Finding Bases of Kernel and Image of $T:M_{2,2} \rightarrow M_{2,2}$.

Question

I am meant to find a basis for the kernel and image of the linear transformation $T:M_{2,2} \rightarrow M_{2,2}$ given by $T(X) = AXA$ where $A = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$

and det$A = 0$. I cannot think of any way to do this other than to expand the matrix product, and find conditions on the elements of $X$. Is there any other way? Expanding doesn't seem to be working, I can't seem to use the fact $ad - bc = 0$.

Edit: The field is not specified in the question, but it is assumed to be either $\mathbb{R}$ or $\mathbb{C}$. Which ever one yields a simpler solution can be chosen.

Answer 1

The matrix of $T$ with respect to the canonical basis of $M_{2,2}$ given by $E_{11}, E_{12}, E_{21}, E_{22}$ is $$ \pmatrix{ a^2 & ac & ab & bc \\ ab & ad & b^2 & bd \\ ac & c^2 & ad & cd \\ bc & cd & bd & d^2 \\ } $$ Now row-reduce this matrix...

Answer 2

Although I haven't carried through the calculations I think this approach will work. At very least it gives the rank and nullity which is something.

The characteristic polynomial of $A$ is $t^2-(a+d)t +0$, so the eigenvalues are $0, a+d$.

There are three cases: (a) $A=O$; (b) the eigenvalues of $A$ are distinct, ($a\not=-d$); (c) the eigenvalues of $A$ are both zero, but $A$ is not diagonalisable.

The case (a) is trivial.

For the case (b) we can find LI eigenvectors of $A$ for the two distinct eigenvalues. To be precise, choose $(d,-c)^T$ or (if $c=d=0$) $(-b,a)^T$ as a $0$- eigenvector, and similarly choose $(a,c)^T$ or (if $a=c=0$) $(b,d)^T$ as an $(a+d)$-eigenvector. Form $S$ with these two columns, and since eigenvectors corresponding to distinct eigenvalues are LI the matrix is non-singular.

Now writing $T(X)=S (S^{-1}AS)Y(S^{-1}AS) S^{-1}$ with $Y=(S^{-1}XS)$ lets us calculate (since $(S^{-1}AS)$ has only one non-zero entry) in terms of $Y$ bases for the image and kernel of $T$; we can then undo the conjugation and find the bases in terms of $X$.

In case (c) we have $a=-d$ and $ad=bc$. We need to calculate a Jordan basis to get $(S^{-1}AS)=\begin{pmatrix}0 & 1\\0 & 0\end {pmatrix}$, and then follow the same strategy.