A real symmetric or complex Hermitian matrix is diagonalizable in an orthonormal basis of eigenvectors.
(Notation)
For a matrix \(A\), \(A^t\) is the transpose and \(A^{\dagger}\) the conjugate transpose, and \(\inner{\cdot}{\cdot}\) is the standard inner product on \(\RR^n\) or \(\CC^n\).
The adjoint of \(A\) is the matrix \(A^{\dagger}\), characterized by \(\inner{A\vector x}{\vector y} = \inner{\vector x}{A^{\dagger} \vector y}\) for all \(\vector x,\vector y\), and \(A\) is self-adjoint if \(A=A^\dagger\).
A real matrix \(A\) is symmetric if \(A = A^t\), and orthogonal if \(A^tA = AA^t = I\).
A complex matrix \(A\) is Hermitian if \(A^{\dagger} = A\), normal if \(AA^{\dagger} = A^{\dagger}A\), and unitary if \(A^{\dagger}A = AA^{\dagger} = I\).
Diagonalizability
Diagonalizable matrix (Tag FD-K6FVX)
Let \(R\) be a commutative ring and \(A\) an \(n\times n\) matrix over \(R\). \(A\) is diagonalizable if it is similar to a diagonal matrix: there exist a diagonal \(n\times n\) matrix \(D\) over \(R\) and \(P\in \GL_n(R)\) such that \(A = PDP^{-1}\).
A linear operator over an algebraically closed field has an eigenvector in every nonzero invariant subspace (Tag L-C5DDK)
Let \(k\) be an algebraically closed field, let \(V\) be a finite-dimensional \(k\)-vector space, and let \(A\colon V\to V\) be \(k\)-linear. If \(W\subseteq V\) is a nonzero subspace with \(A(W)\subseteq W\), then \(W\) contains an eigenvector of \(A\).
The restriction \(A|_W\colon W\to W\) is a linear operator on a nonzero finite-dimensional space, so its characteristic polynomial has positive degree and has a root \(\lambda\in k\). Then \(A|_W-\lambda\id_W\) is not injective, and a nonzero \(w\in W\) with \(Aw=\lambda w\) is an eigenvector of \(A\).
A linear operator on an infinite-dimensional vector space over an algebraically closed field need not have an eigenvector: multiplication by \(x\) on \(V=W=k[x]\) has no eigenvector, since \(xf=\lambda f\) has no nonzero solution \(f\in k[x]\) by comparing degrees.
Diagonalizability via the minimal polynomial (Tag T-6ABNR)
Let \(F\) be a field and \(M\in M_n(F)\) with minimal polynomial \(m_M\in F[x]\). Then \(M\) is diagonalizable over \(F\) if and only if \(m_M\) splits over \(F\) into distinct linear factors.
If \(M=PDP^{-1}\) with \(D\) diagonal whose distinct diagonal entries are \(\lambda_1,\ldots,\lambda_r\), then \(\prod_i(x-\lambda_i)\) annihilates \(D\) and hence \(M\), so \(m_M\) divides a product of distinct linear factors.
Conversely, suppose \(m_M=\prod_{i=1}^r(x-\lambda_i)\) with the \(\lambda_i\in F\) distinct. The factors are pairwise coprime, so by the kernel decomposition \(F^n=\ker m_M(M)=\bigoplus_{i=1}^r\ker(M-\lambda_i I)\). Choosing a basis of each eigenspace \(\ker(M-\lambda_i I)\) gives a basis of \(F^n\) of eigenvectors of \(M\).
A matrix whose minimal polynomial splits over \(F\) need not be diagonalizable over \(F\). Over any field \(F\), \(M=\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}\) has \(m_M = (x-1)^2\), which splits over \(F\), but \(\ker(M-I)\) is one-dimensional, so \(M\) is not diagonalizable.
Characterizations of diagonalizability (Tag FT-BC6S2)
Let \(F\) be a field and let \(M\in\Mat_n(F)\). The following are equivalent:
\(M\) is diagonalizable over \(F\).
The minimal polynomial \(\min_M(x)\) is a product of distinct monic linear factors in \(F[x]\).
There is a basis of \(F^n\) consisting of eigenvectors of \(M\).
Every elementary divisor of \(M\) has degree \(1\).
In particular, if \(M\) has \(n\) distinct eigenvalues in \(F\), equivalently if \(\min_M(x)\) has \(n\) distinct roots in \(F\), then \(M\) is diagonalizable over \(F\).
A matrix whose minimal polynomial splits over \(F\) need not be diagonalizable over \(F\). For \(M=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) over any field \(F\), \(\min_M(x)=(x-1)^2\) splits over \(F\), and \(M\) is not diagonalizable, because its only eigenvalue is \(1\) and \(M\ne I\).
