# 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 $\rm B$.

$A$ 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.

• I have checked that the result is true for $n=2$, by looking at the radical of the polynomial ideal generated by $|A|=0=|B|$, trace$(A)=x$, trace$(B)=y$, $|A+B|=xy$ with $A, B$ symmetric, and verifying that the entries of $AB$ and $BA$ lie in the radical. Could this be the first step for an inductive proof on the size of the matrices? Maybe the case $n=2$ is just too simple... Commented Oct 4, 2017 at 19:20
• @Jose Brox: Interesting that you can get it in a "purely algebraic" way. Commented Oct 5, 2017 at 12:18
• @Jose Brox: See the $2n$ polynomial equations for the entries of $A$, $B$ that come from the characteristic polynomial ( see additions in my answer), and whether you can use some Groebner bases for say $n=3$ or larger, just curious. Commented Oct 5, 2017 at 12:27
• @orangeskid Ah, that's a neat way of putting it! Actually there are $2n-1$ polynomial equations, since the traces equation is always granted. I have produced the 5 polynomials for $n=3$ with Matlab (with charpoly, poly2sym and collect), and run the radical membership function in Maple in an Intel Core i7-4510U at 2.6GHz for 2500s by now. It has consumed about 3.5GB of memory and is still running. If curious, you can take a look at the polynomials here: ibin.co/3clEj7LvtmTQ.jpg Commented Oct 5, 2017 at 17:42
• @orangeskid There was no luck for the case $n=3$: the program halted after 11ks and 5GB of allocated memory (could give it more, but I don't think it would make much a difference). I wouldn't know how to express the problem in such a way that I could use Groebner bases machinery without explicitly computing the corresponding polynomials for each $n$, case by case. Any ideas on the matter? Commented Oct 5, 2017 at 22:17

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 $\operatorname{im}B \subset \ker A$.

• Thanks for your nice answer Commented Sep 25, 2017 at 10:23
• You only used that $A+B$ has $r+s$ non-zero eigenvalues. And you have to show that $AB = 0$. Commented Sep 25, 2017 at 11:55
• @user60589 There is something I don't understand: the rank is determined by the eigenvectors, not the eigenvalues, so the number of eigenvalues is not the important parameter. A priori it could happen that some $\lambda_i$ has a bigger multiplicity for $A$ than for $A+B$, implying there actually is a nonempty intersection (being equal the rest of multiplicities involved). What am I missing? Commented Sep 25, 2017 at 12:35
• @user60589: the second part at least gets towards a solution. However, nowhere is it proved that $AB=0$. From $im (A+B) = im A \oplus im B$ you cannot conclude $AB = 0$. Nowhere is it shown that the two summands are perpendicular. Commented Oct 3, 2017 at 17:11
• @user60589: how do we know that? Yes, $im B \oplus ker B = im B\oplus im A \oplus X$. So? Does it follow right away that $ker B = im A \oplus X$? There are many possible complements of a subspace. Commented Oct 3, 2017 at 17:25

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 $$P_A(\lambda) \cdot P_B(\lambda) = \lambda^n P_{A+B}(\lambda)$$

• How can we see that the condition of the OP equivalent to this identity of traces? Commented Sep 25, 2017 at 13:57
• Also, I don't understand the implication that gives $\sqrt{A}\sqrt{B}=0$. Why is it so? By other arguments I just get to showing that the diagonal of $AB$ is full of zeros. Commented Sep 25, 2017 at 14:16
• @About the $0$ product : $C^{t} C=0$ implies $C=0$ for real matrices. Look at the diagonal elements of the product Commented Sep 25, 2017 at 14:19
• @Jose Brox: I added some detail above. Also, it for $n$ numbers $\nu_i$ we know what the sums $\sum_i \nu_i^k$ are, for $k=1,\ldots n$, then we know the numbers up to a permutation, Commented Sep 25, 2017 at 14:55
• My main concern about the traces thing is with multiplicities: I don't think we can just assume trace$A+$trace$B=$trace$(A+B)$, since we don't know how many times $\lambda_i$ appears in spec$A$ and spec$(A+B)$. [Btw, in your interesting lemma for the $\nu_i$, more conditions are needed. Is positivity of the $\nu_i$ enough?] Commented Sep 25, 2017 at 15:12

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 $AB=0$.

• Yes, that is a good source! Commented Oct 4, 2017 at 23:40
• Yes, indeed, there are only $2n-1$ equations in fact. What are the dimensions of the matrices? I see you have $9+9$ parameters. (Are you using symmetric matrices?). I only have access to Wolfram Alpha at the moment. Commented Oct 5, 2017 at 17:09
• @orangeskid Yes, I immediately noticed after the posting of the previous comment that my matrices where not symmetric in that run! So I erased the comment and started again... Commented Oct 5, 2017 at 17:45
• There is minor error in the published proof. The correction is too long for a comment, so I made a separate answer. Commented Oct 5, 2017 at 21:58
• maybe there isn't a purely algebraic solution... that is, it may be possible that we have the equality and still $AB\ne 0$... of course that should be over the complex.... so if you find a counterexample over the complex numbers for $n=3$, we know for sure it cannot work. in general, over the reals, to prove an implication like that you would need the real radical of an ideal... perhaps harder to work with. it would be interesting to get a counterexample with complex matrices for $n=3$ still, to settle this. Commented Oct 6, 2017 at 8:47

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.