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Let $V$ be a finite dimensional vector space and $A$ be normal operator on $V$ and $B$ is an operator such that $AB=BA$. Show that $BA^*=A^*B$.

I guess that this problem should not be so difficult. I have tried different approaches and I got some identities which do not lead to desired equality.

So I would be thankful if you show the solution to this problem, please!

RFZ
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2 Answers2

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The way to think about this problem is when $B$ is diagonalizable, and $A$ being normal is diagonalizable (over $\mathbb C$) so we can call on simultaneous diagonalizability, recognize that being normal $A^*$ may also be simultaneously diagonalized with $B$ (via the same similarity transform that we'd use on $AB$) which implies that $A^*B = BA^*$. However it is conceivable that $B$ might be defective-- so a more direct argument can be employed to compute the norm of the commutator

$\Big\Vert A^*B - BA^*\big\Vert_F^2$
$=\text{trace}\Big(\big(A^*B - BA^*\big)^*\big(A^*B - BA^*\big)\Big)$
$=\text{trace}\Big(\big(B^*A - AB^*\big)\big(A^*B - BA^*\big)\Big)$
$=\text{trace}\Big(B^*AA^*B\Big) + \text{trace}\Big(AB^*BA^*\Big)- \text{trace}\Big(B^*ABA^*\Big) -\text{trace}\Big(AB^*A^*B\Big) $
$=\text{trace}\Big(AA^*BB^*\Big) + \text{trace}\Big(B^*BA^*A\Big)- \text{trace}\Big(B^*ABA^*\Big) -\text{trace}\Big(BAB^*A^*\Big) $
$=\text{trace}\Big(AA^*BB^*\Big) + \text{trace}\Big(B^*BA^*A\Big) - \text{trace}\Big(B^*BAA^*\Big) -\text{trace}\Big(ABB^*A^*\Big)$
$=\text{trace}\Big(AA^*BB^*\Big) + \text{trace}\Big(B^*BA^*A\Big) - \text{trace}\Big(B^*BAA^*\Big) -\text{trace}\Big(A^*ABB^*\Big)$
$=\text{trace}\Big(AA^*BB^*\Big) + \text{trace}\Big(B^*BA^*A\Big) - \text{trace}\Big(B^*BA^*A\Big) -\text{trace}\Big(AA^*BB^*\Big)$
$=0$

thus by positive definiteness of the (squared) Frobenius norm we have

$\Big\Vert A^*B - BA^*\big\Vert_F^2 = 0 \longrightarrow A^*B - BA^* = \mathbf 0\longrightarrow A^*B = BA^*$

