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here is my question:

Let $n\geqslant 1$ and $A\in M_n(\mathbb{R})$. Show that $$\boxed{\exists \lambda\in\mathbb{R}^*,~ A=\lambda I_n\iff (\forall M,N\in M_n(\mathbb{R}),~ MN=A \Rightarrow ~ NM=A)}$$

The implication $(\Rightarrow )$ is ok since if $MN=\lambda I_n$ with $\lambda\not=0$, then $det(M)det(N)=\lambda^n\not=0$ so $M$ and $N$ are invertible matrix and $N=(\frac{1}{\lambda}M)^{-1}$ so we $(\frac{1}{\lambda}M)N=N(\frac{1}{\lambda}M)=I_n$ and so $NM=\lambda I_n$.

The implication ($\Leftarrow$) is harder. Here is what I tried. Assume that $A$ is not a homothety. Then there exists $x\in \mathbb{R}^n$ such that $x$ and $Ax$ are linearly independent. Take $e_3,\dots,e_n$ such that $(x,Ax,e_3,\dots,e_n)$ is a basis of $\mathbb{R}^n$. Then define $M$ to be the only matrix such that $$\begin{array}{ccc} Mx & = & A^2x\\ M(Ax) & = & Ax\\ M e_3 & = & Ae_3\\ ~ &\vdots&~\\ M e_n & = & Ae_n\\ \end{array}$$ and define $N$ to be the only matrix such that $$\begin{array}{ccc} Nx & = & Ax\\ N(Ax) & = & x\\ N e_3 & = & e_3\\ ~ &\vdots&~\\ N e_n & = & e_n.\\ \end{array}$$

Then we have $MNx=M(Ax)=Ax$, $MN(Ax)=Mx=A^2x=A(Ax)$ and for $i\geqslant 3$, $MNe_i=Me_i=Ae_i$. Therefore the equality $MN=A$ holds. Also $NM(Ax)=N(Ax)=x$ so if $x\not= A^2x$, then $NM\not=A$ and we have our answer. If $A^2x=x$, I change $M$ and $N$ for

$$\begin{array}{ccc} Mx & = & A^2x~(=x)\\ M(Ax) & = & -Ax\\ M e_3 & = & Ae_3\\ ~ &\vdots&~\\ M e_n & = & Ae_n\\ \end{array}$$ and $$\begin{array}{ccc} Nx & = & -Ax\\ N(Ax) & = & x\\ N e_3 & = & e_3\\ ~ &\vdots&~\\ N e_n & = & e_n.\\ \end{array}$$

Again we have $MNx=M(-Ax)=Ax$, $MN(Ax)=Mx=A^2x=A(Ax)$ and for $i\geqslant 3$, $MNe_i=Me_i=Ae_i$. Finally $NM(Ax)=N(-Ax)=-x\not=A^2x=x$ which implies $NM\not=A$.

Is this correct? Also I had this question in an oral exam two years ago and I would like to know if there is an other answer (the examiner seemed surprised, I think he had something else in mind and I would like to know what it is).

Thanks in advance for your help!

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Here is an alternative answer:

Take $N = \begin{bmatrix} \lambda_{1} & 0 & 0 & \dots & 0 \\ 0 & \lambda_{2} & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & \lambda_{n} \end{bmatrix}$ and $M = \begin{bmatrix} \frac{a_{11}}{\lambda_{1}} & \frac{a_{12}}{\lambda_{2}} & \frac{a_{13}}{\lambda_{3}} & \dots & \frac{a_{1n}}{\lambda_{n}} \\ \frac{a_{21}}{\lambda_{1}} & \frac{a_{22}}{\lambda_{2}} & \frac{a_{23}}{\lambda_{3}} & \dots & \frac{a_{2n}}{\lambda_{n}} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac{a_{n1}}{\lambda_{1}} & \frac{a_{n2}}{\lambda_{2}} & \frac{a_{n3}}{\lambda_{3}} & \dots & \frac{a_{nn}}{\lambda_{n}} \end{bmatrix}$ Where $\lambda_i \in \mathbb{R}^+$ and $\lambda_i \neq \lambda_j$ for $i \neq j$

So we have $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} & \dots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \dots & a_{2n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & a_{n3} & \dots & a_{nn} \end{bmatrix}$ and $MN = A$.

But then we must also have $NM = A$. Looking at the elements of $MN - NM = 0$ we see we must have $a_{ij} = 0$ for $i \neq j$. Thus $A$ has nonzero elements only on its diagonal.

Finally, by letting $M$ be a permutation matrix we see that flipping rows $i$ and $j$ of $A$ must be equivalent to flipping columns $i$ and $j$ of $A$, so all the $a_{ii}$ must be equal and so $A$ is a homothety.

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  • $\begingroup$ That's what I was looking for I think, very elegant thanks ! $\endgroup$ – Adam Chalumeau Apr 20 at 17:01

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