91Ó°ÊÓ

Consider the equation $\mathit{\left[}\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]S=S$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 $\mathit{\left[}\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]S=S$as follows.

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

Compare the equation both side as follows.

$$\begin{gathered} a+c=a \\ b+d=b \\ a+c=c \\ b+d=d \end{gathered}$$

Simplify the equation b + d = b as follows.

b + d = b

d = 0

Simplify the equation a + c = a as follows.

a + c = a

c = 0

Simplify the equation a + c = c as follows.

a + c = c

a = 0

Simplify the equation b + d = d as follows.

b + d = d

b = 0

Substitute the values 0 for a, 0 for b, 0 for c and 0 for d in the equation $S=\mathit{\left[}\begin{array}{cc}a& b\\ c& d\end{array}\mathit{\right]}$as follows.

$$\begin{gathered} S=\mathit{\left[}\begin{array}{cc}a& b\\ c& d\end{array}\mathit{\right]} \\ S=\mathit{\left[}\begin{array}{cc}2& 0\\ 0& 0\end{array}\mathit{\right]} \end{gathered}$$

Therefore, any matrix in the form $S=\mathit{\left[}\begin{array}{cc}2& 0\\ 0& 0\end{array}\mathit{\right]}$satisfy the equation $\mathit{\left[}\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]S=S$.

The matrix S is spanned by $Span\left\{\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\right\}$.

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