TBIL-LA Subspace Basis and Dimension (VS7)

Section 2.7 Subspace Basis and Dimension (VS7)

Learning Outcomes

Compute a basis for the subspace spanned by a given set of Euclidean vectors, and determine the dimension of the subspace.

Subsection 2.7.1 Class Activities

Observation 2.7.1.

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.

Figure 19. A linearly dependent set of three vectors

Activity 2.7.2.

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.

Fact 2.7.3.

Let \(S=\{\vec v_1,\dots,\vec v_m\}\text{.}\) The easiest basis describing \(\vspan S\) is the set of vectors in \(S\) given by the pivot columns of \(\RREF[\vec v_1\,\dots\,\vec v_m]\text{.}\)

Put another way, to compute a basis for the subspace \(\vspan S\text{,}\) simply remove the vectors corresponding to the non-pivot columns of \(\RREF[\vec v_1\,\dots\,\vec v_m]\text{.}\) For example, since

\begin{equation*} \RREF \left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & -2 & -2 \\ -3 & 1 & -2 \end{array}\right] = \left[\begin{array}{ccc} \circledNumber{1} & 0 & 1 \\ 0 & \circledNumber{1} & 1 \\ 0 & 0 & 0 \end{array}\right] \end{equation*}

the subspace \(W=\vspan\setList{ \left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\-2\\1\end{array}\right], \left[\begin{array}{c}3\\-2\\-2\end{array}\right] }\) has \(\setList{ \left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}2\\-2\\1\end{array}\right] }\) as a basis.

Activity 2.7.4.

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 \(W\text{.}\)

Activity 2.7.5.

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 \(W\text{.}\)

Activity 2.7.6.

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 \(W\text{.}\)

Activity 2.7.7.

Let

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 \(\vspan T\text{.}\)

Observation 2.7.8.

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*}

Fact 2.7.9.

Any non-trivial real vector space has infinitely-many different bases, but all the bases for a given vector space are exactly the same size.

For example,

\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*}

are all valid bases for \(\IR^3\text{,}\) and they all contain three vectors.

Definition 2.7.10.

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

contains exactly three vectors.

Activity 2.7.11.

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*}

Subsection 2.7.2 Videos

Figure 20. Video: Finding a basis of a subspace and computing the dimension of a subspace

Subsection 2.7.3 Slideshow

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

Exercises 2.7.4 Exercises

Exercises available at https://teambasedinquirylearning.github.io/linear-algebra/2022/exercises/#/bank/VS7/.

Subsection 2.7.5 Mathematical Writing Explorations

Exploration 2.7.12.

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 \(\vec{v}\text{.}\)

Exploration 2.7.13.

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:

\(\displaystyle S = \{1\}\)
\(\displaystyle S = \{1,2\}\)
\(\displaystyle S = \{1,2,\ldots ,n\}\)
\(\displaystyle S = \mathbb{R}\)

Exploration 2.7.14.

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.

Subsection 2.7.6 Sample Problem and Solution

Sample problem Example B.1.11.