Proof concerning matrix composition.

Question

I have a statement which I don't know how to prove. All matrices are real, $n \times n$.

For all $0 < k < n$ the following has to hold.

It is impossible do define a matrix $A$ of rank $\mathbf{k}$ such that the following holds.

$\forall M_1, M_2. A M_1 M_2 = A M_1 A M_2 $

Any hints greatly appreciated.

In case this is (as I suspect) part of a larger body of work which I am not aware of, I would appreciate if you gave me a pointer.

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In case my statement is false, I would like a characterization of admissible $A$s given $k$.

Answer 1

If $A$ is a such matrix then $A=0$:
Proof: Lat $A$ such that $AM_1M_2=AM_1AM_2$ for all $M_1,M_2$:
In particular (for $M_1=M_2=I_n$) we obtain $A=A^2$, hence the polynomial $X^2-X=X(X-1)$ is annihilator of $A$, and it follow that $A$ is diagonalizable. we denote $D=\text{diag}(\lambda_1\ldots,\lambda_n)$, then $A=P^{-1}DP$ for some invertible matrix $P$ and $\lambda_i$ the eigenvalues of $A$ , it is clear that $\lambda_i\in \{0,1\}$ for all $i$. In the equation we can omit $M_2$ (by $M_2=I_n$), now we obtain $AM=AMA$ for all $M$, so $P^{-1}DPM=P^{-1}DPMP^{-1}DP$ for all $M$. also $D(PMP^{-1})=D(PMP^{-1})D$ for all $M$.
We know that the map $M\mapsto P^{-1}MP$ is bijective: now we obtain that $DM=DMD$ for all $M$. If we denote $M=(a_{i,j})$, the last equation is equivalent to $\lambda_ia_{i,j}=\lambda_i\lambda_ja_{i,j}$ that is $\lambda_ia_{i,j}(1-\lambda_j)=0$, since $D$ is not invertible ($A$ is not invertible $\text{rank}(A)<n$ ) there is $j_0$ such that $\lambda_{j_0}=0$, by taking $a_{i,j}=1$ for all $i,j$ we obtain $\lambda_i(1-\lambda_{j_0})=0$ for all $i$ , that is $\lambda_i=0$ for all $i$, Then $D=0$, it follow that $A=0$.

Answer 2

$\newcommand{\M}[0]{\mathcal M}$Let $\M$ be the algebra of $n \times n$ matrices, and consider the map $$ f : \M \to \M, \qquad f(M) = AM. $$ The hypothesis implies that $f$ is a homomorphism of algebras.

Now $\M$ is a simple ring, so that by considering $\ker(f)$ we see that

• either $\ker(f) = \M$, so that $f$ is the zero map - this can only happen of course when $A = 0$, a case that has been excluded,
• or $\ker(f) = \{ 0 \}$, so that $f$ is injective, and thus an isomorphism - this is because $f$ is also a linear map - it follows that $A$ is invertible (there must be $M$ such that $AM = f(M) = I$, the identity matrix), again a case that has been excluded.

In the case when $A$ has rank $n$, we see immediately that we must have $A = I$.

Answer 3

An alternative answer, in the spirit of that of Hamou.

Setting $M_{1} = M_{2} = I$, we see that $A = A^{2}$, and thus we may assume $$ A = \begin{bmatrix} 1\\ &1\\ &&\ddots\\ &&&1\\ &&&&0\\ &&&&&\ddots\\ &&&&&&0 \end{bmatrix},\tag{diag} $$ where there are $k$ ones, with $0 < k < n$ ones, and omitted entries are zero.

Addendum This is because $A$ is a root of $x^{2} - x$. Since $A \ne 0, I$, we have that $x^{2} -x$ is the minimal polynomial of $A$. Since $x^{2} -x$ has the distinct roots $0$ and $1$, $A$ can be put in diagonal form with $0$ and $1$ on the main diagonal, that is, there is an invertible matrix $P$ such that the conjugate $P^{-1} A P$ of $A$ by $P$ has the form (diag). Clearly, conjugating the main identity $$ A M_{1} M_{2} = A M_{1} A M_{2} $$ by $P$ we find that the matrix (diag) still satisfies the main identity, as claimed.

Now choose $M_{2} = I$, and $M_{1} = e_{1n}$, the matrix whose only non-zero entry is a $1$ in the $(1, n)$ position.

Then $AM_{1} = M_{1} \ne 0 = A M_{1} A$, a contradiction