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Let $V$ be a finite dimensional inner-product space, and suppose that $T_1$, $T_2$ are normal operators on $V$ that commute. How to show that $T_1+T_2$ and $T_1T_2$ are then normal?

It is clear if $T_1$ commutes with $T_2^*$ and $T_2$ with $T_1^*$, but is it true? If yes how to prove it?

Edit: It is an exercise in Schaum's Outlines of Linear Algebra.

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5 Answers 5

up vote 4 down vote accepted

My preferred answer was given by Robert Israel, as it is a great opportunity to meet the Fuglede-Putnam-Rosenblum Theorem.

Here is an interpolation argument in the finite-dimensional case where you stand. I will first prove a lemma for matrices. Of course, the same holds for operators on a finite-dimensional complex inner-product space.

Lemma: a matrix $A\in M_n(\mathbb{C})$ is normal if and only if there exists a polynomial $P\in \mathbb{C}[X]$ such that $A^*=P(A)$.

Proof: the implication $\Leftarrow$ is obvious. So assume $A$ is normal. Equivalently, there exists a unitary matrix $U$ and a diagonal matrix $D=\mbox{Diag}\{\lambda_1,\ldots,\lambda_n\}$ such that $A=UDU^*$. Then $A^*=UD^*U^*$ where $D^*=\mbox{Diag}\{\overline{\lambda_1},\ldots,\overline{\lambda_n}\}$. By Lagrange interpolation, we can find a polynomial $P\in\mathbb{C}[X]$ such that $P(\lambda_j)=\overline{\lambda_j}$ for $j=1,\ldots,n$. Hence $D^*=P(D)$ and it follows that $$P(A)=P(UDU^*)=UP(D)U^*=UD^*U^*=A^*.$$ QED.

Now assume that $T$ is normal and commutes with $S$. Then $S$ commutes with every power of $T$, hence with every polynomial in $T$. By the above, $S$ commutes with $T^*$.

The facts you want to prove follow easily.

Note: this remains true in $\mathbb{R}$. But note that we need to be a little bit careful as real normal matrices are not diagonalizable in $M_n(\mathbb{R})$ in general.

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Thank you very much, your answer is enlightening. –  Spenser Apr 21 '13 at 22:33
@Spenser You're welcome. I always liked this lemma. –  1015 Apr 21 '13 at 22:53
I tried to answer my question. I would be very grateful if you could look at it. Many thanks. –  Spenser Apr 21 '13 at 23:25
(+1) by the way. –  Spenser Apr 22 '13 at 0:07

See the Fuglede-Putnam-Rosenblum theorem (e.g. in Rudin, "Functional Analysis", thm 12.16).

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Hmm... It is an exercise in Schaum's outline of linear algebra. There should be a simpler proof. I know nothing about functional analysis. Note that I said finite dimensional. –  Spenser Apr 21 '13 at 21:18
Like I said, that's my preferred answer, +1. –  1015 Apr 21 '13 at 22:17

Here is an attempt to answer my own question. Please feel free to comment and point out mistakes if you see some.

(Edit: I am learning this stuff right now, so I would really appreciate if an expert could confirm or refute my arguments.)

Lemma: Let $S,T$ be normal operator on a finite dimensional inner-product space, and suppose that $ST=TS$. Then, there is an orthonormal basis for which both $S$ and $T$ are diagonal.

Proof: Since $S$ is normal, $S$ is diagonalizable, so its minimal polynomial factors into distinct linear terms and thus, by the Primary Decomposition Theorem, $$V=E_{\lambda_1}\oplus \cdots\oplus E_{\lambda_k}$$ where $\lambda_i$ are the distinct eigenvalues of $S$ and $E_{\lambda_i}=\ker(S-\lambda_i I)$. Now, if $v\in E_{\lambda_i}$, then $$(S-\lambda_i I)T(v)=ST(v)-\lambda_i T(v)=TS(v)-\lambda_i T(v)=\lambda_i T(v)-\lambda_i T(v)=0$$ so $E_{\lambda_k}$ is $T$-invariant, and thus $T_i:=T|_{E_{\lambda_i}}$ is diagonalizable. Hence, $$E_{\lambda_i}=E_{\mu^i_1}\oplus\cdots\oplus E_{\mu^i_{n_i}}$$ where $\mu^i_1,\ldots,\mu^i_{n_i}$ are the distinct eigenvalues of $T_i$, and $E_{\mu^i_j}=\ker(T_i-\mu^i_j I)$.

Let $B^i_j$ be an orthonormal basis for $E_{\mu^i_j}$. Since $T$ is normal and, for fixed $i$, $\mu^i_1,\ldots,\mu^i_{n_i}$ are distinct, we have $E_{\mu^i_1}\bot\cdots\bot E_{\mu^i_{n_i}}$, so $B^i_1\cup\cdots\cup B^i_{n_i}$ is orthonormal. Moreover, since $S$ is normal and $\lambda_1,\ldots,\lambda_k$ are distinct, $B:=\cup B^i_j$ is still orthonormal. Now, $B$ is an orthonormal basis of eigenvectors for both $S$ and $T$.


Now the result is clear. We have $$[ST^*]_B=[S]_B[T]^*_B=[T]^*_B[S]_B=[T^*S]_B$$ so $ST^*=T^*S$, and similarly, $S^*T=TS^*$.





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That's almost perfect, +1. You need a little extra justification. As you said, since $S$ and $T$ commute, each $E_{\lambda_i}$ is invariant under $T$. But you need it to be invariant under $T^*$ too if you want to claim that $T_i$ is diagonalizable in an orthonormal basis. But $E_{\lambda_i}=\mbox{Ker}(S^*-\overline{\lambda_i})$ is also an eigenspace of $S^*$. And since $T^*$ commutes with $S^*$, it is indeed invariant under $T^*$. So $T_i^*$ is the restriction of $T^*$ and it follows that $T_i$ is normal. The rest should be ok. –  1015 Apr 22 '13 at 2:16
@julien Thank you so much julien. I learned a lot with this exercice. I understand the problem, I will edit it with your recommendations soon. –  Spenser Apr 22 '13 at 2:27

If the operators are normal, then they can be diagonalised, even by a unitary matrix. If they commute, they are simultaneously diagonalisable. I am almost certain that they can be diagonalised by the same unitary matrix: if they have all different eigenvalues then the unitary matrix is unique, so this is clear, and if not then you can use a continuity argument. Since similarity with use of unitary matrix behaves well with the adjoint, you may assume both operators are diagonal. Thus, you are done!

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Denote normal operators: $$\mathcal{N}(\mathcal{H}):=\{N\in\mathcal{B}(\mathcal{H}):N^*N=NN^*\}$$

Denote bounded calculus: $$\mathcal{N}(N):=\{\eta(N)\in\mathcal{B}(\mathcal{H}):\|\eta\|_\infty<\infty\}$$

For commuting ones: $$MN=NM:\quad\mathcal{N}(M)\cup\mathcal{N}(N)\subseteq\mathcal{N}(M\wedge N)$$

Especially one has: $$MN\in\mathcal{N}(M\wedge N):\quad(MN)^*(MN)=(MN)(MN)^*$$

(No boundedness needed.)


For more details see: Superalgebra

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