# Quiz 9. Find a Basis of the Subspace Spanned by Four Matrices ## Problem 349

Let $V$ be the vector space of all $2\times 2$ real matrices.
Let $S=\{A_1, A_2, A_3, A_4\}$, where
$A_1=\begin{bmatrix} 1 & 2\\ -1& 3 \end{bmatrix}, A_2=\begin{bmatrix} 0 & -1\\ 1& 4 \end{bmatrix}, A_3=\begin{bmatrix} -1 & 0\\ 1& -10 \end{bmatrix}, A_4=\begin{bmatrix} 3 & 7\\ -2& 6 \end{bmatrix}.$ Then find a basis for the span $\Span(S)$. Add to solve later

## Solution.

Let $B=\{E_{11}, E_{12}, E_{21}, E_{22}\}$ be the standard basis of $V$, where
$E_{11}=\begin{bmatrix} 1 & 0\\ 0& 0 \end{bmatrix}, E_{12}=\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix}, E_{21}=\begin{bmatrix} 0 & 0\\ 1& 0 \end{bmatrix}, E_{22}=\begin{bmatrix} 0 & 0\\ 0& 1 \end{bmatrix}.$ We find the coordinate vectors of $A_1, A_2, A_3, A_4$ with respect to the basis $B$.

For the matrix $A_1$, we can write it as a linear combination of basis vectors of $B$ as follows.
We have
$A_1=1\cdot E_{11}+2E_{12}+(-1)E_{21}+3E_{22}.$ Hence the coordinate vector of $A_1$ with respect to the basis $B$ is
$[A_1]_B=\begin{bmatrix} 1 \\ 2 \\ -1 \\ 3 \end{bmatrix}.$ Similarly, we obtain
$[A_2]_B=\begin{bmatrix} 0 \\ -1 \\ 1 \\ 4 \end{bmatrix}, [A_3]_B=\begin{bmatrix} -1 \\ 0 \\ 1 \\ -10 \end{bmatrix}, [A_4]_B=\begin{bmatrix} 3 \\ 7 \\ -2 \\ 6 \end{bmatrix}.$

Let
\begin{align*}
T&=\{[A_1]_B, [A_2]_B, [A_3]_B, [A_4]_B\}\6pt] &= \left \{ \,\begin{bmatrix} 1 \\ 2 \\ -1 \\ 3 \end{bmatrix}, \begin{bmatrix} 0 \\ -1 \\ 1 \\ 4 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \\ -10 \end{bmatrix}, \begin{bmatrix} 3 \\ 7 \\ -2 \\ 6 \end{bmatrix} \, \right\}. \end{align*} Then T is a subset in \R^4. We find a basis of \Span(T) by the leading 1-method. We form a matrix whose columns are vectors in T and reduce it by elementary row operations as follows. \begin{align*} \begin{bmatrix} 1 & 0 & -1 & 3 \\ 2 &-1 & 0 & 7 \\ -1 & 1 & 1 & -2 \\ 3 & 4 & -10 & 6 \end{bmatrix} \xrightarrow{\substack{R_2-2R_1 \\ R_3+R_1 \\ R_4-3R_1}} \begin{bmatrix} 1 & 0 & -1 & 3 \\ 0 &-1 & 2 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 4 & -7 & -3 \end{bmatrix}\\[6pt] \xrightarrow{R_2 \leftrightarrow R_3} \begin{bmatrix} 1 & 0 & -1 & 3 \\ 0 & 1 & 0 & 1 \\ 0 &-1 & 2 & 1 \\ 0 & 4 & -7 & -3 \end{bmatrix} \xrightarrow{\substack{R_3+R_2 \\R_4-4R_2}} \begin{bmatrix} 1 & 0 & -1 & 3 \\ 0 &1 & 0 & 1 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & -7 & -7 \end{bmatrix}\\[6pt] \xrightarrow{\substack{\frac{1}{2}R_3 \\ -\frac{1}{7} R_4}} \begin{bmatrix} 1 & 0 & -1 & 3 \\ 0 &1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix} \xrightarrow{\substack{R_1+R_3 \\ R_4-R_3}} \begin{bmatrix} 1 & 0 & 0 & 4 \\ 0 &1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}. \end{align*} The last matrix is in reduced row echelon form and the first three columns contain the leading 1. Thus the first three vectors of T is a basis of \Span(T). Hence \[\{[A_1]_B, [A_2]_B, [A_3]_B\} is a basis of $\Span(T)$.
This yields that
$\{A_1, A_2, A_3\}$ is a basis of $\Span(S)$ by the correspondence of coordinate vectors.

## Comment.

These are Quiz 9 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.

### List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017

There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions. Add to solve later

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