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.


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 $\lambda = 0$.

  • $\begingroup$ 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

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 $$ \lambda_k=\langle v_k,Bv_k\rangle\in\mathbb R. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.