The Spectral Theorem for Normal Operators

Definition Let $V$ be a finite-dimensional inner product space over $\mathbb{F}$, where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$, and let $T: V \to V$ be linear with adjoint $T^{*}$, defined by $\langle Tv, w \rangle = \langle v, T^{*}w \rangle$ for all $v, w \in V$. The operator $T$ is normal if

A square matrix $A \in \mathbb{C}^{n \times n}$ is normal if $AA^{*} = A^{*}A$, where $A^{*} = \overline{A}^{\mathsf{T}}$. For a real matrix this reads $AA^{\mathsf{T}} = A^{\mathsf{T}}A$. In an orthonormal basis the matrix of $T^{*}$ is the conjugate transpose of the matrix of $T$, so $T$ is normal exactly when its matrix in an orthonormal basis is a normal matrix. The adjoint depends on the inner product, and so does normality. In a basis that is not orthonormal, the conjugate transpose of the matrix need not represent $T^{*}$, and the matrix test can give the wrong answer. Hermitian operators, with $T^{*} = T$, and unitary operators, with $T^{*}T = I$, are normal, but the normal operators form a larger class.

Definition Let $V$ be a finite-dimensional complex inner product space and let $T: V \to V$ be linear. Then $V$ has an orthonormal basis consisting of eigenvectors of $T$ if and only if $T$ is normal. In matrix form, a matrix $A \in \mathbb{C}^{n \times n}$ is normal if and only if

for some unitary $Q$ and some diagonal $\Lambda$. The proof rests on a lemma. An upper triangular normal matrix $R \in \mathbb{C}^{n \times n}$ is diagonal. To see this, compare the first diagonal entries of $RR^{*}$ and $R^{*}R$. The first is the squared length of the first row of $R$. The second is the squared length of the first column, which is only $\lvert r_{11} \rvert^2$ because $R$ is upper triangular. Equality leaves

and a sum of nonnegative real numbers is zero only when every term is zero. So the first row has no entries off the diagonal, $R$ splits into the $1 \times 1$ block $r_{11}$ and an upper triangular block $R_2$ of size $n - 1$, and comparing the lower blocks of $RR^{*}$ and $R^{*}R$ shows that $R_2$ is normal. Induction on $n$ finishes the lemma, since every $1 \times 1$ matrix is diagonal. For the theorem, Schur decomposition gives $A = QRQ^{*}$ with $Q$ unitary and $R$ upper triangular. Then $RR^{*} = Q^{*}AA^{*}Q$ and $R^{*}R = Q^{*}A^{*}AQ$, so $R$ is normal when $A$ is, and the lemma makes $R$ diagonal. Conversely, if $A = Q\Lambda Q^{*}$, then $AA^{*} = Q\Lambda\Lambda^{*}Q^{*}$ and $A^{*}A = Q\Lambda^{*}\Lambda Q^{*}$, and these agree because both diagonal products have entries $\lvert \lambda_j \rvert^2$. The field matters. Schur decomposition needs the complex eigenvalues that the fundamental theorem of algebra supplies, and over $\mathbb{R}$ the quarter turn is normal but has no real eigenvector.

Definition A matrix $A \in \mathbb{C}^{n \times n}$ is unitarily diagonalizable if $A = Q\Lambda Q^{*}$, or equivalently $Q^{*}AQ = \Lambda$, for some unitary $Q$ and some diagonal $\Lambda = \operatorname{diag}(\lambda_1, \dots, \lambda_n)$. Writing $q_1, \dots, q_n$ for the columns of $Q$, this says exactly that the $q_j$ are orthonormal and $Aq_j = \lambda_j q_j$. It also gives the spectral decomposition

