Why is the set of matrices with determinant zero not a subspace?

Question

I'm reading my linear algebra textbook, and it says word for word: The following set is not a subspace:

the set of all $2\times 2$ matrices $B$ such that $\det(B)=0$.

I just need help trying to understand what that means.

Answer 1

Let $\mathcal M_n$ be the vector space of $n\times n$ matrices and let $$ S_n=\{A\in \mathcal M_n:\det A=0\} $$ Then $S_n$ is not a subspace of $\mathcal M_n$. Indeed, let \begin{align*} A&= \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & 0 \end{bmatrix} & B &= \begin{bmatrix} 0 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & 1 \end{bmatrix} \end{align*} Then $A,B\in S$ but $\det(A+B)=1$ so $A+B\notin S_n$.

In particular, when $n=2$, the subset $S_2$ is not a subspace of $\mathcal M_2$ because \begin{align*} A&=\begin{bmatrix}1&0\\0&0\end{bmatrix} & B &= \begin{bmatrix}0&0\\0&1\end{bmatrix} \end{align*} are elements of $S_2$ but $\det(A+B)=1$ so $A+B\notin S_2$.

Extra Credit. Is the subset $$ \mathfrak{sl}_n=\{A\in\mathcal M_n:\DeclareMathOperator{trace}{trace}\trace A=0\} $$ of $\mathcal M_n$ a subspace of $\mathcal M_n$?