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Question:

Let $A,B\in M_{n\times n}(\mathbb{R})$ be real symmetric matrices. Suppose that we have $\begin{pmatrix}A+B&O\\O&O\end{pmatrix}$ is similar to $\begin{pmatrix}A&O\\O&B\end{pmatrix}$. Prove that $AB=O$.

My attempt:

As we know that real symmetric matrix is diagonalizable, so the condition is equivalent to that $\sigma(A+B)\cup\{0\}^n=\sigma(A)\cup\sigma(B)$. Then $\text{tr}(A+B)^2=\text{tr}\ A^2+\text{tr}\ B^2$, i.e. $\text{tr}\ AB=0$.

Consider $A$ and $B$ as self-adjoint operators on Hilbert space $H=\mathbb{R}^n$. Then we have orthogonal decomposition $H=\ker (A+B)\oplus\text{im}(A+B)$. Note that $\text{rank}(A+B)=\text{rank}(A)+\text{rank}(B)$, then we have $\text{im}(A+B)=\text{im}(A)\oplus\text{im}(B)$ and $\ker(A+B)=\ker A\cap\ker B$. Therefore $A$ and $B$ are self-adjoint operators on $H'=\text{im}(A+B)$ and they vanish on $\ker(A+B)$. All in all, we can suppose that $A+B$ is invertible.

Now we have orthogonal decomposition $H=\ker A\oplus\text{im}\ A$ and $H=\ker B\oplus\text{im}\ B$.

MatrixBi
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    What have you tried? – Zoe Allen Mar 12 '23 at 11:25
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    @MatrixBi Note that askers are expected to provide context for their questions, as is explained here. Please [edit] your question to tell us where you encountered this problem, what you have tried so far, and any other relevant thoughts you have. – Ben Grossmann Mar 12 '23 at 13:36
  • @MatrixBi For example, are you able to show that the result holds in the case that $A,B$ are positive semidefinite? – Ben Grossmann Mar 12 '23 at 14:08
  • @MatrixBi How do you go from $\text{rank}(A+B)=\text{rank}(A)+\text{rank}(B)$ to $\text{im}(A+B)=\text{im}(A)\oplus\text{im}(B)$? – Ben Grossmann Mar 12 '23 at 14:41
  • $(A+B)v=Av+Bv$, so $\text{im}(A+B)\subset\text{im}(A)+\text{im}(B)$. – MatrixBi Mar 12 '23 at 14:44
  • I agree with you there. However, you seem to be saying that we have an orthogonal decomposition $\text{im}(A+B)=\text{im}(A)\oplus\text{im}(B)$. Note that this orthogonal decomposition is not implied from the rank equality alone; for example, consider $$ A = \pmatrix{1&0\0&0}, \quad B = \pmatrix{1&1\1&1}. $$ – Ben Grossmann Mar 12 '23 at 14:52
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    @MatrixBi Again, it would help if you could tell us where you encountered this question. For instance, if this is from a certain textbook, then there may be some relevant theorem that the creator of the question had in mind. – Ben Grossmann Mar 12 '23 at 14:56
  • @MatrixBi Also, note that because $A$ and $B$ are symmetric, $AB = 0$ is equivalent to the statement that the subspaces ${\rm im}(A),{\rm im}(B)$ are orthogonal. – Ben Grossmann Mar 12 '23 at 14:58
  • +1 for showing your efforts – TShiong Mar 12 '23 at 17:38
  • duplicate of https://math.stackexchange.com/questions/87982/old-amm-problem – user8675309 May 28 '24 at 21:34

1 Answers1

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Let $a=\operatorname{rank}(A)$ and $b=\operatorname{rank}(B)$. Since $A$ and $B$ are real symmetric, we may write $$ A=UDU^T \quad\text{and}\quad B=V\Lambda V^T $$ for some nonsingular diagonal matrices $D\in GL_a(\mathbb R),\,\Lambda\in GL_b(\mathbb R)$ and some tall matrices $U\in M_{n,a}(\mathbb R),\,V\in M_{n,b}(\mathbb R)$ with orthonormal columns. Then $$ A+B=\underbrace{\pmatrix{U&V}\pmatrix{D\\ &\Lambda}}_X\ \underbrace{\pmatrix{U^T\\ V^T}}_Y=XY. $$ By Sylvester's secular theorem, it has the same multi-set of nonzero eigenvalues as $$ YX=\pmatrix{I_a&U^TV\\ V^TU&I_b}\pmatrix{D\\ &\Lambda}. $$ However, $(A+B)\oplus0$ is similar to $A\oplus B$, which in turn is similar to $D\oplus\Lambda\oplus0$. Therefore, the nonzero eigenvalues of $YX$ are precisely the eigenvalues of the nonsingular diagonal matrix $D\oplus\Lambda$. Since the two matrices have the same size, we in turn obtain $\det(YX)=\det(D\oplus\Lambda)$. It follows that $$ 1=\det\pmatrix{I_a&U^TV\\ V^TU&I_b} =\det(I_b-V^TUU^TV). $$ As $U$ and $V$ have orthonormal columns, $0\preceq V^TUU^TV\preceq V^T(I_n)V=I_b$. So, for the determinantal equality above to hold, we must have $V^TUU^TV=0$ . Therefore $U^TV=0$ and $AB=0$.

user1551
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