\(\impliedby\): If \(\min_A\) is a product of distinct linear factors, then every invariant factor of \(A\) divides \(\min_A\), so every elementary divisor is linear, and the Jordan form of \(A\) is diagonal.
(Criterion)
\(A\) is diagonalizable over \(F\) if and only if \(\min_A\) is a product of distinct linear factors over \(F\). Hence a polynomial identity \(f(A) = 0\) with \(f\) a product of distinct linear factors over \(F\), such as \(A^2 = A\), implies that \(A\) is diagonalizable; see Find the canonical form.
The spectral theorem
Spectral theorem for Hermitian and real symmetric matrices (Tag T-WQHMA)
If \(A\in\Mat_n(\CC)\) is Hermitian, that is, \(A^{\dagger} = A\), then \(A\) is diagonalizable over \(\CC\); more precisely, there is a unitary matrix \(U\) such that \(U^{\dagger}AU\) is diagonal with real entries.
If \(A\in\Mat_n(\RR)\) is symmetric, that is, \(A^t = A\), then \(A\) is diagonalizable over \(\RR\); more precisely, there is an orthogonal matrix \(Q\) such that \(Q^tAQ\) is diagonal.
A real matrix \(A\) is diagonalizable by an orthogonal matrix if and only if \(A\) is symmetric: if \(Q^tAQ = D\) with \(Q\) orthogonal and \(D\) diagonal, then \(A = QDQ^t\) and \(A^t = QD^tQ^t = A\). A complex matrix is diagonalizable by a unitary matrix if and only if it is normal.
(of the spectral theorem)
Let \(A\) be Hermitian, and let \(\lambda\) be an eigenvalue of \(A\) with unit eigenvector \(\vector v\), which exists over \(\CC\). Then \[ \lambda = \lambda\inner{\vector v}{\vector v} = \inner{A\vector v}{\vector v} = \inner{\vector v}{A\vector v} = \overline{\lambda}, \] so \(\lambda \in \RR\). A real symmetric matrix is Hermitian, so its eigenvalues are real, and it has a real unit eigenvector.
Let \(W = \ts{\vector v}^\perp\). For \(\vector w \in W\), \[ \inner{\vector v}{ A \vector w} = \inner{A \vector v}{\vector w} = \lambda \inner{\vector v}{\vector w} = 0, \] so \(A(W)\subseteq W\), and the restriction of \(A\) to \(W\) is again self-adjoint.
By induction on the dimension, \(W\) has an orthonormal basis of eigenvectors of \(A\), and adjoining \(\vector v\) gives an orthonormal basis of eigenvectors of \(A\) for the whole space.
Simultaneous diagonalization
Simultaneous diagonalizability of commuting diagonalizable operators (Tag PR-GV5CF)
Let \(V\) be a finite-dimensional vector space over a field \(k\), and let \(\theset{A_i}_{i \in I}\) be a family of diagonalizable linear operators on \(V\). The \(A_i\) pairwise commute if and only if they are simultaneously diagonalizable: there is a basis of \(V\) in which every \(A_i\) is diagonal.
The implication from simultaneous diagonalizability to commuting holds for any family of operators, since diagonal matrices commute.
Commuting operators need not be simultaneously diagonalizable without the diagonalizability hypothesis: \(\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}\) commutes with itself and is not diagonalizable.
Suppose \(A_1,\ldots,A_n\) are pairwise commuting diagonalizable operators on \(V\); we induct on \(n\).
Since \(A_n\) is diagonalizable, \(V = \bigoplus_i E_i\), where \(E_i\) is the eigenspace of \(A_n\) for the eigenvalue \(\mu_i\).
If \(A_jA_n = A_nA_j\) and \(\vector v\in E_i\), then \(A_n(A_j\vector v) = A_jA_n\vector v = \mu_iA_j\vector v\), so each \(E_i\) is \(A_j\)-invariant.
The minimal polynomial of the restriction \(A_j|_{E_i}\) divides that of \(A_j\), so \(A_j|_{E_i}\) is diagonalizable.
By induction, \(A_1|_{E_i},\ldots,A_{n-1}|_{E_i}\) are simultaneously diagonalizable, and every vector of \(E_i\) is an eigenvector of \(A_n\).
The union over \(i\) of these bases of \(E_i\) is a basis of \(V\) in which every \(A_j\) is diagonal.
(Noncommuting diagonalizable operators)
The matrices \(A = \matt{1}{0}{0}{2}\) and \(B = \matt{0}{1}{1}{0}\) are diagonalizable over \(\RR\) and do not commute. They are not simultaneously diagonalizable: the eigenvectors of \(A\) are the multiples of \(e_1\) and of \(e_2\), and neither \(e_1\) nor \(e_2\) is an eigenvector of \(B\).