# Proving that normal operators can be diagonalized

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."

• 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 $V$. Apr 2 at 2:21

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.