Non-inductive proof of the spectral theorem (for normal matrices)

Here we present a non-inductive proof of the spectral theorem for normal matrices (which doesn’t use, for instance, proposition 8.6.4 in Artin’s Algebra). (But it does seem to be the same as the proof in Herstein’s Topics in algebra.) It is motivated by a similar direct proof (presented in my class) for Hermitian operators with {n} distinct eigenvalues.

We work in matrix form, so we need to prove that there exists an orthonormal basis of {\mathbb{C}^n} consisting of eigenvectors of a normal matrix {A}.

First we show that {A} has no generalized eigenvectors of order {2} (or higher). Fix {\lambda\in\mathbb{C}}, and define {A' = A-\lambda I}. Since {I = I^*} commutes with every {n\times n} matrix, {A'} is normal. It suffices to show that {A'X=0} whenever {A'^2 X = 0} (here {X\in\mathbb{C}^n} is a vector). Suppose the latter holds. Since {A',A'^*} commute,

\displaystyle \lvert A'^*A' X\rvert^2 = [X^* A'^*A'] [A'^*A' X] = [X^*(A'^*)^2] [A'^2 X] = 0 \implies A'^*A' X = 0,

from which it similarly follows that

\displaystyle \lvert A'X \rvert^2 = [X^*A'^*] [A'X] = X^* (A'^*A' X) = 0 \implies A'X = 0,

as desired.

Hence {A} has all Jordan blocks of size {1} ({A} is diagonalizable). Let {\lambda_1,\ldots,\lambda_k} be the distinct eigenvalues of {A}, with corresponding eigenspaces {E_{\lambda_1},\ldots,E_{\lambda_k}}, so that {\mathbb{C}^n = \bigoplus_{i=1}^{k} E_{\lambda_i}}.

We now prove that the {E_{\lambda_i}} are pairwise orthogonal. Let {r,s} be two distinct indices and take {X\in E_{\lambda_r}}, {Y\in E_{\lambda_s}}. Let {A' = A-\lambda_r I}, so {A'} is normal as before. Then {A'X = 0} (by definition), so

\displaystyle \lvert A'^* X \rvert^2 = [X^*A'] [A'^*X] = [X^*A'^*] [A'X] = 0\implies A'^*X = 0\implies A^* X = \overline{\lambda_r} X.


\displaystyle X^* (\lambda_s Y) = X^* (AY) = (A^* X)^* Y = (\lambda_r X^*)Y \implies (\lambda_r-\lambda_s) X^*Y = 0,

so {\lambda_r\ne\lambda_s} forces {X^*Y = 0}, establishing the desired orthogonality.

Finally, we can just choose arbitrary orthonormal bases for the (nondegenerate) individual {E_{\lambda_i}}, and merge the resulting {k} bases into a single orthonormal basis of eigenvectors for {A}, so we’re done.

2 thoughts on “Non-inductive proof of the spectral theorem (for normal matrices)”

  1. It would be interesting to see if there’s a direct proof that the minimal polynomial of a normal matrix is squarefree.

    Also, there’s a nice perturbation approach to the spectral theorem (due to von Neumann-Wigner; see

    For the operator version, need to look at more carefully, but see


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s