Diagonalizing Real-Symmetric Matrices

The spectral theorem for finite-dimensional operators state that a normal matrix is unitarily diagonalizable into a diagonal matrix with complex eigenvalues and a self-adjoint matrix into that with purely real eigenvalues over the complex number field. This can be proved using Schur’s decomposition, which states that any matrix over the complex numbers can be unitarily decomposed into an upper-triangular matrix.

In this post, I present a particular result about real-symmetric matrices.

Theorem 1. For any real-symmetric matrix $A \in \mathbb R^{n \times n}$ , there exist an $n$ -by- $n$ matrix $S$ such that

$S^\intercal B S = \Lambda$

where $\Labmda$ is a diagonal matrix.

Before presentation of a proof, we need a technical lemma.

Lemma. Let $B$ be an self-adjoint operator over a $n$ -dimensional Hilbert Space that is nonzero, i.e. there exist some vector $\bm u$ such that $B \bm u \neq 0$ . Then, there exist a set of unit vectors $\{\bm v, \bm e_2, \dots, \bm e_{n}\}$ such that $\bm v^\intercal , B\bm e_j = 0$ for all $2 \leq j \leq n$ .

proof. Choose a basis such that $B_{11} \neq 0$ . Such a choice is guaranteed since $B$ is a nonzero operator. It is straightforward to show that given an upper-triangular matrix $T \in \mathbb R^{n \times n}$ defined as

$T_{ij} \ =\ \delta_{ij} + \delta_{i1} \delta_{j1} - \delta_{i1} \frac{B_{j1}}{B_{11}}$

we have

$(T^\intercal B T)_{i1} = (T^\intercal B T)_{1i} = 0$

for any $1 < i \leq n$ . Set vector $\bm v$ the unit vector parallel to the first column and $\bm e_2, \dots \bm e_n$ to be the remaining column vectors. By construction, we have that

$\bm v^\intercal B \bm e_{j} = 0$

for any $1 < j \leq n$ . □

This lemma basically tells us that given an operator on a finite-dimensional Hilbert space, it is possible to choose a vector and some subspace such that for the dot product $\langle \cdot, \cdot \rangle_{B } := \langle \cdot, B \cdot \rangle$ vanishes over the choice of the special vector and any other vector in the subspace.

Now, we are ready to present a proof of our theorem.

proof of Theorem 1. We show that there exist $n$ column vectors $\bm v_1, \dots \bm v_n$ such that $\bm v_i^\intercal B \bm v_j = 0$ for any $i, j \in [n]$ where $i \neq j$ . For $n = 1$ , the theorem is trivially true by choosing $S = (1)$ . Now, for $n > 1$ , we apply the Lemma to obtain $\bm v_1$ and $(\bm e_j)_{j = 2}^n$ such that $\bm v ^\intercal B \bm e_j = 0$ for all $2 \leq j \leq n$ . By the inductive hypothesis, within the restriction of $B$ to $\span(\bm e_2, \dots, \bm e_n)$ , we obtain $n - 1$ vectors $(v_j)_{j = 2}^n$ such that $\bm v_j^\intercal B v_k = \lambda_j \delta_{jk}$ for some $\lambda_j \in \mathbb R$ . Also, these new matrices $(\bm v_{j})_{j = 2}^n$ are linear combination of $e_j$ ‘s, and therefore by linearity,

$\bm v_1 ^\intercal B \bm v_j = 0$

for any $1 < j \leq n$ . Choosing $S$ to be the collection of column vectors $(\bm v_j)_{j = 1}^n$ , we obtain

$S^\intercal B S = \Lambda$

where $\Lambda$ a diagonal matrix. □

It appears that a clever choice of matrix $T$ will allow $S$ to be orthogonal. However, Claude claims that such a choice for $n > 5$ is known to not exist. Proof would require a different approach.

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