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Over a field of characteristic not $2$ , every symmetric matrix is congruent to a diagonal matrix

Statement

If $A \in M_{n} (F)$ is symmetric and $char F \neq 2$ , then there is an invertible $P$ such that $P^{T} A P$ is diagonal.

Facts & Assumptions

Given: A symmetric matrix $A \in M_{n} (F)$ over a field of characteristic not $2$ .

[L1]

The standard coordinate vectors form a basis of $F^{n}$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[L2]

Every symmetric bilinear form in the stated characteristic has an orthogonal basis (Every symmetric bilinear form on a finite-dimensional space over a field of characteristic not $2$ has an orthogonal basis).

[L3]

A basis change by $P$ sends a bilinear-form matrix $A$ to $P^{T} A P$ (A basis change by $P$ changes the matrix of a bilinear form from $A$ to $P^{T} A P$ ).

Proof

technique · direct

1.1L1givenalgebra

In the standard basis [L1], let $B (u, v) = u^{T} A v$ . Symmetry of $A$ makes $B$ symmetric.

1.2L2choose

Choose an orthogonal basis for $B$ by [L2], and let $P$ have those basis vectors as its columns in standard coordinates. Then $P$ is invertible and the matrix of $B$ in that basis is diagonal.

2.1step 1.2L3∎

By [L3], this diagonal matrix is $P^{T} A P$ . When $n = 0$ , the empty matrix is already diagonal and the same conclusion holds.

Depends on

Dependency tree · 2 levels

Sources

K. Conrad, Bilinear Forms, §5

Over a field of characteristic not 22, every symmetric matrix is congruent to a diagonal matrix