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11.3: Normal operators and the spectral decomposition

( \newcommand{\kernel}{\mathrm{null}\,}\)

Recall that an operator TL(V) is diagonalizable if there exists a basis B for V such that B consists entirely of eigenvectors for T. The nicest operators on V are those that are diagonalizable with respect to some orthonormal basis for V. In other words, these are the operators for which we can find an orthonormal basis for V that consists of eigenvectors for T. The Spectral Theorem for finite-dimensional complex inner product spaces states that this can be done precisely for normal operators.

Theorem 11.3.1. Let V be a finite-dimensional inner product space over C and TL(V). Then T is normal if and only if there exists an orthonormal basis for V consisting of eigenvectors for T.

Proof. ("⟹") Suppose that T is normal.
Combining Theorem7.5.3~??? and Corollary9.5.5~???, there exists an orthonormal basis e=(e1,,en) for which the matrix M(T) is upper triangular, i.e.,
M(T)=[a11a1n0ann].
We will show that M(T) is, in fact, diagonal, which implies that the basis elements e1,,en are eigenvectors of T.

Since M(T)=(aij)ni,j=1 with aij=0 for i>j, we have Te1=a11e1 and Te1=nk=1¯a1kek. Thus, by the Pythagorean Theorem and Proposition11.2.3~???,
|a11|2=a11e12=Te12=Te12=nk=1¯a1kek2=nk=1|a1k|2,
from which it follows that |a12|==|a1n|=0. Repeating this argument, Tej2=|ajj|2 and Tej2=nk=j|ajk|2 so that aij=0 for all 2i<jn. Hence, T is diagonal with respect to the basis
e, and e1,,en are eigenvectors of T.

("⟸") Suppose there exists an orthonormal basis (e1,,en) for V that consists of eigenvectors for T. Then the matrix M(T) with respect to this basis is diagonal. Moreover, M(T)=M(T) with respect to this basis must also be a diagonal matrix.
It follows that TT=TT since their corresponding matrices commute:
M(TT)=M(T)M(T)=M(T)M(T)=M(TT).

The following corollary is the best possible decomposition of a complex vector space V into subspaces that are invariant under a normal operator T. On each subspace null(TλiI), the operator T acts just like multiplication by scalar λi. In other words,

T|null(TλiI)=λiInull(TλiI).

Corollary 11.3.2. Let TL(V) be a normal operator, and denote by λ1,,λm the distinct eigenvalues of T.

  1. V=null(Tλ1I)null(TλmI).
  2. If ij, then null(TλiI)null(TλjI).

As we will see in the next section, we can use Corollary 11.3.2 to decompose the canonical matrix for a normal operator into a so-called “unitary diagonalization”.


This page titled 11.3: Normal operators and the spectral decomposition is shared under a not declared license and was authored, remixed, and/or curated by Isaiah Lankham, Bruno Nachtergaele, & Anne Schilling.

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