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Section 2.7 Subspace Basis and Dimension (V7)

Observation 2.7.1.

Recall 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 2.7.2. A linearly dependent set of three vectors

Activity 2.7.3.

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.

Activity 2.7.5.

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 2.7.6.

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 2.7.7.

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 2.7.8.

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

Observation 2.7.9.

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*}

Definition 2.7.11.

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.

Activity 2.7.12.

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

\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*}
\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*}
\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*}
\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 2.7.1 Videos

Figure 2.7.13. Video: Finding a basis of a subspace and computing the dimension of a subspace

Exercises 2.7.2 Exercises

Exercises available at checkit.clontz.org 1 .

https://checkit.clontz.org/#/banks/tbil-la/V7/