Spectral theorem for real symmetric matrices

We first record a lemma: every eigenvalue of a real symmetric matrix is real. If $A\mathbf{v} = \lambda \mathbf{v}$ with $\mathbf{v} \neq \mathbf{0}$ possibly complex, consider $\mathbf{v}^{*}A\mathbf{v}$ where $\mathbf{v}^{*}$ is the conjugate transpose. Since $A$ is real and symmetric, $\mathbf{v}^{*}A\mathbf{v} = (\mathbf{v}^{*}A\mathbf{v})^{*}$, so this quantity is real; but it also equals $\lambda \, \mathbf{v}^{*}\mathbf{v}$, and $\mathbf{v}^{*}\mathbf{v} > 0$ is a positive real number, forcing $\lambda$ itself to be real.

We now prove the theorem by induction on $n$. The case $n = 1$ is trivial: any $1 \times 1$ matrix is already diagonal, with $Q = (1)$.

For the inductive step, assume the theorem holds for all real symmetric matrices of size $n - 1$. Since $A$ is $n \times n$ real symmetric, by the lemma its characteristic polynomial has a real root $\lambda_1$; choose a corresponding eigenvector and normalize it to a unit vector $\mathbf{v}_1$.

Let $W$ be the orthogonal complement of the line spanned by $\mathbf{v}_1$, an $(n-1)$-dimensional subspace. We claim $W$ is invariant under $A$: for any $\mathbf{w} \in W$ (so $\mathbf{w}^\top \mathbf{v}_1 = 0$), we compute $(A\mathbf{w})^\top \mathbf{v}_1 = \mathbf{w}^\top A^\top \mathbf{v}_1 = \mathbf{w}^\top A \mathbf{v}_1 = \lambda_1 \mathbf{w}^\top \mathbf{v}_1 = 0$, using symmetry of $A$ in the middle step. So $A\mathbf{w}$ is again orthogonal to $\mathbf{v}_1$, i.e. $A\mathbf{w} \in W$.

Choose an orthonormal basis of $W$; in this basis, the restriction of $A$ to $W$ is represented by an $(n-1) \times (n-1)$ matrix $A'$, and $A'$ is symmetric because $A$ is (restricting a symmetric bilinear form to a subspace, in an orthonormal basis, keeps it symmetric). By the induction hypothesis, $A'$ has an orthonormal basis of eigenvectors $\mathbf{v}_2, \dots, \mathbf{v}_n$ inside $W$, with real eigenvalues $\lambda_2, \dots, \lambda_n$; since $W$ is $A$-invariant, these are also genuine eigenvectors of $A$ itself.

Collecting $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n$ gives an orthonormal basis of $\mathbb{R}^n$ made entirely of eigenvectors of $A$. Assembling them as the columns of a matrix $Q$ makes $Q$ orthogonal, and $AQ = Q\Lambda$ where $\Lambda = \mathrm{diag}(\lambda_1, \dots, \lambda_n)$; since $Q^{-1} = Q^\top$ for an orthogonal matrix, this rearranges exactly to $A = Q \Lambda Q^\top$, completing the induction.