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Mathematics LibreTexts

11.3 Normal operators and the spectral decomposition

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”.

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