppose you have the vector space $$V = \left\{\left(\begin{array}{c}x\\y\\z\end{array}\right)\in \mathbb{R}^3: x + y + z = 1\right\}$$ with the operations $$\left(\begin{array}{c}x_1\\y_1\\z_1\end{array}\right) \oplus \left(\begin{array}{c}x_2\\y_2\\z_2\end{array}\right) = \left(\begin{array}{c}x_1 + x_2 - 1\\y_1 + y_2\\z_1+z_2\end{array}\right) \mbox{ and } \alpha\odot\left(\begin{array}{c}x_1\\y_1\\z_1\end{array}\right) = \left(\begin{array}{c}\alpha x_1 - \alpha +1\\\alpha y_1\\\alpha z_1\end{array}\right).$$ Find a basis for $$V$$ and determine it's dimension.

### Subsection2.7.6Sample Problem and Solution

Sample problem Example B.1.11.