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

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.

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

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

• 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. Commented Jul 25, 2023 at 15:04
• @user1551 I have edited my question now. Can it be proven now? Commented Jul 25, 2023 at 15:22
• Yes. To begin with, show that $\operatorname{range}(M)\perp\operatorname{range}(M^T)$. Commented Jul 25, 2023 at 15:52
• @user1551 I have tried that. Does it mean if I show this then the inequality becomes a equality? Commented Jul 25, 2023 at 16:20

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

finish:
$$\implies 2r = \text{rank}\big(BA\big) = \text{rank}\big(AB\big)=\text{rank}\big(M+M^T\big)$$
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 $$\text{rank}(A+B)=\text{rank}(A)+\text{rank}(B)=2\operatorname{rank}(M)$$.