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let A $\in \mathcal{M}_{4,3}(\mathbb{R})$ and B $\in \mathcal{M}_{3,4}(\mathbb{R})$ such as BA$ = \begin{pmatrix} 0&1&1 \\ 1&0&1 \\ 1&1&0 \end{pmatrix}$

show that AB is diagonalizable

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    Note that if $v$ is an eigenvector of $BA$ corresponding to a non zero eigenvalue then $Av$ is an eigenvector of $AB$ corresponding to the same eigenvalue. – copper.hat Nov 25 '20 at 19:08

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we have $\:\:\:\:p_{BA}(t) = \begin{vmatrix} t&-1&-1 \\ -1&t&-1 \\ -1&-1&t \end{vmatrix} = \begin{vmatrix} t-2&-1&-1 \\ t-2&t&-1 \\ t-2&-1&t \end{vmatrix} =\:(t-2) \begin{vmatrix} 1&0&0 \\ 1&t+1&0 \\ 1&-2&t+1 \end{vmatrix} =(t-2)(t+1)^{2}$ $BA+I_3= \begin{pmatrix} 1&1&1 \\ 1&1&1 \\ 1&1&1 \end{pmatrix}$ we have clearly $\:\:\:\:rg(BA+I_3)=1$ thus dim$(Ker((BA+I_3))=2 $ and BA is diagonalisable.

using the relation $t^4p_{BA}(t)=t^3p_{AB}(t)\:\:\:$we have $p_{AB}(t)=t(t-2)(t+1)^{2}$ thus $0,(-1),(-1)\:$and$\:2$ are the four eigenvalues of AB thus $\mathbb{R^4} = Ker(AB) ⊕ Ker(AB-2I_4)⊕Ker((AB+I_4)^2)\\ dim(ker(AB)) = dim(Ker(AB-2I_4))=1\\ (the\:dimention\:of\: the\: eigenspace \:is\: majorised\: by\: the\: multiplicity\: of \:the \:associated \:eigenvalue) \:thus\: dim(Ker(AB+I_4)^2)=2$

I would like to show that dim$(Ker(AB+I_4)) =2$

$edit: \\ $ On one site, we have $dim(Ker(AB+I_4)) \le 2 (same \: reason\: cf\: above)$

On the other side $\\ \begin{array}{ccccc} Φ & : & Ker(BA+I_3) & \to & Ker(AB+I_4) \\ & & v & \mapsto & Av \\ \end{array}$

Φ is well defined and clearly linear $\:\:\:\:\:\:let \:v \in Ker(BA+I_3) \\ ABAv =A*-1*v=-1*A*v$

Φ is injective $let \:x \in Ker(Φ) \\ Ax = 0 \:\: ie \:\: x \in Ker(A)\cap Ker(BA+I_3) \\(BA+I_3)x = BAx +I_3x =0 = I_3x\\ so \: x = 0 \:since \:I_3 \: is \:invertible \:thus\: ker(Φ)= {0}$

As a consequence $dim(Ker(BA+I_3))\leq dim(Ker(AB+I_4)) \:ie\: 2\leq dim(Ker(AB+I_4))$

We have dim$(Ker(AB+I_4)) =2$ thus $Ker(AB+I_4) = Ker((AB+I_4)^2)$ thus $ \mathbb{R^4} = Ker(AB) ⊕ Ker(AB-2I_4)⊕Ker(AB+I_4)$ thus AB is diagonalisable.

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