Let $V = \{ A \in M_{2}(\mathbb{R}) \colon tr(A) = 0\}$. If $f \in V^{\circ}$ then prove that $f(I) = 0 \Rightarrow f = 0.$

Question

Let $V = \{ A \in M_{2}(\mathbb{R}) \colon tr(A) = 0\}$. If $f \in V^{\circ}$(annihilator of $V$) then prove that $$f(I) = 0 \Rightarrow f = 0.$$

I have concluded that $B = \{ \begin{bmatrix}1&0\\0&-1\end{bmatrix} ,\begin{bmatrix}0&1\\0&0\end{bmatrix},\begin{bmatrix}0&0\\1&0\end{bmatrix} \}$ is the base for V and it can easily be complemented to base for $M_{2}(\mathbb{R})$ by adding matrice (for example) $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ to $B$. Then, we can get the base for $V^{\circ} = \{-d\}$, but after this I am pretty much stuck. I am still new in theory of annihilator so any hint helps!

Answer 1

If $A\in M_2(\mathbb R)$, it can be written as a linear combination of$$\operatorname{Id},\ \overbrace{\begin{bmatrix}1&0\\0&-1\end{bmatrix},\ \begin{bmatrix}0&1\\0&0\end{bmatrix},\text{ and }\begin{bmatrix}0&0\\1&0\end{bmatrix}}^{\phantom V\in V}.$$If $f(\operatorname{Id})=0$ and $f\in V^\circ$, then $f$ maps each of them to $0$, and therefore $f(A)=0$.

Answer 2

For any $A$ we have $f(A)=f(A-cI+cI)=f(A-cI)+cf(I)=0+0=0$ when $c=\frac 1 2 Tr(A)$ because $A-cI \in V$.