I'm trying to prove it by using invariant subspaces and this theorem, which is a final touch in my proof. I don't understand the theorem though. Why is invariance needed? "$A:V \longrightarrow V$, A-normal operator, $U$ is a subspace of $V$ and is invariant under the operator $A, A^*$. Then $A$ on $U$ is normal."

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    $\begingroup$ If $U$ is not an invariant subspace, then the restriction of $A$ to $U$ is not an endomorphism. That is, we can't write $A|_U:U \to U$, we can only write $A|_U:U \to V$. For a map $A$ that has a domain distinct from its codomain, it is impossible to have $A^*A = AA^*$ since for $A:U \to V$, $A^*A$ is a map over $U$ while $AA^*$ is a map over $V$. $\endgroup$ Apr 2 at 2:21

1 Answer 1


Proof that normal matrices are diagonalizable: By Schur's decomposition $A=UTU^H$ (https://en.wikipedia.org/wiki/Schur_decomposition) where $T$ is a upper triangular matrix, $U$ is a unitary matrix. Now $AA^H = A^HA \implies UTT^HU^H = UT^HTU^H \implies TT^H=T^HT$. Note that $T$ is upper triangular and satisfies $TT^H = T^HT$, Hence $T$ is diagonal. So Normal operators are diagonalizable.

In your case they must be demonstrating more information on normal matrices. Please write your question in more detail. Probably if you prove that in a subspace $U$ the matrix $A$ is normal then you can apply induction on dimension on the space and then conclude the proof by saying that $A$ is diagonalizable when restricted to $U$ as $\dim(U) < \dim(V)$ and induction can be applied for $A$ when restricted to $U$. Thats probably the point of your invariance theorem.


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