# $A=\lambda I_n\iff (\forall M,N\in M_n(\mathbb{R}),~ MN=A \Rightarrow ~ NM=A)$

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).

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.