Prove $\rm AB = BA = 0$ if the set of nonzero eigenvalues of $\rm A + B$ is union of set of nonzero eigenvalues of $\rm A$ and $\rm B$.

Question

$A$ and $B$ are $n×n$ symmetric matrices, their non-zero eigenvalues are $(\lambda_{1},\ldots,\lambda_{r} )$，$( \mu_{1},\ldots,\mu_{s} )$. If the nonzero eigenvalues of $A + B$ are $(\lambda_{1},\ldots,\lambda_{r}，\mu_{1},\ldots,\mu_{s} )$, show that $AB = BA = 0$

$\bf{Note}$ that the (non-zero) eigenvalues of a matrix form a multiset, so the above is a union of multisets.

Answer 1

Edit: As stated this is not true. Take $$A=\begin{bmatrix}2 &0 &0\\0&1 &0 \\ 0 & 0&0\end{bmatrix} \text{ and } B= \begin{bmatrix}0 &0 &0\\0&1 &0 \\ 0 & 0&1\end{bmatrix}$$ then $AB \ne 0$, but $A,B, A+B$ fulfill the hypothesis. So you can either assume that $A,B$ are positive semi-definite as orangeskid did or assume that the multiplicities add up as I will do:

What do we know about a diagonalizable matrix $M$? By looking at eigenspaces with non zero eigenvalue and the eigenspace with eigenvalue zero we get the decomposition $$k^n= \operatorname{im} M \oplus \ker M$$

They are an isomorphism when restricted to their image.

By counting eigenvalues we know that $$\operatorname{rank}{A}+ \operatorname{rank}{B} = \operatorname{rank}{(A+B)}.$$

This gives us $\dim\text{im}{A} + \dim \text{im}{B} =\dim \text{im}{(A+B)}$ by looking at the dimension of subspaces we know that the intersection must be $0$ and thus $$\operatorname{im}{A} \oplus \operatorname{im}{B} =\operatorname{im}{(A+B)}.$$

In other word we get a decomposition $$k^n = \operatorname{im}{A} \oplus \operatorname{im}{B} \oplus \ker (A+B).$$

How does $\ker (A+B)$ looks like? We know that $\ker(A) \cap \ker(B) \subset \ker (A+B)$, but since $\operatorname{im}{A} $ and $\operatorname{im}{B} $ only intersect at zero we get equiality.

Thus $$k^n = \operatorname{im}{A} \oplus \operatorname{im}{B} \oplus( \ker(A) \cap \ker(B)),$$

hence we can assume that $A+B$ is an isomorphism and diagonalize $A$ and $B$ on their images.

Let $O$ be orthonormal such that $A$ is diagonalized on $\operatorname{im} A$ and let $U$ be orthonormal such that $B$ is diagonalized on $\operatorname{im} B$

Choose appropriate $M,N,C$ such that $$A+B=\begin{bmatrix}M &C\\ C^T& N\end{bmatrix} $$

then

$$T= \begin{bmatrix}O^T &0\\ 0& U^T\end{bmatrix} \begin{bmatrix}M &C\\ C^T& N\end{bmatrix} \begin{bmatrix}O &0\\ 0& U\end{bmatrix} = \begin{bmatrix}O^T M O & O^T CU\\ U^T C^TO& U^TNU \end{bmatrix} . $$

Now let us consider the trace of the square of $T$ \begin{align*} \operatorname{tr}(T^2) &= \operatorname{tr} (O^T M^2O + O^TC C^TO)+ \operatorname{tr} (U^T N^2U + U^TC^T CU) \\ &= \sum \lambda_i + \operatorname{tr}(CC^T) + \sum \mu_i + \operatorname{tr}(C^TC) \end{align*}

where $$ \operatorname{tr}(T^2) = \operatorname{tr}(A+B)^2 = \sum \lambda_i + \sum \mu_i $$ and $\operatorname{tr}(CC^T) >0,$ thus we can conclude that $$ C=0.$$

Hence $T$ is diagonal, thus we have $\operatorname{im}A \subset \ker B$ and $\operatorname{im}B \subset \ker A$.

Answer 2

Just some thoughts and a partial answer:

The condition is equivalent to $$\operatorname{Trace} A^k + \operatorname{Trace}B^k = \operatorname{Trace}(A+B)^k$$ for all $k\ge 1$.

For $k=2$ this implies $\operatorname{Trace}AB = 0$.

If we had the extra hypothesis $A$, $B$ positive semi-definite this would imply $AB = 0$. Indeed, $$ \operatorname{Trace} (\sqrt{B} \sqrt{A})(\sqrt{A} \sqrt{B})=\operatorname{Trace}AB =0$$ implies $\sqrt{A} \sqrt{B} = 0$ and so $AB = 0$.

