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.


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 $f$.

  • $\begingroup$ My understanding of what a diagonal matrix is must be flawed. Since there needs to exist a matrix $P$ verifying $D = P^{-1} A P$ so that a matrix $A$ is diagonalizable, I thought $X = A^{-1} X A$ implied that X is diagonal and is, in fact, identic to its diagonalized form. $\endgroup$ – mar Jan 2 at 22:35
  • $\begingroup$ If $X$ is any square matrix whatsoever and $P$ is the identity matrix, then $X=P^{-1}AP$, right? But I suppose that you will not deduce from this that any square matrix is diagonalizable. $\endgroup$ – José Carlos Santos Jan 2 at 22:38
  • $\begingroup$ You're right, thank you. $\endgroup$ – mar Jan 2 at 22:41
  • $\begingroup$ In order to get a basis for the image, you looked at the image of two matrices that form a 2-dimensional basis in the space of squared matrices of order $2$. Could you have chosen any two matrices as long as they form a basis, or is the choice constrained by the basis found for the kernel? $\endgroup$ – mar Jan 2 at 22:57
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    $\begingroup$ If my answer was useful, perhaps that you could mark it as the accepted one. $\endgroup$ – José Carlos Santos Jan 3 at 7:42

Hint: Observe \begin{align} AX-XA =&\ \begin{pmatrix} 1 & 2\\ 2 & -1 \end{pmatrix} \begin{pmatrix} x & y\\ z & w \end{pmatrix} - \begin{pmatrix} x & y\\ z & w \end{pmatrix} \begin{pmatrix} 1 & 2\\ 2 & -1 \end{pmatrix}\\ =& \begin{pmatrix} x+2z & y+2w\\ 2x-z & 2y-w \end{pmatrix} - \begin{pmatrix} x+2y & 2x-y\\ z+2w & 2z-w \end{pmatrix}\\ =&\ \begin{pmatrix} 2(z-y) & 2(w-x)+2y\\ 2(x-w)-2z & 2(y-z) \end{pmatrix}. \end{align}


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