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Consider the equation $\mathit{\left[}\begin{array}{cc}3& 2\\ 4& 5\end{array}\right]S=S$where $S=\mathit{\left[}\begin{array}{cc}a& b\\ c& d\end{array}\mathit{\right]}$.

Substitute the value $\mathit{\left[}\begin{array}{cc}a& b\\ c& d\end{array}\mathit{\right]}$for S in the equation $\mathit{\left[}\begin{array}{cc}3& 2\\ 4& 5\end{array}\right]S=S$as follows.

$$\begin{gathered} \mathit{\left[}\begin{array}{cc}3& 2\\ 4& 5\end{array}\right]S=S \\ \mathit{\left[}\begin{array}{cc}3& 2\\ 4& 5\end{array}\right]\mathit{\left[}\begin{array}{cc}a& b\\ c& d\end{array}\mathit{\right]}=\mathit{\left[}\begin{array}{cc}a& b\\ c& d\end{array}\mathit{\right]} \\ \mathit{\left[}\begin{array}{cc}3a\mathit{+}\mathit{2}c& 3b\mathit{+}\mathit{2}d\\ 4a\mathit{+}\mathit{5}c& 4b\mathit{+}\mathit{5}d\end{array}\mathit{\right]}=\mathit{\left[}\begin{array}{cc}a& b\\ c& d\end{array}\mathit{\right]} \end{gathered}$$

Compare the equation both side as follows.

$$\begin{gathered} 3a+2c=a \\ 3b+2d=b \\ 4a+5c=c \\ 4b+5d=d \end{gathered}$$

Simplify the equation 3a +2c = a as follows.

3a +2c = a

2c = -2a

c = -a

Simplify the equation 3b + 2d = b as follows.

3b + 2d = b

2d = -2b

d = -b

Simplify the equation 4a + 5c = c as follows.

4a + 5c = c

4a + 5c = c

4a = -4c

a = -c

Simplify the equation 4b + 5d = d as follows.

4b + 5d = d

4b = -4d

b = -4d

Substitute the values -a for c and -b for d in the equation$S=\mathit{\left[}\begin{array}{cc}a& b\\ c& d\end{array}\mathit{\right]}$as follows.

localid="1659414511746" $$\begin{gathered} S=\mathit{\left[}\begin{array}{cc}a& b\\ c& d\end{array}\mathit{\right]} \\ S=\mathit{\left[}\begin{array}{cc}a& b\\ -a& -b\end{array}\mathit{\right]} \end{gathered}$$

Therefore, any matrix in the form$S=\mathit{\left[}\begin{array}{cc}a& b\\ -a& -b\end{array}\mathit{\right]}$satisfy the equation $\mathit{\left[}\begin{array}{cc}3& 2\\ 4& 5\end{array}\right]S=S$.

Simplify the equation$S=\mathit{\left[}\begin{array}{cc}a& b\\ a& -b\end{array}\mathit{\right]}$as follows.

$$\begin{gathered} S=\mathit{\left[}\begin{array}{cc}a& b\\ a& -b\end{array}\mathit{\right]} \\ S=\mathit{\left[}\begin{array}{cc}a& 0\\ a& 0\end{array}\mathit{\right]}+\mathit{\left[}\begin{array}{cc}0& b\\ 0& -b\end{array}\mathit{\right]} \\ S=a\mathit{\left[}\begin{array}{cc}1& 0\\ 1& 0\end{array}\mathit{\right]}\mathit{+}b\mathit{\left[}\begin{array}{cc}0& 1\\ 0& -1\end{array}\mathit{\right]} \end{gathered}$$

where $\mathit{\left[}\begin{array}{cc}1& 0\\ 1& 0\end{array}\mathit{\right]}\mathit{}\mathrm{and}\mathit{}\mathit{\left[}\begin{array}{cc}0& 1\\ 0& -1\end{array}\mathit{\right]}$are linear independent.

Therefore, the matrix S is spanned by $Span\left\{\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ 0& -1\end{array}\right]\right\}$.

Hence, the dimension of S is 2 and spanned by $Span\mathit{\left\{}\mathit{\left[}\begin{array}{cc}\mathit{1}& \mathit{0}\\ \mathit{1}& \mathit{0}\end{array}\mathit{\right]}\mathit{,}\mathit{\left[}\begin{array}{cc}\mathit{0}& \mathit{1}\\ \mathit{0}& \mathit{-}\mathit{1}\end{array}\mathit{\right]}\mathit{\right\}}$.