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Let $A$ be a real $n \times n$ symmetric matrix with distinct eigenvalues (that is, $A$ has no repeated eigenvalue). Let $B$ be a real $n \times n$ matrix that commutes with $A.$ Then show that the eigenvalues of $B$ are all real.

I couldn't quite able to do this problem. Could anyone please give me some small hint? I started with some like that $:$

Let $\lambda$ be an eigenvalue of $B$ corresponding to an eigenvector $x$ then we have $$\overline {\lambda}\ \|x\|^2 = \langle Bx, x \rangle = \langle x, B^t x \rangle.$$ But since $A$ is symmetric and $B$ commutes with $A$ hence so is $B^t.$ Using commutativity of $A$ and $B^t$ one can easily show that if $x$ is eigenvector of $A$ corresponding to some eigenvalue $\mu$ then $B^t x$ is also an eigenvector of $A$ corresponding to the same eigenvalue. But eigenspaces of $A$ are all one dimensional and hence this proves that $x$ is also an eigenvector of $B^t.$ Now if we can show that $\lambda$ is an eigenvalue of $B^t$ corresponding to the eigenvector $x$ we are through. This is where I got stuck.

Thanks a bunch.

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Hint: By the spectral theorem, all eigenvalues of $A$ are real. Let $x \in \Bbb R^n$ be an eigenvector of $A$ with associated eigenvalue $\lambda$. Show that $x$ must also be an eigenvector of $B$. You might find it helpful to separately consider the cases where $\lambda \neq 0$ and $\lambda = 0$.

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  • $\begingroup$ BenGrossman$:$ please have a look at my edit. $\endgroup$ Sep 15 at 20:08
  • $\begingroup$ The fact you mentioned is easy to prove. But then how do I proceed? $\endgroup$ Sep 15 at 20:12
  • $\begingroup$ @Rabin Your proof is mostly fine. However, it is not necessarily true that $B^t x$ is an eigenvector; it is also possible that $B^t x = 0$; this needs to be addressed. Also, there is no reason to go from $B$ to $B^t$; you could make the same argument using $B$ instead. $\endgroup$ Sep 15 at 20:12
  • $\begingroup$ @Rabin If $x \in \Bbb R^n$ is an eigenvector of a real matrix $B$, then it must hold that the associated eigenvalue is real. $\endgroup$ Sep 15 at 20:13
  • $\begingroup$ Why do you need spectral theorem? Eigenvalues of symmetric matrices are all real is an one line argument. $\endgroup$ Sep 15 at 20:13
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If $A$ is real symmetric with distinct eigenvalues, it is diagonalizable with real eigenvalues and orthogonal eigenvectors. So $$ A=U^TDU $$ where $U\in\mathbb R^{n\times n}$ and $U^TU=I$ and $D$ is diagonal with distinct real elements. Hence $$ AB=BA \quad\Longleftrightarrow\quad U^TDUB=BU^TDU $$ and hence $$ DUBU^T=UBU^TD $$ If $D=\mathrm{diag}(d_1,\ldots,d_n)$, $L=UBU^T=(L_{ij})$, then $LD=DL$, and for $i\ne j$, $$ d_iLe_i=L(d_ie_i)LDe_i=DLe_i $$ and hence $Le_i$ is a multiple of $e_i$. (Here $e_1,\ldots,e_n$ is the standard basis of $\mathbb R^n$.) Say $Le_i=\lambda_i e_i$. This means that $$ L=\mathrm{diag}(\lambda_1,\ldots,\lambda_n). $$ Hence $B=U^TLU$, If $v_k\in\mathbb R^n$ is the $k-$column of $U^T$, then $$ Bv_k=U^TLUv_k=U^TLe_k=\lambda_k U^Te_k=\lambda_kv_k $$ and therefore $$ \lambda_k=\langle v_k,Bv_k\rangle\in\mathbb R. $$

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