91影视

Consider the equation $\mathit{\left[}\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]S=S\mathit{\left[}\begin{array}{cc}2& 0\\ 0& 0\end{array}\mathit{\right]}$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 $\mathit{\left[}\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]S=S\mathit{\left[}\begin{array}{cc}2& 0\\ 0& 0\end{array}\mathit{\right]}$ as follows.

role="math" localid="1659408717313" $$\begin{gathered} \mathit{\left[}\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]S=S\mathit{\left[}\begin{array}{cc}2& 0\\ 0& 0\end{array}\mathit{\right]} \\ \mathit{\left[}\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\mathit{\left[}\begin{array}{cc}a& b\\ c& d\end{array}\mathit{\right]}=\mathit{\left[}\begin{array}{cc}\mathit{a}& \mathit{b}\\ \mathit{c}& \mathit{d}\end{array}\mathit{\right]}\mathit{\left[}\begin{array}{cc}\mathit{2}& \mathit{0}\\ \mathit{0}& \mathit{0}\end{array}\mathit{\right]} \\ \left[\begin{array}{cc}a+c& b+d\\ a+c& b+d\end{array}\right]=\mathit{\left[}\begin{array}{cc}2\mathrm{a}& 0\\ 2\mathrm{c}& 0\end{array}\mathit{\right]} \end{gathered}$$

Compare the equation both side as follows.

$$\begin{gathered} a+c=2a \\ b+d=0 \\ a+c=2c \end{gathered}$$

Simplify the equation b + d = 0 as follows.

$$\begin{gathered} b+d=0 \\ b=-d \end{gathered}$$

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

$$\begin{gathered} a+c=2a \\ c=a \end{gathered}$$

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.

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

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

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

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

where $\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]\mathrm{and}\left[\begin{array}{cc}0& 1\\ 0& -1\end{array}\right]$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 2and spanned by $Span\left\{\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right],\left[\begin{array}{cc}0& 1\\ 0& -1\end{array}\right]\right\}$.