I have the following question here:
Let $X=\begin{pmatrix}1 &-1\\ 1 & 1 \end{pmatrix}$, and let $U=\{A \in M_2(\mathbb{R})\mid \mathrm{Tr}(XA)=0\}$ , a subspace of $M_2{\mathbb{(R)}}$.
a) Show that the matrices $$A_1=\begin{pmatrix}1 &1\\ 1 & -1 \end{pmatrix},A_2=\begin{pmatrix}1 &-1\\ -1 & -1 \end{pmatrix},A_3=\begin{pmatrix}1 &-1\\ 1 & 1 \end{pmatrix}$$are all in $U$.
That was easy. Just just the conditions given. $\mathrm{Tr}(XA)=0$ in all cases.
b) Decide whether $\{A_1,A_2,A_3\}$ is a basis for $U$.
I think I have to use the rank nullity theorem here somehow but I am not sure. Can someone help out here? Thanks!
Those matrices are linearly independent, because, if $\alpha,\beta,\gamma\in\Bbb R$, then\begin{align}\alpha A_1+\beta A_2+\gamma A_3=\begin{bmatrix}0&0\\0&0\end{bmatrix}&\iff\begin{bmatrix}\alpha+\beta+\gamma&\alpha-\beta-\gamma\\\alpha-\beta+\gamma&-\alpha-\beta+\gamma\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}\\&\iff\left\{\begin{array}{l}\alpha+\beta+\gamma=0\\\alpha-\beta-\gamma=0\\\alpha-\beta+\gamma=0\\-\alpha-\beta+\gamma=0\end{array}\right.\\&\iff\alpha=\beta=\gamma=0.\end{align}If $\{A_1,A_2,A_3\}$ was not a basis of $U$, then $\dim U>3$. But $U$ is a subspace of $M_2(\Bbb R)$, whose dimension is $4$. So, you would have $U=M_2(\Bbb R)$. But this is impossible, since $X^T\notin U$.