# Does $AB+BA=0$ imply $AB=BA=0$ when $A$ is real semidefinite matrix?

This is a general question that came to my mind while listening to a lecture(although its framing may make it look like a textbook question).

Suppose that $A$ and $B$ be real matrices. $A$ is symmetric and positive semi-definite$(x^tAx\ge0\ \ \forall x\in \mathbb{R}^n)$.

Let $AB+BA=0$. Does this imply $AB=BA=0$? If yes can you give me some example?

• Perhaps trace would be helpful. For $C$ symmetrical (or skew), we have $$\text{tr}(C^2)=0\implies C=0.$$ – Vim May 1 '16 at 17:29
• The title and first paragraph are useless. Why not just state the question? – zyx May 1 '16 at 17:59
• @zyx: I didn't want to unnecessarily add the self study tag, which other wise I would have to add to such a bookish question – Qwerty May 1 '16 at 18:51
• Normally I don't click on questions on the front page that say "just asking!" or "neat question" or other vagueness. And when the top paragraph doesn't contain any details of the question, it defeats the short summary on the Questions page. Then we are left only with tags, which is better than nothing. But why not have something with AB+BA=0 in the title? – zyx May 1 '16 at 18:58
• Edited the title. (Notifying in case you prefer to reverse the edit.) – zyx May 1 '16 at 19:31

Let $v$ be an eigenvector of $A$ with eigenvalue $\lambda$. Then $$0 = A(Bv) + B(Av) = A(Bv)+\lambda Bv\Longrightarrow A(Bv) = -\lambda Bv$$ what this means is that $Bv$ is also an eigenvector of $A$ with eigenvalue $-\lambda$. But since $A$ is positive semi-definite all eigenvalues of $A$ are $\geq 0$. If $\lambda=0$, so we have two possiblities:

• Either $v$ is an eigenvector with eigenvalue zero, in which case $A(Bv) = 0$, This means $ABv = BAv = 0$ for all eigenvectors corresponding to eigenvalue zero.
• Or $v$ is an eigenvector with eigenvalue $\lambda>0$. Suppose $Bv\neq 0$, then $Bv$, by above observation is an eigenvector of $A$ with negative eigenvalue! This is impossible, so $Bv=0$. Therefore no only $ABv = 0$, we also have $BAv = \lambda Bv = 0$.

Since all vectors can be written as a linear composition of these eigenvectors, we have $AB=BA = 0$ in these circumstances.

Extra: If you are also interested in what this $B$ can actually be if $AB+BA = 0$, note that if $w$ is any vector, and $P_0$ is the porjection linear transformation which sends $w$ to the subspace of vectors with eigenvalue zero. Then we know the following:

Write $w = P_0w + (1-P_0)w$, then by second observation above $B(1-P_0)w = 0$. So in the basis in which $A$ is diagonal (since $A$ is symmetric it is diagonalizable), $B$ is necessarily of the form $$B=\begin{pmatrix} B_0 & 0\\ 0 & 0 \end{pmatrix}$$ In other words $B$ satisfies $AB+BA=0$ if and only if $P_0BP_0=B$.

Proposition. Let $A,B\in M_n(\mathbb{C})$; if $A$ is diagonalizable with non-negative eigenvalues and $AB+BA=0$, then

1. $AB=BA=0$.

2. If moreover $A$ is hermitian, then $A,B$ are unitarily simultaneously similar to $A_1=diag(D_p,0_{n-p}),B_1=\begin{pmatrix}0_p&0\\0&T_{n-p}\end{pmatrix}$, where $D$ is diagonal $> 0$ and $T$ is upper triangular.

Proof. For 1. cf. the Hamed's answer.

For 2. Using the syeh_106's proof, $A,B$ are unitarily simultaneously similar to $A_1=diag(D_p,0_{n-p})),B'=\begin{pmatrix}0_p&0\\0&U_{n-p})\end{pmatrix}$, where $D$ is diagonal $> 0$. It remains to triangularize $U$, using Schur's method.

An alternative proof is as follows. Since $A$ is psd, $A=UDU^*$, where $U$ is unitary and $D$ is diagonal. Suppose $A$ has $k$ distinct eigenvalues, $\lambda_1 < \lambda_2 <...< \lambda_k$. Then $D$ may be chosen to be $D=\lambda_1I_{m_1}\oplus\lambda_2I_{m_2}\oplus...\oplus\lambda_kI_{m_k}$, where $m_i$ denotes the multiplicity of $\lambda_i$, and $I_{m_i}$ is an $m_i$-by-$m_i$ identity matrix. (I.e. $D$ is block-diagonal, consisting of $k$ blocks.)

Now, note that $AB+BA=0 \Rightarrow UDU^*B=-BUDU^* \Rightarrow DU^*BU=-U^*BUD.$ Letting $\hat B\triangleq U^*BU$, we have $D\hat B=-\hat B D$. Given the form of $D$, it's easy to verify that $\hat B$ must be block-diagonal, i.e. $\hat B=\hat B_1\oplus\hat B_2\oplus...\oplus\hat B_k$, where $\hat B_i$ is of the same size as $I_{m_i}$ Moreover, since $\lambda_2, ..., \lambda_k$ must be positive, we conclude that $\hat B_2, ..., \hat B_k$ must all be zero matrices. If $\lambda_1$ is also positive (i.e. $A$ is p.d.), $\hat B_1$ must also be zero. So $\hat B=0$ and $D\hat B=\hat B D=0$. On the other hand, if $\lambda_1=0$ (i.e. $A$ is p.s.d.), a direct multiplication clearly also shows $D\hat B=\hat B D=0$.

Since $AB=UD\hat BU^*$ and $BA=U\hat BDU^*$, it follows that $AB=BA=0$.