I am struggling with an algebra problem here is what we got :
$$ T\begin{bmatrix} a &b \\ c&d \end{bmatrix}= a+d\;\; \; \; and \; \; \;\; L\begin{bmatrix} a &b \\ c&d \end{bmatrix}=\begin{bmatrix} 2a+d &a+b+d \\ a+c+d & -2a-d \end{bmatrix} $$
$$ T : Mat_{2\times 2}\rightarrow R \; \; and \; \; L : Mat_{2\times 2}\rightarrow Mat_{2\times 2}$$
I have to find a basis for ker(T) and the elements in the basis B1 , B2 , B3
and then show that ker(T) = im(L) ,
i'm stuck here if you could help me ,
thanks in advance
Best Answer
Note that $$T\begin{pmatrix} a & b \\ c & d\end{pmatrix}=a+d=0\Leftrightarrow a=-d$$ So we quickly find that $$A_{1}=\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix},\ A_{2}=\begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix},A_{3}=\text{ and} \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$$ are in the kernel of $T$. Note that they are linearly independent, $\ker(T)$ has at least dimension 3. Note that if $\ker(T)$ has dimension 4, then $T=0$ which is a contradiction, so $\{A_{1},A_{2},A_{3}\}$ is a basis for the kernel of $T$.
Also note that $LA_{1}=A_{1},\ LA_{2}=A_{2}$ and $LA_{3}=A_{3}$ and $$L\begin{pmatrix} 1 & 1 \\ 1 & -2\end{pmatrix}=\begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix}$$ So $\text{im}(L)$ has dimension 3 and $A_{1},A_{2},A_{3}\in\text{im}(L)$. It follows that $\{A_{1},A_{2},A_{3}\}$ is a basis for the image of $L$. So $\ker(T)=\text{im}(L)$.