Prove if $M^2 =0$, then $\operatorname{rank}(M+M^T)=2\operatorname{rank}(M)$.

I have used the rank nullity theorem and the fact that rank of matrix and its transpose is same to prove it but I am not sure if it correct or not. I do not how to exactly use the fact of $M^2=0$. I can think of it means that $M$ is nilpotent and square matrix. In my answer I get less than or equal to but it I have to prove it as equal. Help would be highly appreciated. I am using real square matrices.

$\operatorname{rank}(M + M^T)\le\operatorname{rank}(M) +\operatorname{rank}(M^T)$

$\operatorname{rank}(M + M^T)\le2\operatorname{rank}(M)$

  • $\begingroup$ Please add your question to your post. $\endgroup$ Commented Jul 25, 2023 at 15:01
  • 9
    $\begingroup$ The statement is not true in general. Consider $A=\pmatrix{1&1\\ 1&1}$ over $GF(2)$ for instance. If you are considering real square matrices, please specify that in your question. $\endgroup$
    – user1551
    Commented Jul 25, 2023 at 15:04
  • $\begingroup$ @user1551 I have edited my question now. Can it be proven now? $\endgroup$ Commented Jul 25, 2023 at 15:22
  • 2
    $\begingroup$ Yes. To begin with, show that $\operatorname{range}(M)\perp\operatorname{range}(M^T)$. $\endgroup$
    – user1551
    Commented Jul 25, 2023 at 15:52
  • $\begingroup$ @user1551 I have tried that. Does it mean if I show this then the inequality becomes a equality? $\endgroup$ Commented Jul 25, 2023 at 16:20

3 Answers 3


Split $\mathbb R^n$ into the orthogonal sum $V\oplus W$, where $V=\operatorname{range}(M)$ and $W=V^\perp=\ker(M^T)$. Since $M^2=0$, we have $\langle M^Tx,My\rangle=\langle x,M^2y\rangle=0$ for all $x,y\in\mathbb R^n$. Hence $$ \operatorname{range}(M^T)\perp\operatorname{range}(M). $$ In turn, $V=\operatorname{range}(M)\subseteq\operatorname{range}(M^T)^\perp=\ker(M)$. It follows that $$ MV=0. $$ By the definition of $W$, we also have $$ M^TW=0. $$ Therefore $$ \begin{aligned} \operatorname{range}(M+M^T) &=(M+M^T)(V+W)\\ &=(M+M^T)V+(M+M^T)W\\ &=M^TV+MW\\ &=M^T(V+W)+M(V+W)\\ &=\operatorname{range}(M^T)\oplus\operatorname{range}(M)\\ \end{aligned} $$ and the conclusion follows.

Remark. In terms of block matrices, the proof above essentially says that up to a change of orthonormal basis, $M$ can be expressed in the form of $$ \pmatrix{0&X_{r\times(n-r)}\\ 0&0} $$ where $r$ is the rank of $M$ and $X$ is an $r\times (n-r)$ matrix of full row rank. Therefore $M+M^T=\pmatrix{0&X\\ X^T&0}$ has rank $2r$.


Making use of the fact that the row space of $M$ is orthogonal to the column space of $M$:
Let $\text{rank}\big(M\big) =r$ and using SVD write $M=U\Sigma_r V^T=\sum_{k=1}^r \sigma_k\mathbf u_k\mathbf v_k^T$. Note SVD is not unique but we may choose to write this so $\Sigma_r$ is invertible and $U$ and $V$ each have $r$ columns.

$M + M^T =\Big(\sum_{k=1}^r \sigma_k\mathbf u_k\mathbf v_k^T\Big) +\Big(\sum_{k=1}^r \mathbf v_k(\sigma_k\mathbf u_k)^T\Big)$
$= \bigg[\begin{array}{c|c|c|c} \sigma_1\mathbf u_1 & \cdots & \sigma_r\mathbf u_{r}&\mathbf v_{1}& \cdots &\mathbf v_{r} \end{array}\bigg]\begin{bmatrix} \mathbf v_1^T \\ \vdots\\ \mathbf v_{r}^T\\ \sigma_1\mathbf u_{1}^T\\ \vdots\\ \sigma_r\mathbf u_{r} \end{bmatrix}= AB $
$\implies \begin{bmatrix} \mathbf 0 & I_r \\ \Sigma_r^2 &\mathbf 0 \end{bmatrix}=BA$ which is invertible

$\implies 2r = \text{rank}\big(BA\big) = \text{rank}\big(AB\big)=\text{rank}\big(M+M^T\big)$
since $BA$ and $AB$ have the same non-zero eigenvalues, which specifies the rank of $AB$ since it is real symmetric (diagonalizable).

alternative finish:
$\implies 2r= \text{rank}\big(BA\big)=\text{rank}\big(B(AB)A\big)\leq \text{rank}\big(AB\big)=\text{rank}\big(M+M^T\big)\leq 2r$
where the RHS holds by sub-additivity of rank (and is stated in OP)


Using $\text{rank}(A^tA)=\text{rank}(A)$ for all real square matrix $A$, we have $$\text{rank}(M+M^t)=\text{rank}(M+M^t)^2=\text{rank}(MM^t+M^tM)$$

Note that $A:=MM^t$ and $B:=M^tM$ are both positive semi-definite, and $AB=BA=O$ hence commute, we can simultaneously diagonalize them, and without loss of generality assume $A, B$ are already diagonal. Now the number of non-zero elements on the diagonal of $A$ (or $B$) is the same as $\text{rank}(A)=\text{rank}(B)=\text{rank}(M)$. If there is a common index $i$ such that neither of $A_{ii}, B_{ii}$ is $0$, then $AB$ would not be $O$, therefore the nonzero elements of $A$ and $B$ don't mix, hence $\text{rank}(A+B)=\text{rank}(A)+\text{rank}(B)=2\operatorname{rank}(M)$.


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