## Elementary Linear Algebra 7th Edition

$S$ does not span $M_{2,2}$.
The set $S=\left\{\left[\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right],\left[\begin{array}{ll}{0} & {1} \\ {1} & {0}\end{array}\right],\left[\begin{array}{ll}{1} & {1} \\ {0} & {0}\end{array}\right]\right\}$ is not a basis for $M_{2,2}$ because $S$ does not span $M_{2,2}$. For example, consider the matrix $\left[\begin{array}{ll}{1} & {0} \\ {0} & {0}\end{array}\right]$ as a linear combination of the elements of $S$ as follows \begin{align*} \left[\begin{array}{ll}{1} & {0} \\ {0} & {0}\end{array}\right]&=a\left[\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right]+b\left[\begin{array}{ll}{0} & {1} \\ {1} & {0}\end{array}\right]+c\left[\begin{array}{ll}{1} & {1} \\ {0} & {0}\end{array}\right]\\ &=\left[\begin{array}{ll}{a+c} & {b+c} \\ {b} & {a}\end{array}\right]. \end{align*} We have the following system of equations \begin{align*} a+c&=1\\ b+c&=0\\ a&=0\\ b&=0. \end{align*} It is clear that the above system is in-consistent. Hence, $S$ does not span $M_{2,2}$.