11.3: Normal operators and the spectral decomposition
( \newcommand{\kernel}{\mathrm{null}\,}\)
Recall that an operator T∈L(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 T∈L(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)=[a11⋯a1n⋱⋮0ann].
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 T∗e1=∑nk=1¯a1kek. Thus, by the Pythagorean Theorem and Proposition11.2.3~???,
|a11|2=‖a11e1‖2=‖Te1‖2=‖T∗e1‖2=‖n∑k=1¯a1kek‖2=n∑k=1|a1k|2,
from which it follows that |a12|=⋯=|a1n|=0. Repeating this argument, ‖Tej‖2=|ajj|2 and ‖T∗ej‖2=∑nk=j|ajk|2 so that aij=0 for all 2≤i<j≤n. 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∗=T∗T since their corresponding matrices commute:
M(TT∗)=M(T)M(T∗)=M(T∗)M(T)=M(T∗T).
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(T−λiI), the operator T acts just like multiplication by scalar λi. In other words,
T|null(T−λiI)=λiInull(T−λiI).
Corollary 11.3.2. Let T∈L(V) be a normal operator, and denote by λ1,…,λm the distinct eigenvalues of T.
- V=null(T−λ1I)⊕⋯⊕null(T−λmI).
- If i≠j, then null(T−λiI)⊥null(T−λjI).
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”.
Contributors
- Isaiah Lankham, Mathematics Department at UC Davis
- Bruno Nachtergaele, Mathematics Department at UC Davis
- Anne Schilling, Mathematics Department at UC Davis
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