6
$\begingroup$

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.

$\endgroup$
14
$\begingroup$

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 $\mathcal M_n$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.