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# 11.3 Normal operators and the spectral decomposition

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Recall that an operator $$T \in \mathcal{L}(V)$$ 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 $$\mathbb{C}$$ and $$T\in\mathcal{L}(V)$$. Then $$T$$ is normal if and only if there exists an orthonormal basis for $$V$$ consisting of eigenvectors for $$T$$.

Proof. $$( "\Longrightarrow" )$$ Suppose that $$T$$ is normal.
Combining Theorem7.5.3~\ref{thm:ComplexLinearMapsUpperTriangularWrtSomeBasis} and Corollary9.5.5~\ref{thm:ComplexLinearMapsUpperTriangularWrtSomeOrthonormalBasis}, there exists an orthonormal basis $$e=(e_1,\ldots,e_n)$$ for which the matrix $$M(T)$$ is upper triangular, i.e.,
\begin{equation*}
M(T) = \begin{bmatrix} a_{11} & \cdots & a_{1n}\\ &\ddots& \vdots \\ 0&& a_{nn} \end{bmatrix}.
\end{equation*}
We will show that $$M(T)$$ is, in fact, diagonal, which implies that the basis elements $$e_1,\ldots,e_n$$ are eigenvectors of $$T$$.

Since $$M(T)=(a_{ij})_{i,j=1}^n$$ with $$a_{ij}=0$$ for $$i>j$$, we have $$Te_1=a_{11}e_1$$ and $$T^*e_1=\sum_{k=1}^n \overline{a}_{1k} e_k$$. Thus, by the Pythagorean Theorem and Proposition11.2.3~\ref{prop:adjoint norm},
\begin{equation*}
|a_{11}|^2 = \norm{a_{11}e_1}^2 = \norm{Te_1}^2 = \norm{T^*e_1}^2
=\norm{\sum_{k=1}^n \overline{a}_{1k} e_k}^2 = \sum_{k=1}^n |a_{1k}|^2,
\end{equation*}
from which it follows that $$|a_{12}| = \cdots = |a_{1n}| = 0$$. Repeating this argument, $$\norm{Te_j}^2=|a_{jj}|^2$$ and $$\norm{T^*e_j}^2 = \sum_{k=j}^n |a_{jk}|^2$$ so that $$a_{ij}=0$$ for all $$2\le i<j\le n$$. Hence, $$T$$ is diagonal with respect to the basis
$$e$$, and $$e_1,\ldots,e_n$$ are eigenvectors of $$T$$.

$$( "\Longleftarrow" )$$ Suppose there exists an orthonormal basis $$(e_1,\ldots,e_n)$$ 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:
\begin{equation*}
M(TT^*) = M(T)M(T^*) = M(T^*)M(T) = M(T^*T).
\end{equation*}

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 $$\kernel(T-\lambda_i I)$$, the operator $$T$$ acts just like multiplication by scalar $$\lambda_i$$. In other words,

$T|_{\kernel(T-\lambda_i I)} = \lambda_{i}I_{\kernel(T-\lambda_i I)}.$

Corollary 11.3.2. Let $$T\in\mathcal{L}(V)$$ be a normal operator, and denote by $$\lambda_1,\ldots,\lambda_m$$ the distinct eigenvalues of $$T$$.

1. $$V = \kernel(T-\lambda_1 I) \oplus \cdots \oplus \kernel(T-\lambda_m I)$$.
2. If $$i\neq j$$, then $$\kernel(T-\lambda_i I)\bot \kernel(T-\lambda_j I)$$.

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

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