Subspace Basis and Dimension (V7)

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.

\begin{tikzpicture}[x={(210:0.8cm)}, y={(0:1cm)}, z={(90:1cm)},scale=0.4]
  \draw[->] (0,0,0) -- (6,0,0);
  \draw[->] (0,0,0) -- (0,6,0);
  \draw[->] (0,0,0) -- (0,0,6);
  \draw[fill=purple!20,fill opacity=0.5]
    (-2,-2,2) -- (6,-2,-2) -- (2,2,-2) -- (-6,2,2) -- (-2,-2,2);
  \draw[thick,blue,->] (0,0,0) -- (1,-1,0);
  \draw[thick,red,->] (0,0,0) -- (-2,0,1);
  \draw[thick,purple,->] (0,0,0) -- (1,1,-1);
\end{tikzpicture}

Activity 2.7.1.

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 2.7.2.

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 2.7.2.

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 2.7.3.

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 2.7.4.

Let \(W\) be the subspace of \(M_{2,2}\) 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\}. \end{equation*}

Find a basis for \(W\text{.}\)

Activity 2.7.5.

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 2.7.3.

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 2.7.4.

Any non-trivial 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 2.7.5.

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 2.7.6.

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*}

Exercises 2.7.1 Exercises

1.

Consider the subspace

\begin{equation*} W=\operatorname{span} \left\{ \left[\begin{array}{c} -4 \\ -4 \\ 1 \\ -2 \end{array}\right] , \left[\begin{array}{c} -3 \\ -2 \\ 3 \\ -3 \end{array}\right] , \left[\begin{array}{c} -7 \\ -6 \\ 4 \\ -5 \end{array}\right] , \left[\begin{array}{c} -4 \\ -2 \\ 3 \\ 3 \end{array}\right] \right\} . \end{equation*}

Explain how to find a basis of \(W\text{.}\)
Explain how to find the dimension of \(W\text{.}\)

Answer

\begin{equation*} \operatorname{RREF} \left[\begin{array}{cccc} -4 & -3 & -7 & -4 \\ -4 & -2 & -6 & -2 \\ 1 & 3 & 4 & 3 \\ -2 & -3 & -5 & 3 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}

A basis of \(W\) is \(\left\{ \left[\begin{array}{c} -4 \\ -4 \\ 1 \\ -2 \end{array}\right] , \left[\begin{array}{c} -3 \\ -2 \\ 3 \\ -3 \end{array}\right] , \left[\begin{array}{c} -4 \\ -2 \\ 3 \\ 3 \end{array}\right] \right\} \text{.}\)
The dimension of \(W\) is \(3 \text{.}\)

2.

Consider the subspace

\begin{equation*} W=\operatorname{span} \left\{ \left[\begin{array}{c} -2 \\ 2 \\ 3 \\ -4 \end{array}\right] , \left[\begin{array}{c} 2 \\ -3 \\ -2 \\ -3 \end{array}\right] , \left[\begin{array}{c} 0 \\ -1 \\ -2 \\ 3 \end{array}\right] , \left[\begin{array}{c} -2 \\ 2 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} 4 \\ -8 \\ -8 \\ 0 \end{array}\right] \right\} . \end{equation*}

Explain how to find a basis of \(W\text{.}\)
Explain how to find the dimension of \(W\text{.}\)

Answer

\begin{equation*} \operatorname{RREF} \left[\begin{array}{ccccc} -2 & 2 & 0 & -2 & 4 \\ 2 & -3 & -1 & 2 & -8 \\ 3 & -2 & -2 & 0 & -8 \\ -4 & -3 & 3 & 1 & 0 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & 0 \end{array}\right] \end{equation*}

A basis of \(W\) is \(\left\{ \left[\begin{array}{c} -2 \\ 2 \\ 3 \\ -4 \end{array}\right] , \left[\begin{array}{c} 2 \\ -3 \\ -2 \\ -3 \end{array}\right] , \left[\begin{array}{c} 0 \\ -1 \\ -2 \\ 3 \end{array}\right] , \left[\begin{array}{c} -2 \\ 2 \\ 0 \\ 1 \end{array}\right] \right\} \text{.}\)
The dimension of \(W\) is \(4 \text{.}\)

3.