in which each $q_j q_j^{*}$ is the orthogonal projection onto the line spanned by $q_j$. The diagonal entries of $\Lambda$ are the eigenvalues of $A$ repeated according to algebraic multiplicity, so $\Lambda$ is unique up to the order of its entries. The matrix $Q$ is not unique. Each column can be multiplied by a complex number of modulus one, and the columns that belong to an eigenspace of dimension two or more can be replaced by any other orthonormal basis of that eigenspace. Since $A - \lambda I = Q(\Lambda - \lambda I)Q^{*}$, the eigenspace for $\lambda$ is spanned by the columns $q_j$ with $\lambda_j = \lambda$. Hence eigenvectors of a normal matrix for distinct eigenvalues are automatically orthogonal, and a unitary diagonalization is found by choosing an orthonormal basis of each eigenspace and putting these bases together. Ordinary diagonalization $A = P\Lambda P^{-1}$ with an invertible $P$ is weaker. By the spectral theorem, $P$ can be chosen unitary only when $A$ is normal.

Definition Let $V$ be a finite-dimensional complex inner product space. A linear operator $T: V \to V$ is Hermitian, or self-adjoint, if $T^{*} = T$, that is, if $\langle Tv, w \rangle = \langle v, Tw \rangle$ for all $v, w \in V$. A matrix $A \in \mathbb{C}^{n \times n}$ is Hermitian if $A^{*} = A$, so its diagonal entries are real and $a_{ji} = \overline{a_{ij}}$. Every eigenvalue of a Hermitian operator is real. If $Tv = \lambda v$ with $v \ne 0$, then

using $T^{*} = T$ for the middle equality, linearity of the inner product in the second argument and conjugate-linearity in the first. Since $\lVert v \rVert^2 > 0$, this gives $\lambda = \overline{\lambda}$. Together with the spectral theorem, this shows that $T$ is Hermitian if and only if it has an orthonormal eigenbasis whose eigenvalues are all real. In matrix form, $A$ is Hermitian if and only if $A = Q\Lambda Q^{*}$ with $Q$ unitary and $\Lambda$ real diagonal, where the converse holds because a real $\Lambda$ gives $A^{*} = Q\Lambda^{*}Q^{*} = Q\Lambda Q^{*} = A$. A real symmetric matrix, meaning $A \in \mathbb{R}^{n \times n}$ with $A^{\mathsf{T}} = A$, is Hermitian, so its eigenvalues are real, and it even has a real orthonormal eigenbasis, $A = Q\Lambda Q^{\mathsf{T}}$ with $Q$ real orthogonal. The reason is that the real and imaginary parts of a complex eigenvector for a real eigenvalue solve the same real equation, and at least one of them is nonzero. If $Au = \lambda u$ with $u$ real, symmetry gives $\langle u, Aw \rangle = \langle Au, w \rangle = \lambda \langle u, w \rangle$, so $A$ maps the real orthogonal complement of $u$ into itself, the restriction there is again symmetric, and induction on dimension finishes the proof.

Definition Let $V$ be a finite-dimensional complex inner product space. A linear operator $T: V \to V$ is unitary if $T^{*}T = I$. In finite dimension this makes $T$ invertible with $T^{-1} = T^{*}$, so $TT^{*} = I$ as well, and every unitary operator is normal. The condition $T^{*}T = I$ holds exactly when $T$ preserves inner products, because

for all $v, w \in V$. In particular $\lVert Tv \rVert = \lVert v \rVert$ for every $v$. A matrix $U \in \mathbb{C}^{n \times n}$ is unitary when $U^{*}U = I$, which says that its columns are orthonormal, and a real matrix with $Q^{\mathsf{T}}Q = I$ is called orthogonal. Every eigenvalue of a unitary operator has modulus one. If $Tv = \lambda v$ with $v \ne 0$, then $\lVert v \rVert = \lVert Tv \rVert = \lvert \lambda \rvert \lVert v \rVert$, and dividing by $\lVert v \rVert$ gives $\lvert \lambda \rvert = 1$. Together with the spectral theorem, this shows that $T$ is unitary if and only if it has an orthonormal eigenbasis whose eigenvalues all lie on the unit circle. In matrix form the converse holds because $\Lambda^{*}\Lambda = I$ for such eigenvalues, so $A = Q\Lambda Q^{*}$ with $Q$ unitary gives $A^{*}A = Q\Lambda^{*}\Lambda Q^{*} = I$.