Normal endomorphism is unitary diagonalizable

Let $$V$$ be some finite-dimensional unitary vector space and $$f\in L(V,V)$$ be some normal endomorphism. Show that $$f$$ is unitary diagonalizable.

Hint: Use the following proposition: If $$f_1,f_2\in L(V,V)$$ are commuting endomorphisms, then $$f_1$$ and $$f_2$$ are simultaneously diagonalizable.

I think, what I have to show is that

$$UM(f;b)U^*=D$$ where $$M(f;b)$$ is the transformation matrix corresponding to $$f$$ (with respect to some basis $$b$$), and $$U$$ is an unitary matrix, and $$D$$ is a diagonal matrix.

I am not sure if I am allowed to use the spectral theorem for normal endomorphisms which tells me that, since $$f$$ is normal, there is an orthonormal basis $$e$$ consisting of eigenfunctions of $$f$$ such that the corresponding $$M(f;e)$$ is diagonal.

• The hint your were given is strange and doesn't provide obvious help with the problem. I strongly suspect that you are supposed to use some version of the spectral theorem... what do you mean by "I am not sure if I am allowed"? Is this for a class? Can you ask your instructor? Commented Jan 17, 2023 at 18:49
• I also had the impression that the given hint is not very helpful. This is an exercise in class and I am allowed to use the spectral theorem. However, I am not sure how to use it. Commented Jan 17, 2023 at 18:52
• Got it, thanks for clarifying Commented Jan 17, 2023 at 18:54

The precise approach here depends on what exactly your definition of "unitarily diagonalizable" is; it's clear what the definition should be for a complex matrix, but it's not clear what the definition should be for a transformation over an arbitrary (finite-dimensional) unitary space.

What I suspect is meant is the following: $$f$$ is unitarily diagonalizable iff there exists a unitary transformation $$U:\Bbb C^n \to V$$ (where $$n = \dim(V)$$) such that $$U^* \circ f \circ U$$ (a linear map over $$\Bbb C^n$$) is a diagonal matrix (or more precisely, corresponds to multiplication by a diagonal matrix). With that in mind, let $$\lambda_1,\dots,\lambda_n$$ be the eigenvalues of $$f$$, and $$v_1,\dots,v_n$$ a corresponding orthonormal set of eigenvectors. Show that the map $$U(x_1,\dots,x_n) = x_1 v_1 + \cdots+ x_n v_n$$ has this property. It is helpful to note/show that its adjoint has the form $$U^*(v) = (\langle v,v_1\rangle , \dots, \langle v,v_n\rangle).$$

Perhaps the definition that you would prefer is that $$f$$ is unitarily diagonalizable iff for an orthonormal basis $$\mathcal B$$ (i.e. for at least one such basis), the matrix $$[f]_{\mathcal B} = [f]^{\mathcal B}_{\mathcal B}$$ (of $$f$$ relative to $$\mathcal B$$) is diagonalizable.

We can show that this holds with the help of the proof outlined above. Let $$\mathcal A$$ denote the standard basis of $$\Bbb C^n$$. Let $$D$$ denote the diagonal matrix corresponding to $$U^* \circ f \circ U$$ (i.e. $$D = \operatorname{diag}(\lambda_1,\dots, \lambda_n)$$). We can argue that $$[U^* \circ f \circ U]^{\mathcal A}_{\mathcal A} = D \implies\\ [U^*]_{\mathcal A}^{\mathcal B} [f]^{\mathcal B}_{\mathcal B} [U]^{\mathcal A}_{\mathcal B} = D \implies\\ [U]_{\mathcal B}^{\mathcal A *} [f]_{\mathcal B} [U]^{\mathcal A}_{\mathcal B} = D.$$ So, $$[f]_{\mathcal B}$$ is indeed unitarily diagonalizable with the unitary change of basis matrix $$[U]^{\mathcal A}_{\mathcal B}$$.

Note: $$[U]^{\mathcal A}_{\mathcal B}$$ denotes the matrix of a transformation relative two separate bases, $$\mathcal A$$ being the basis of the domain and $$\mathcal B$$ the basis of the codomain.

All that said, if it really is enough to find one orthonormal basis $$\mathcal B$$ such that $$[f]_{\mathcal B}$$ is diagonalizable, we might as well choose the basis that diagonalizes $$f$$, i.e. it basis of eigenvectors. In particular, if $$\mathcal B = \{v_1,v_2,\dots,v_n\}$$, then we simply have $$[f]_{\mathcal B} = D$$.

• Seems to me that $U$ is the change of basis transformation from one orthonormal basis to another. One of these to orthonormal basis is the one which exists by the spectral theorem. Which is the other? Commented Jan 17, 2023 at 20:02
• @Bridge $U$ is not a change of basis transformation: a change of basis is a map from $\Bbb C^n$ to $\Bbb C^n$. Commented Jan 17, 2023 at 20:50
• @BridgeTYH See my latest edit. Commented Jan 17, 2023 at 21:01
• I recommend that you review your notes for the precise wording of the definition of "diagonalizable" in this context. Commented Jan 17, 2023 at 21:08