Let $\Phi:M_2(\mathbb R)\to\mathcal P_3$ be a linear operator s.t. $$\left(\Phi A\right)(x)=\operatorname{trace}(AB+BA)+\operatorname{trace}(AB-BA)x+\operatorname{trace}(A+A^{\tau})x^2$$ $$B=\begin{bmatrix} 3&-2\\2&1\end{bmatrix}$$ Find $r(\Phi):=\dim Im(\Phi),d(\Phi):=\dim Ker(\Phi)$ and some base for the kernell.
My attempt:
Since $\operatorname{trace}:M_n(\mathbb R)\to\mathbb R$ is a linear operator in general: $$\operatorname{trace}(\alpha A+\beta B)=\alpha\operatorname{trace}(A)+\beta\operatorname{trace}(B),\forall\alpha,\beta\in\mathbb R,\forall A,B\in M_n(\mathbb R)$$ $$\implies\operatorname{trace}(AB+BA)=\operatorname{trace}(AB)+\operatorname{trace}(BA)=2\operatorname{trace}(AB)$$ $$\operatorname{trace}(AB-BA)=\operatorname{trace}(AB)-\operatorname{trace}(BA)=0$$ $$\operatorname{trace}(A+A^{\tau})=\operatorname{trace}(A)+\operatorname{trace}(A^{\tau})=2\operatorname{trace}(A)$$ $$\implies (\Phi A)(x)=2(\operatorname{trace}(AB)+\operatorname{trace}(A)x^2)$$ Let $$A=\begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix}$$ $$\operatorname{trace}(AB)=3a_{11}+2a_{12}-2a_{21}-a_{22}$$ $$\operatorname{trace}(A)=a_{11}+a_{22}$$ $$\implies \Phi A=4(2a_{11}+a_{12}-a_{21})$$ Finding the kernell: $$\dim M_2(\mathbb R)=4$$ $$2a_{11}+a_{12}-a_{21}=0\iff a_{21}=2a_{11}+a_{12}$$
Then we have: $$Ker(\Phi)\ni A=\begin{bmatrix}a_{11}&a_{12}\\ 2a_{11}+a_{12}&a_{22}\end{bmatrix}=a_{11}\begin{bmatrix}1&0\\ 2&0\end{bmatrix}+a_{12}\begin{bmatrix}0&1\\ 1&0\end{bmatrix}+a_{22}\begin{bmatrix}0&0\\ 0&1\end{bmatrix}$$ Base for the kernell: $$\left\{\begin{bmatrix}1&0\\ 2&0\end{bmatrix},\begin{bmatrix}0&1\\ 1&0\end{bmatrix},\begin{bmatrix}0&0\\ 0&1\end{bmatrix}\right\}\implies d(\Phi)=3$$ By the rank-nullity theorem: $$r(\Phi)=\dim M_2(\mathbb R)-d(\Phi)=1$$
May I ask if this is correct? Thank you in advance!
No, it is not correct.
Yes, $(\Phi A)(x)=2\bigl(\operatorname{tr}(AB)+\operatorname{tr}(A)x^2\bigr)$. But, if $A=\left[\begin{smallmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{smallmatrix}\right]$, then this is equal to$$2\bigl(3a_{11}+2a_{12}-2a_{21}-a_{22}+(a_{11}+a_{22})x^2\bigr).$$So\begin{align}A\in\ker\Phi&\iff\left\{\begin{array}{l}3a_{11}+2a_{12}-2a_{21}-a_{22}=0\\a_{11}+a_{22}=0\end{array}\right.\\&\iff A=\begin{bmatrix}a_{11}&a_{12}\\2a_{11}+a_{12}&-a_{11}\end{bmatrix}.\end{align}So, $\dim\ker\Phi=2$. Can you take it from here?