# Find a Basis for a Subspace of the Vector Space of $2\times 2$ Matrices

## Problem 152

Let $V$ be the vector space of all $2\times 2$ matrices, and let the subset $S$ of $V$ be defined by $S=\{A_1, A_2, A_3, A_4\}$, where

\begin{align*}

A_1=\begin{bmatrix}

1 & 2 \\

-1 & 3

\end{bmatrix}, \quad

A_2=\begin{bmatrix}

0 & -1 \\

1 & 4

\end{bmatrix}, \quad

A_3=\begin{bmatrix}

-1 & 0 \\

1 & -10

\end{bmatrix}, \quad

A_4=\begin{bmatrix}

3 & 7 \\

-2 & 6

\end{bmatrix}.

\end{align*}

Find a basis of the span $\Span(S)$ consisting of vectors in $S$ and find the dimension of $\Span(S)$.

## Proof.

Let $B=\{E_{11}, E_{12}, E_{21}, E_{22}\}$ be the standard basis for the vector space $V$, 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}.\]
With respect to the basis $B$, we find the coordinate vectors for $A_1, A_2, A_3, A_4$ as follows.

Since we have

\[A_1=E_{11}+2E_{12}-E_{21}+3E_{22},\]
the coordinate vector for $A_1$ is

\[[A_1]_B=\begin{bmatrix}

1 \\

2 \\

-1 \\

3

\end{bmatrix}.\]
Similarly, we have

\[[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}.\]
To find a basis for $\Span(S)$ among vectors in $S$, we first find a basis for $\Span(T)$ among vectors in

\[T=\{[A_1]_B, [A_2]_B, [A_3]_B, [A_4]_B\}.\]
Let form a matrix whose columns are vectors in $T$. That is,

\[\begin{bmatrix}

1 & 0 & -1 & 3 \\

2 &-1 & 0 & 7 \\

-1 & 1 & 1 & -2 \\

3 & 4 & -10 & 6

\end{bmatrix}.\]

We apply the elementary row operations as follows and obtain a reduced row echelon form matrix.

\begin{align*}

\begin{bmatrix}

1 & 0 & -1 & 3 \\

2 &-1 & 0 & 7 \\

-1 & 1 & 1 & -2 \\

3 & 4 & -7 & 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}

\xrightarrow{\substack{R_3+R_2 \\ R_4+4R_2}}

\begin{bmatrix}

1 & 0 & -1 & 3 \\

0 &-1 & 2 & 1 \\

0 & 0 & 2 & 2 \\

0 & 0& 1 & 1

\end{bmatrix}\\[6pt]
\xrightarrow{\substack{R_1+R_4 \\ R_2-2R_4 \\ R_3-2R_4}}

\begin{bmatrix}

1 & 0 & 0 & 4 \\

0 &-1 & 0 & -1 \\

0 & 0 & 0 & 0 \\

0 & 0& 1 & 1

\end{bmatrix}

\xrightarrow{\substack{1R_2\\ R_3 \leftrightarrow R_4}}

\begin{bmatrix}

1 & 0 & 0 & 4 \\

0 &1 & 0 & 1 \\

0 & 0 & 1 & 1 \\

0 & 0& 0 & 0

\end{bmatrix}.

\end{align*}

The the first three columns of the reduced row echelon form contains the leading 1’s.

Thus by, what I call, the leading $1$ method, it follows that

\[\{[A_1]_B, [A_2]_B, [A_3]_B\}\]
is a basis for $\Span(T)$.

Therefore, by the correspondence of coordinate vectors, we obtain that

\[\{A_1, A_2, A_3\}\]
is a basis of $\Span(S)$.

Since the basis $\{A_1, A_2, A_3\}$ consists of three vectors, the dimension of $\Span(S)$ is $3$.

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