Can somebody tell me if my solution on the following problem is right?
Show that set is a vector subspace $$ L:\mathbb{R}^{3\times3}\to\mathbb{R}^{3\times3}$$ $$\{A \in \mathbb{R}^{3\times3} \mid L(A)=0 \}$$ $$A \to L(A):=A +A^T $$
Solution
- Let $A$ be the zero matrix, then
$$L(0)=0.$$- Let $A$ and $B$ be two 3x3 matrices, then $$L(A + B) = L(A) + L(B) = 0+0=0.$$
- Let $c$ be a scalar in $\mathbb{R}$, then $$L(cA)= c\cdot L(A) = c\cdot0 =0.$$
So the set is a subspace? However I am not sure if it's meant to be a subspace of $\mathbb{R}^{3\times3}$ as in the problem is only that mentioned.
I am asked also to give the basis of $L$ and the dimension but there is not a given matrix only this set here, does someone know I can do that? Or just the Problem is not somehow false written or so?
Your set is the kernel of the linear map $L$. The fact that $L(A)=A+A^T$ is irrelevant here. The kernel of any linear map is always a vector subspace (for the reason that you mentioned).
On the other hand, in order to determine a basis of $\ker L$, you need to know which map $L$ is. Since$$\ker L=\{A\,|\,A=-A^T\},$$a basis of $\ker L$ is$$\left\{\begin{bmatrix}0&1&0\\-1&0&0\\0&0&0\end{bmatrix},\begin{bmatrix}0&0&1\\0&0&0\\-1&0&0\end{bmatrix},\begin{bmatrix}0&0&0\\0&0&1\\0&-1&0\end{bmatrix}\right\},$$for instance.