# Kernel and image of a product of two rectangular matrices

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.

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

• Thank you for taking your time to answer. Should the third vector be (1 1 -1) (typo)?
– Axel
Apr 27 '20 at 6:15
• That is a nice approach. So what do you suggest for $\mathrm{Im}(BA)$ then?
– Axel
Apr 27 '20 at 6:26
• 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 $\mathrm{Im}(BA) = (BA(1 \; 0), BA(0 \; 1))$
– Axel
Apr 27 '20 at 9:55
• Do you think we can do something knowing $(AB)^2$ and that $AB$ is diagonalizable though?
– Axel
Apr 27 '20 at 9:57
• 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.
– Axel
Apr 27 '20 at 10:00