Simultaneous Diagonalization

Definition Let $V$ be a finite-dimensional vector space over $\mathbb{F}$, where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$, and let $T_1, \dots, T_k: V \to V$ be linear operators. The family is simultaneously diagonalizable if some basis of $V$ consists of common eigenvectors, that is, vectors that are eigenvectors of every $T_j$ at once. For matrices $A_1, \dots, A_k \in \mathbb{F}^{n \times n}$ this means that one invertible $P \in \mathbb{F}^{n \times n}$ makes every $P^{-1}A_jP$ diagonal. A common eigenvector $v$ may have a different eigenvalue for each operator, $T_jv = \lambda_j v$. When $V$ is an inner product space and the common basis can be chosen orthonormal, such a basis is called a common orthonormal eigenbasis. For matrices acting on $\mathbb{F}^n$ with the standard inner product, there is then one unitary $Q$, real orthogonal when $\mathbb{F} = \mathbb{R}$, such that every $Q^{*}A_jQ$ is diagonal.

Definition Two linear operators $S, T: V \to V$ on the same space commute if $ST = TS$. A family $T_1, \dots, T_k$ is pairwise commuting if $T_iT_j = T_jT_i$ for all $i$ and $j$. In any basis, two operators commute exactly when their matrices commute. Commuting is necessary for simultaneous diagonalization. If $P^{-1}A_iP = D_i$ and $P^{-1}A_jP = D_j$ with $D_i$ and $D_j$ diagonal, then

because diagonal matrices commute. If the common basis is orthonormal over $\mathbb{C}$, so that $A_j = QD_jQ^{*}$ with $Q$ unitary, each $A_j$ is also normal by the complex spectral theorem. So a family with a common orthonormal eigenbasis is pairwise commuting and consists of normal operators.

Definition Let $S, T: V \to V$ be linear operators on a vector space $V$ over $\mathbb{F}$ with $ST = TS$, and let $\lambda$ be an eigenvalue of $T$ with eigenspace $E_\lambda(T) = \ker(T - \lambda I)$. Then $S$ maps $E_\lambda(T)$ into itself. Indeed, if $Tv = \lambda v$, then

so $Sv$ lies in $E_\lambda(T)$, possibly as the zero vector. Now let $V$ be a finite-dimensional complex inner product space and $T$ normal, with distinct eigenvalues $\lambda_1, \dots, \lambda_r$. By the spectral theorem, $V$ is the orthogonal direct sum of the eigenspaces $E_{\lambda_1}(T), \dots, E_{\lambda_r}(T)$. The commuting operator $S$ preserves every one of them. The orthogonal complement of any single eigenspace is spanned by the remaining eigenspaces, so $S$ preserves that complement too. Each eigenspace of $T$ is therefore a reducing subspace for $S$, and in an orthonormal basis adapted to the decomposition the matrix of $S$ is block diagonal. Here $S$ itself need not be normal.

Definition Let $V$ be a finite-dimensional inner product space over $\mathbb{F}$ and let $S: V \to V$ be linear. A subspace $U$ reduces $S$ if $S(U) \subseteq U$ and $S(U^{\perp}) \subseteq U^{\perp}$, equivalently if $U$ is invariant under both $S$ and $S^{*}$. In that case the restriction $S|_U: U \to U$, with the inner product of $V$ restricted to $U$, has adjoint

and if $S$ is normal, then $S|_U$ is normal. To see this, take an orthonormal basis of $U$ followed by an orthonormal basis of $U^{\perp}$. The matrix of $S$ is block diagonal with a block $B_1$ for $U$ and a block $B_2$ for $U^{\perp}$, and the matrix of $S^{*}$ is block diagonal with blocks $B_1^{*}$ and $B_2^{*}$. The equation $SS^{*} = S^{*}S$ then holds block by block, and its first block says $B_1B_1^{*} = B_1^{*}B_1$. In the same way, if two operators both map $U$ into itself, the restriction of their product is the product of their restrictions, so restrictions of commuting operators to a common invariant subspace still commute.

Definition Let $V$ be a finite-dimensional complex inner product space and let $T_1, \dots, T_k: V \to V$ be a finite family of linear operators. Then $V$ has an orthonormal basis consisting of common eigenvectors of $T_1, \dots, T_k$ if and only if every $T_j$ is normal and $T_iT_j = T_jT_i$ for all $i$ and $j$. In matrix form, matrices $A_1, \dots, A_k \in \mathbb{C}^{n \times n}$ are normal and pairwise commuting exactly when one unitary $Q$ makes every $Q^{*}A_jQ$ diagonal. The conditions are necessary because diagonal matrices are normal and commute. Sufficiency is proved by induction on $k$, for all spaces at once. For $k = 1$ it is the complex spectral theorem. For the step, split $V$ into the orthogonal eigenspaces of $T_1$. Each of them reduces $T_2, \dots, T_k$, so on each eigenspace $E$ the restrictions of these operators are normal and still pairwise commute. By the induction hypothesis $E$ has an orthonormal basis of common eigenvectors of those $k - 1$ restrictions, and $T_1$ acts on $E$ as a scalar, so the same vectors are eigenvectors of $T_1$ too. Eigenspaces of $T_1$ for distinct eigenvalues are orthogonal, so the bases of all the eigenspaces together form an orthonormal basis of $V$. If $T_1$ is a scalar multiple of $I$, the split leaves $V$ whole, but the number of operators still drops, so the induction never needs the dimension to shrink. The common basis is not unique in general. Over $\mathbb{R}$, pairwise commuting real symmetric matrices have a common real orthonormal eigenbasis by the same argument. Real normal matrices need not, since the quarter turn is real normal, commutes with $I$, and has no real eigenvector.

Definition Let $A, B \in \mathbb{C}^{n \times n}$ with $AB = BA$. If $A$ is not normal, the complex spectral theorem says that $A$ has no orthonormal eigenbasis, so $A$ and $B$ have no common orthonormal eigenbasis, however simple $B$ is. Commuting alone does not even give a common eigenbasis. A matrix that is not diagonalizable has no eigenbasis of its own, yet it commutes with $I$ and with every polynomial in itself. A diagonalizable matrix that is not normal can share an eigenbasis with a commuting partner, but never an orthonormal one.