# Let $M_{2\times 2}$ be the vector space of all $2\times 2$ matrices. Show that the set of non-singular matrices is NOT a subspace.

I am working on problems from my textbook. However, I am lost as to how to show this.

A. Let $M_{2\times 2}$ be the vector space of all $2\times 2$ matrices. Show that the set of non-singular $2\times 2$ matrices is NOT a subspace of $M_{2\times 2}$.

B. Let $M_{2\times 2}$ be the vector space of all $2\times 2$ matrices. Show that the set of singular $2\times 2$ matrices is NOT a subspace of$M_{2\times 2}$.

C. Describe the smallest subspace of $M_{2\times 2}$ that contains matrices $$\begin{pmatrix}2 & 1 \\ 0 & 0\end{pmatrix},\ \ \begin{pmatrix}1 & 0 \\ 0 & 2\end{pmatrix}\ \ \text{and}\ \ \ \begin{pmatrix}0 & -1 \\ 0 & 0\end{pmatrix}$$

Find the dimension of the subspace.

I think I can prove that addition for A and B is not closed, thus disproving the potential for subspace. Though, I am not sure about C.

For item A, I'd like you to consider the following matrices: $A = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$ and $B = \begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix}$. Both $A,B \in M_{2\times2}$. These two matrices are both non-singular since $|A|=|B|= 1\ne 0$. but their sum, $A + B = 0_{2\times2}$ is singular. Thus, because of violation of closure, the set of non-singular $2\times2$ matrices is not a subspace of $M_{2\times2}$.
For item B, now consider, $C = \begin{bmatrix}0 & 0\\0 & 1\end{bmatrix}$ and $D = \begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}$. For both matrices, we have a row of zeroes, and consequently, since $|C|=|D| = 0$, both matrices are singular. But the sum, $C + D = I_{2\times2}$ which is non-singular. Again, in violation of closure, the set of singular $2\times2$ matrices is not a subspace of $M_{2\times2}$.
Now for item C, notice that they are all of the form, $E = \begin{bmatrix}a & b\\0 & d\end{bmatrix}$. Thus, we can show that $E = a\begin{bmatrix}1 & 0\\0 & 0\end{bmatrix} + b\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix} + d\begin{bmatrix}0 & 0\\0 & 1\end{bmatrix}$. You can show that $E_1 = \begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}$, $E_2 = \begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$, $E_3 = \begin{bmatrix}0 & 0\\0 & 1\end{bmatrix}$ forms a linearly independent set and will span $E$, and the set $\{E_1, E_2, E_3\}$ forms a basis for $E$. Thus, the dimension of $E$ is 3.