$\bf{Added:}$ Another possible approach inspired by the source given by @Jose Brox: is to use the equivalent condition for the characteristic polynomials $$P_A(\lambda) \cdot P_B(\lambda) = \lambda^n P_{A+B}(\lambda)$$

Answer 3

I have finally found the answer in the paper A simple proof of the generalized Craig-Sakamoto theorem by Jin Zhang and Jikun Yi:

a) If $r+s=n$ they cleverly write $A+B$ in terms of $A$ and $B$, as follows:

$$A=\left(\array{A_1 & 0 \\ 0 & 0}\right), B=T\left(\array{0 & 0 \\ 0 & B_1}\right)T'$$ with $A_1=$diag$(\lambda_1,\ldots,\lambda_r)$, $B_1=$diag$(\mu_1,\ldots,\mu_s)$, $T=\left(\array{T_1 & T_2 \\ T_3 & T_4}\right)$ orthogonal, so that $$A+B=\left(\array{I & T_2 \\ 0 & T_4}\right)\left(\array{A_1 & 0 \\ 0 & B_1}\right)\left(\array{I & 0 \\ T_2' & T_4'}\right).$$

Then, by hypothesis, $$|A_1||B_1|=|A||B|=|A+B|=|A_1||B_1||T_4'T_4|,$$ which implies $|T_4'T_4|=1$. But $1=|I_s|=|T_2'T_2+T_4'T_4|\geq |T_2'T_2|+|T_4'T_4|$ since the matrices are positive semidefinite (why?), hence $|T_2'T_2|=0$, so that $T_2=0$. This implies $AB=0$.

b) If $r+s<n$, then as rank$(A+B)=$rank$(A)+$rank$(B)$ while rank$(A+B)\leq$rank$\left(\array{A \\ B}\right)\leq$rank$(A+B)$, we get equality of the three ranks. Therefore there exists an orthogonal matrix $P=(p_1 \ldots p_n)$ such that $Ap_i=Bp_i=0$ for every $i\in\{1,\ldots,n-r-s\}$. This matrix serves to give a common "block-triangularization" $$P'AP=\left(\array{0 & 0 \\ 0 & A^*}\right), P'BP=\left(\array{0 & 0 \\ 0 & B^*}\right)$$ in which $A^*,B^*$ satisfy the hypotheses of point a), hence $A^*B^*=0$, hence $AB=0$.

Answer 4

The paper A simple proof of the generalized Craig-Sakamoto theorem by Jin Zhang and Jikun Yi cited by Jose Brox contains a minor error. This answer will correct the error. By the way, the error is actually in the original published paper, and was not introduced by Jose Brox.

The cited paper uses the following inequality: If $X$ and $Y$ are positive semidefinite of the same size $s$, then $\det(X + Y) \ge \det(X) + \det(Y)$. This is applied as follows: we are given $X$ and $Y$ positive semidefinite such that $\det(Y) = \det(X+Y) \ge \det(X) + \det(Y)$. The desired conclusion is that $X = 0$, but we cannot conclude this, only that $\det(X) = 0$. We can obtain the desired conclusion using a sharper inequality: If $X$ and $Y$ are positive semidefinite, $Y$ is invertible and $X \ne 0$, then $\det(X +Y) > \det(X) + \det(Y)$. Proof:
$$ X + Y = Y^{1/2}( Y^{-1/2} X Y^{-1/2} + I) Y^{1/2} $$ Then $Z =Y^{-1/2} X Y^{-1/2}$ is non-zero positive semidefinite, with non-negative eigenvalues, not all zero, $(z_1, \dots, z_s)$. Then $\det(I + Z) = \prod_j (1 + z_j) = 1 + \det(Z) + \sum_J \prod_{j \in J} z_j$, where the sum is over non-empty proper subsets $J$ of $\{1, \dots, s\}$. Since not all $z_j$ are zero, the sum $\sum_J \prod_{j \in J} z_j$ is strictly positive. Thus, $\det(I + Z) > 1 + \det(Z)$. It follows that $\det(X+Y) > \det(X) + \det(Y)$.

In the application, we have $1 = \det(Y) = \det(X+Y)$. If $X \ne 0$, we would have $1 = \det(Y) = \det(X+Y) > \det(X) + \det(Y) \ge \det(Y)$, a contradiction. Hence $X = 0$. Apply this with $X = T_2' T_2$ and $Y = T_4' T_4$, using the notation in the answer by Jose Brox.