I have been trying to solve this, but I can't. I am given the endorphism: $$f(X) = A\cdot X – X\cdot A$$
Where $X$ is any real-valued $2 \times 2$ matrix and $$A = \begin{pmatrix}1 & 2 \\ 2 & -1\end{pmatrix}$$
And I have to find a basis of the kernel and a basis of the image of $f$. Finding a basis for the kernel was easy. By definition, the basis is the set of matrices $X$ that verify $f(X) = (0)$, therefore:
$$A\cdot X – X \cdot A = 0$$
$$A\cdot X = X \cdot A$$
$$X = A^{-1}\cdot X \cdot A$$
So, since X must be a diagonal matrix with its two elements equal to its eigenvalues, we know it must be a matrix of the form: $$X =\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$$
And since both $\lambda_1$ and $\lambda_2$ must verify $A-\lambda I_2 = 0$. Writing out the just mentioned equation in matrix form, one gets a matrix with only one non-zero entry for each eigenvalue, being the result in both cases that $\lambda_1= \lambda_2$. Therefore, a basis for the kernel could be: $$B_{ker(f)} = \Big{\{} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\Big{\}}$$
But I am stuck here. Since this is an endomorphism on the space of square matrices of order $2$, the dimension equation tells us that $$\dim(Im(f)) + \dim(ker(f)) = \dim(Im(f)) + 1 = 4 \Rightarrow \dim(Im(f)) = 4$$
I am not sure I am right and I don't know how to find a basis for the image of $f$, though.
Best Answer
Actually, the condition $X=A^{-1}.X.A$ does not mean that $X$ is diagonal.
If $X=\left[\begin{smallmatrix}x_1&x_2\\x_3&x_4\end{smallmatrix}\right]$, then\begin{align}A.X=X.A&\iff A.X-X.A=0\\&\iff\left\{\begin{array}{l}2(-x_3+x_2)=0\\2(-x_1+x_2+x_4)=0\\2(x_1-x_3-x_4)=0\\2(x_2-x_3)=0\end{array}\right.\\&\iff X=\begin{bmatrix}x_1&x_2\\x_2&x_1-x_2\end{bmatrix}\end{align}So, this gives you the kernel, which is $2$-dimensional. Therefore, by the rank-nullity theorem, the dimension of the image of $f$ is also $2$.
Now, since$$f\left(\begin{bmatrix}1&0\\0&0\end{bmatrix}\right)=\begin{bmatrix}0&2\\-2&0\end{bmatrix}\text{ and }f\left(\begin{bmatrix}0&1\\0&0\end{bmatrix}\right)=\begin{bmatrix}2&-2\\0&-2\end{bmatrix}$$and since $\left[\begin{smallmatrix}0&2\\-2&0\end{smallmatrix}\right]$ and $\left[\begin{smallmatrix}2&-2\\0&-2\end{smallmatrix}\right]$ are linearly independent, they span the image of $f$.