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 $\Span(S)$.

 
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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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