Consider the subspace

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

Explain how to find a basis of \(W\text{.}\)
Explain how to find the dimension of \(W\text{.}\)

Answer

\begin{equation*} \operatorname{RREF} \left[\begin{array}{cccccc} 3 & 0 & -1 & 3 & -1 & 1 \\ -3 & -2 & 2 & -3 & -3 & 3 \\ -2 & 0 & 3 & -2 & 0 & -2 \\ 2 & 1 & -4 & 2 & -4 & 2 \end{array}\right] = \left[\begin{array}{cccccc} 1 & 0 & 0 & 1 & 0 & \frac{1}{43} \\ 0 & 1 & 0 & 0 & 0 & -\frac{76}{43} \\ 0 & 0 & 1 & 0 & 0 & -\frac{28}{43} \\ 0 & 0 & 0 & 0 & 1 & -\frac{12}{43} \end{array}\right] \end{equation*}

A basis of \(W\) is \(\left\{ \left[\begin{array}{c} 3 \\ -3 \\ -2 \\ 2 \end{array}\right] , \left[\begin{array}{c} 0 \\ -2 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 2 \\ 3 \\ -4 \end{array}\right] , \left[\begin{array}{c} -1 \\ -3 \\ 0 \\ -4 \end{array}\right] \right\} \text{.}\)
The dimension of \(W\) is \(4 \text{.}\)

4.

Consider the subspace

\begin{equation*} W=\operatorname{span} \left\{ \left[\begin{array}{c} 3 \\ 3 \\ 1 \\ -3 \end{array}\right] , \left[\begin{array}{c} -1 \\ -3 \\ 0 \\ -2 \end{array}\right] , \left[\begin{array}{c} -5 \\ -9 \\ -1 \\ -1 \end{array}\right] , \left[\begin{array}{c} 15 \\ 27 \\ 3 \\ 3 \end{array}\right] , \left[\begin{array}{c} 1 \\ -3 \\ 1 \\ -7 \end{array}\right] \right\} . \end{equation*}

Explain how to find a basis of \(W\text{.}\)
Explain how to find the dimension of \(W\text{.}\)

Answer

\begin{equation*} \operatorname{RREF} \left[\begin{array}{ccccc} 3 & -1 & -5 & 15 & 1 \\ 3 & -3 & -9 & 27 & -3 \\ 1 & 0 & -1 & 3 & 1 \\ -3 & -2 & -1 & 3 & -7 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & -1 & 3 & 1 \\ 0 & 1 & 2 & -6 & 2 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}

A basis of \(W\) is \(\left\{ \left[\begin{array}{c} 3 \\ 3 \\ 1 \\ -3 \end{array}\right] , \left[\begin{array}{c} -1 \\ -3 \\ 0 \\ -2 \end{array}\right] \right\} \text{.}\)
The dimension of \(W\) is \(2 \text{.}\)

5.

Consider the subspace

\begin{equation*} W=\operatorname{span} \left\{ \left[\begin{array}{c} 0 \\ 0 \\ -2 \\ -2 \end{array}\right] , \left[\begin{array}{c} -4 \\ 1 \\ -1 \\ -1 \end{array}\right] , \left[\begin{array}{c} 12 \\ -3 \\ -1 \\ -1 \end{array}\right] , \left[\begin{array}{c} -4 \\ 1 \\ -1 \\ -1 \end{array}\right] , \left[\begin{array}{c} 2 \\ 2 \\ 3 \\ 2 \end{array}\right] , \left[\begin{array}{c} 16 \\ -4 \\ 10 \\ 10 \end{array}\right] \right\} . \end{equation*}

Explain how to find a basis of \(W\text{.}\)
Explain how to find the dimension of \(W\text{.}\)

Answer

\begin{equation*} \operatorname{RREF} \left[\begin{array}{cccccc} 0 & -4 & 12 & -4 & 2 & 16 \\ 0 & 1 & -3 & 1 & 2 & -4 \\ -2 & -1 & -1 & -1 & 3 & 10 \\ -2 & -1 & -1 & -1 & 2 & 10 \end{array}\right] = \left[\begin{array}{cccccc} 1 & 0 & 2 & 0 & 0 & -3 \\ 0 & 1 & -3 & 1 & 0 & -4 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}

A basis of \(W\) is \(\left\{ \left[\begin{array}{c} 0 \\ 0 \\ -2 \\ -2 \end{array}\right] , \left[\begin{array}{c} -4 \\ 1 \\ -1 \\ -1 \end{array}\right] , \left[\begin{array}{c} 2 \\ 2 \\ 3 \\ 2 \end{array}\right] \right\} \text{.}\)
The dimension of \(W\) is \(3 \text{.}\)

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