3
$\begingroup$

I cannot find any concise and effective answer to the following problem :

Problem : given two matrices $A$ and $B$ of size $3\times 2$ and $2\times 3$ such that $AB = \left(\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1\\ 1 & 1 & 2 \end{matrix}\right)$ find the kernel and the image of $AB$ and $BA$.

What I have found so far : Ker$(AB)=$ Vect$\left(\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \right)$ and Im$(AB) = $ Vect$\left(\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} , \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \right)$. Furthermore $AB \in S_3(\mathbb{R})$ so AB is diagonalizable, and I have $\chi_{AB}=X(X+1)(X-3)$ so :

$\mathrm{Sp}(AB) = \{-1,0,3 \}$ and $E_{-1}(AB) = \mathrm{Vect}\left( \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \right) \quad E_{0}(AB) = \mathrm{Vect}\left( \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \right) \quad E_{3}(AB) = \mathrm{Vect}\left( \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \right) $

Then I had the idea of saying if $X$ is an eigenvector of $AB$ then there exists $\lambda \in \{-1,0,3\}$ such as $ABX = \lambda X$ so $BABX = \lambda BX$ and thus $BX$ is either null or an eigenvector of $BA$ for the eigenvalue $\lambda$. Furthermore $BA$ is a square matrix of size 2 so $BA$ cannot have $3$ distinct eigenvalues, it has at most $2$ eigenvalues that can be found within $\{-1,0,3\}$.

I have stopped here. I don't expect any complete answer, just any idea will be welcomed.

$\endgroup$
2
$\begingroup$

Notice that rank($AB$) $=2$

This implies $A$ and $B$ are full rank, i.e. $B$ is surjective and $A$ is injective.

Since $A$ is injective, Ker $(AB) = $ Ker $B$

Since $B$ is surjective, Im $(AB) = $ Im $A$

Since $\left(\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} , \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \right)$ are linearly independent, we see that

Ker $B$ $\cap$ Im $A =$ Ker $AB$ $\cap$ Im $AB = 0$

Thus,

Ker $BA = $ Ker $A = 0$ so that $BA$ is invertible since it is square

$\endgroup$
7
  • $\begingroup$ Thank you for taking your time to answer. Should the third vector be (1 1 -1) (typo)? $\endgroup$
    – Axel
    Apr 27 '20 at 6:15
  • $\begingroup$ That is a nice approach. So what do you suggest for $\mathrm{Im}(BA)$ then? $\endgroup$
    – Axel
    Apr 27 '20 at 6:26
  • $\begingroup$ Ok I got it, as $BA$ is invertible so $BA$ is bijective (isomorphism) and as $((1 \; 0),(0 \; 1))$ is a basis of $\mathbb{R}^2$ then $\mathrm{Im}(BA) = (BA(1 \; 0), BA(0 \; 1))$ $\endgroup$
    – Axel
    Apr 27 '20 at 9:55
  • $\begingroup$ Do you think we can do something knowing $(AB)^2$ and that $AB$ is diagonalizable though? $\endgroup$
    – Axel
    Apr 27 '20 at 9:57
  • $\begingroup$ Could you explain simply how you can prove "This implies A and B are full rank, i.e. B is surjective and A is injective.", I could see why it works, but do you have any more rigourous explanation please? I will validate your answer afterwards. $\endgroup$
    – Axel
    Apr 27 '20 at 10:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.