Determine whether or not the indicated set of 3 x 3 matrices is a subspace of M33.
The set of all non singular 3 x 3 matrices?

Question

kirbykirby · Answer

I think you can find an easy counter example for this. A non-singular matrix means it's invertible, so it also means that its determinant $\ne 0$. 

Consider:
$$ A=\left[\begin{matrix}1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1\end{matrix}\right]$$
$$ B=\left[\begin{matrix}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{matrix}\right]$$
Then, $$A+B =\left[\begin{matrix}1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1\end{matrix}\right]$$
Notice that $\det(A)=-1$, $\det(B)=1$ but $\det(A+B)=0$ so $A+B$ is not closed under addition

anonymous · Answer

Hey thank you very much so I'm guessing if it is the case where it is a singular matrix means it is not invertible. So if I was asked the same question but for singular matrix's.
So...
$$A = \left[\begin{matrix}1 & 0 & 1\ 0 & 1 & 0 \ 0 & 0 & 0\end{matrix}\right]$$
$$B = \left[\begin{matrix}0 & 0 & 0 \ 0 & 0 & 0\ 0 & 0 & 1\end{matrix}\right]$$
Then
$$A + B = \left[\begin{matrix}1 & 0  & 1\ 0 & 1&0\0&0&1\end{matrix}\right]$$
and $$\det(A) = 0 , \det(B) = 0 , \det(A+B)=1$$ so it is  not closed under addition?
I guess the big  question I have is could I show this with $$cA$$?

kirbykirby · Answer

Yeah your example works :) It's not closed under addition, so the set of singular matrices wouldn't be a subspace either. 

You don't actually have to show the case for $cA$ (closed under scalar multiplication). Just showing either  not closed under addition, or not closed under scalar multiplication is enough to show it's not a subspace. (In fact, it's possible that it may be closed for scalar mult. and not closed for addition, or vice-versa, and it's still not a subspace... both operations have to be closed)

anonymous · Answer

Thank you so much |dw:1425437474080:dw|