Spectral Theorem for Real Symmetric Matrices
Claim
Every real symmetric matrix \(A \in \mathbb{R}^{n \times n}\) has an orthonormal basis of \(\mathbb{R}^n\) consisting of eigenvectors of \(A\), with real eigenvalues (Trefethen and Bau 1997). Equivalently, there exist an orthogonal matrix \(Q \in \mathbb{R}^{n \times n}\) and a real diagonal matrix \(\Lambda = \operatorname{diag}(\lambda_1, \ldots, \lambda_n)\) such that
\[ A = Q \Lambda Q^\top. \]
Proof
The proof proceeds in four steps: the eigenvalues of \(A\) are real, \(A\) has at least one real eigenvector, the orthogonal complement of any eigenvector is invariant under \(A\), and induction on dimension assembles the orthonormal eigenbasis.
Step 1: Eigenvalues are real.
Let \(\lambda \in \mathbb{C}\) be a root of the characteristic polynomial \(\det(A - \lambda I) = 0\). Since \(A - \lambda I\) is singular over \(\mathbb{C}\), there exists a nonzero \(\mathbf{v} \in \mathbb{C}^n\) with \(A\mathbf{v} = \lambda \mathbf{v}\). Let \(\mathbf{v}^*\) denote the conjugate transpose. Then
\[ \mathbf{v}^* A \mathbf{v} = \mathbf{v}^* (\lambda \mathbf{v}) = \lambda \|\mathbf{v}\|^2. \]
On the other hand, because \(A\) is real and symmetric, \(A^* = A^\top = A\), so
\[ (\mathbf{v}^* A \mathbf{v})^* = \mathbf{v}^* A^* \mathbf{v} = \mathbf{v}^* A \mathbf{v}. \]
Thus \(\mathbf{v}^* A \mathbf{v}\) is real. Combined with \(\|\mathbf{v}\|^2 > 0\), this forces \(\lambda \in \mathbb{R}\).
Step 2: A real eigenvector exists.
Because \(\lambda\) is real, the matrix \(A - \lambda I\) is a singular real matrix, so its null space contains a nonzero real vector \(\mathbf{v} \in \mathbb{R}^n\). After normalization we may assume \(\|\mathbf{v}\| = 1\), and \(A\mathbf{v} = \lambda \mathbf{v}\).
Step 3: The orthogonal complement is invariant.
Let \(W = \{\mathbf{w} \in \mathbb{R}^n : \mathbf{w}^\top \mathbf{v} = 0\}\). For any \(\mathbf{w} \in W\),
\[ (A\mathbf{w})^\top \mathbf{v} = \mathbf{w}^\top A^\top \mathbf{v} = \mathbf{w}^\top A \mathbf{v} = \lambda \mathbf{w}^\top \mathbf{v} = 0, \]
so \(A\mathbf{w} \in W\). The restriction \(A|_W : W \to W\) is well-defined.
Step 4: Induction on dimension.
The base case \(n = 1\) is immediate: \(A = (a)\) has eigenvalue \(a\) with eigenvector \(1\).
For \(n > 1\), \(W\) has dimension \(n - 1\). Pick any orthonormal basis \(\mathbf{q}_2, \ldots, \mathbf{q}_n\) of \(W\) and let \(Q_W = [\mathbf{q}_2 \;\cdots\; \mathbf{q}_n] \in \mathbb{R}^{n \times (n-1)}\). The matrix representation of \(A|_W\) in this basis is
\[ B = Q_W^\top A Q_W \in \mathbb{R}^{(n-1) \times (n-1)}, \]
which is symmetric: \(B^\top = Q_W^\top A^\top Q_W = Q_W^\top A Q_W = B\). By the induction hypothesis, \(B\) has an orthonormal eigenbasis \(\mathbf{c}_2, \ldots, \mathbf{c}_n \in \mathbb{R}^{n-1}\) with real eigenvalues \(\lambda_2, \ldots, \lambda_n\).
Pulling these back into \(\mathbb{R}^n\), set \(\mathbf{u}_i = Q_W \mathbf{c}_i\) for \(i = 2, \ldots, n\). Each \(\mathbf{u}_i \in W\) is a unit vector and
\[ A \mathbf{u}_i = A Q_W \mathbf{c}_i = Q_W B \mathbf{c}_i = Q_W (\lambda_i \mathbf{c}_i) = \lambda_i \mathbf{u}_i, \]
using \(A Q_W = Q_W B\) which holds because \(W\) is \(A\)-invariant. The vectors \(\mathbf{u}_2, \ldots, \mathbf{u}_n\) are orthonormal (as \(Q_W\) has orthonormal columns and \(\mathbf{c}_i\) are orthonormal) and lie in \(W\), so they are orthogonal to \(\mathbf{v}\).
Setting \(\mathbf{u}_1 = \mathbf{v}\) and \(\lambda_1 = \lambda\), the vectors \(\mathbf{u}_1, \ldots, \mathbf{u}_n\) form an orthonormal eigenbasis of \(\mathbb{R}^n\) with real eigenvalues \(\lambda_1, \ldots, \lambda_n\).
Matrix form.
Let \(Q = [\mathbf{u}_1 \;\cdots\; \mathbf{u}_n]\) and \(\Lambda = \operatorname{diag}(\lambda_1, \ldots, \lambda_n)\). The eigenvector equations \(A\mathbf{u}_i = \lambda_i \mathbf{u}_i\) are equivalent to \(AQ = Q\Lambda\), and since \(Q\) is orthogonal,
\[ A = Q \Lambda Q^\top. \qquad \square \]
Corollary: Positive Semidefinite Case
If in addition \(A\) is positive semidefinite — that is, \(\mathbf{x}^\top A \mathbf{x} \geq 0\) for all \(\mathbf{x}\) — then every eigenvalue satisfies \(\lambda_i \geq 0\). Indeed, \(\lambda_i = \mathbf{u}_i^\top A \mathbf{u}_i \geq 0\).
This is the form invoked in the proof that every matrix has a singular value decomposition, applied to the positive semidefinite matrix \(A^\top A\).