user8675309
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  • I am not so familiar with norm of an operator. Could you explain what is Frobenius norm? And why it can be calculated as a trace? Would be nice if you can give details about that – RFZ Apr 06 '20 at 15:44
  • Am I right that $\Big\Vert A \Big\Vert_F^2=\Big\Vert C^{-1}AC\Big\Vert_F^2$ where $C$ is non-singular matrix? If its true how to prove it? – RFZ Apr 06 '20 at 16:27
  • your second question is incorrect, just try a few example. Your first question... I assume that you know what the 2 norm of a vector is -- the Frobenius norm is the natural generalization of said 2 norm to matrices if you view matrices as living in a vector space. In any case the Frobenius norm is induced by an inner product and easy to work with. $\mathbf A = \bigg[\begin{array}{c|c|c|c|c} \mathbf a_1 & \mathbf a_2 &\cdots & \mathbf a_{n-1} & \mathbf a_{n} \end{array}\bigg]$. – user8675309 Apr 06 '20 at 18:24
  • if the norm of operator depends on the basis how are you defining the norm then? Could you give more details, please? – RFZ Apr 06 '20 at 18:28
  • $\Big \Vert A\Big \Vert_F = \big(\sum_{i=1}^m\sum_{j=1}^n \vert a_{i,j}\vert^2 \big)^\frac{1}{2} $ $= \big(\sum_{k=1}^n \big \Vert \mathbf a_k\big \Vert_2^2\big)^\frac{1}{2} $ $=\Big(\text{trace}\big(A^A\big)\Big)^\frac{1}{2}$. To be honest I find it extremely hard to believe that you are unfamiliar with the Frobenius norm of a matrix, because you are asking about normal matrices* and a common way of defining them is $\big \Vert \mathbf A \big \Vert_F^{2} \geq \sum_{i = 1}^{n} \big \vert \lambda_i\big \vert ^2$ and a matrix is normal iff that is met with equality (due to Schur) – user8675309 Apr 06 '20 at 18:28
  • it is common to choose an orthonormal basis – user8675309 Apr 06 '20 at 18:29
  • So the norm of operator depends on the basis? I am a bit confused. – RFZ Apr 06 '20 at 18:30
  • For purposes of this problem --it's actually the other way around -- a matrix is isn't normal under arbitrary change of basis. Your questions echo this one: https://math.stackexchange.com/questions/3542129/are-unitary-matrices-still-unitary-under-similarity-transformations/3542169 where unitary matrices are a special case of normal. To be honest the way 'choice/change of basis' is being used is abusive here... being normal corresponds much more with bilinear forms than linear maps (i.e. effecting congruence transforms instead of similarity transforms). I think you have some reading to do. – user8675309 Apr 06 '20 at 18:34
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    I guess that I got you reasoning almost completely. Am I right that when you wrote $\text{trace}(B^AA^B)=\text{trace}(AA^BB^)$ you used that $\text{trace}(CD)=\text{trace}(DC)$, right? – RFZ Apr 06 '20 at 19:05
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    right. The idea here is we have 4 matrices and the manipulations allowed are (i) cyclic property of trace, (ii)$AB = BA$, (iii)$B^A^ = A^B^$, (iv) $AA^* = A^*A$... we always have (i) but ${\text{ii, iii, iv}}$ are special features of your problem. – user8675309 Apr 06 '20 at 19:10
  • Thanks a lot! I got your solution completely! Really nice reasoning! I realized that Frobenius norm is the norm generated from usual inner product on the space of matrices, i.e. $(A,B):=\text{trace}(A^*B)$. I accept your answer as the best! How did you come up with this? – RFZ Apr 06 '20 at 19:22
  • another way of characterizing normality of $A$ is $\text{trace}\Big(\big(A^* A\big)^2 \Big) \geq \text{trace}\Big(\big(A^2\big)^* \big(A^2\big) \Big)$ with equality iff $A$ is normal. It's a worthwhile exercise to prove this to yourself; (a) one approach is compute $\big \Vert \mathbf A^* \mathbf A - \mathbf {AA}^* \big \Vert_F^2 $ (b) a nicer approach comes from Cauchy-Schwarz (try it). For your problem I kept trying to prove the result via use of Cauchy-Schwarz in the spirit of (b) but I hit too many dead ends and decided instead to imitate (a) – user8675309 Apr 06 '20 at 19:40
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This is Fuglede's theorem. From the spectral theorem, $A$ can be written as

$$A=\sum_{i=1}^{n}\lambda_iP_i,$$

and $A^{*}$ can be expressed similarly by replacing $\lambda_i$ by its conjugate. Now, notice that for $p_i(x)=x\prod_{j : \lambda_j \neq \lambda_i}(x-\lambda_j)$

$$p_i(A)=\sum_{j=1}^{n}p_i(\lambda_j)P_j=\lambda_i\prod_{j : \lambda_j \neq \lambda_i}(\lambda_i-\lambda_j)\sum_{j : \lambda_j=\lambda_i}P_j.$$ (The first equality is a generalization of the fact that $A^k = \sum_{i=1}^{n}\lambda_i^kP_i$.) This implies that $$\lambda_i\sum_{j : \lambda_j=\lambda_i}P_j=\frac{p_i(A)}{\prod_{j : \lambda_j \neq \lambda_i}(\lambda_i-\lambda_j)}.$$ Given that A and B commute, we have that $p_i(A)$ and $B$ also commute. So, $$\lambda_i\sum_{j : \lambda_j=\lambda_i}BP_j=\lambda_i\sum_{j : \lambda_j=\lambda_i}P_jB,$$ and by replacing $\lambda_i$ with its conjugate and summing over all distinct eigenvalues we get that $BA^{*}=A^{*}B$.

gmou3
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  • You applied spectral theorem to $A$. But in my lectures spectral theorem is proved for self-adjoint operators. Our $A$ may not be self-adjoint. – RFZ Apr 06 '20 at 16:39
  • The spectral theorem applies more generally to normal matrices. – gmou3 Apr 06 '20 at 23:36