The Spectral Theorem: Orthogonal Diagonalization of Symmetric Matrices

Every real symmetric matrix can be orthogonally diagonalized — we prove this spectral theorem and its complex generalization for normal operators. The spectral decomposition A = sum lambda_i P_i into orthogonal projections is derived, and we apply it to classify quadratic forms via Sylvester's law of inertia.

Folio Official

March 24, 2026

1. Motivation: From Diagonalization to Orthogonal Diagonalization

In earlier chapters we established when a linear operator can be diagonalized: the algebraic and geometric multiplicities of every eigenvalue must agree. When this holds, there exists an invertible matrix $P$ such that $P^{- 1} A P$ is diagonal. But the matrix $P$ is not unique, and in general the columns of $P$ need not be orthogonal.

On an inner product space, we can ask for something stronger. We say that a real $n \times n$ matrix $A$ is orthogonally diagonalizable if there exists an orthogonal matrix $Q$ (that is, $Q^{T} Q = I$ ) such that $Q^{T} A Q$ is diagonal. Equivalently, $A$ possesses an orthonormal basis of eigenvectors.

Why does this matter? Orthogonal diagonalization preserves lengths and angles, so the eigenvector decomposition respects the geometry of the inner product. This is essential in applications ranging from principal component analysis and the classification of conic sections to the study of vibrations in mechanical systems.

The central question of this chapter is: which matrices are orthogonally diagonalizable? The answer, given by the spectral theorem, is remarkably clean: the real symmetric matrices are precisely the orthogonally diagonalizable matrices.

2. The Adjoint of a Linear Map

Definition 1 (Adjoint operator).

Let

V

be a finite-dimensional inner product space over

R

(or

C

), and let

T : V \to V

be a linear operator. The adjoint of

T

is the unique linear operator

T^{*} : V \to V

satisfying

⟨ T v, w ⟩ = ⟨ v, T^{*} w ⟩ for all v, w \in V .

Theorem 2 (Existence and uniqueness of the adjoint).

Let

V

be a finite-dimensional inner product space and

T : V \to V

a linear operator. Then

T^{*}

exists and is unique.

Proof.

Fix

w \in V

and consider the map

φ_{w} : V \to R

defined by

φ_{w} (v) = ⟨ T v, w ⟩

. This is a linear functional on

V

. By the Riesz representation theorem (which follows from the theory of orthogonal complements developed in Chapter 10), there exists a unique vector

w^{*} \in V

such that

φ_{w} (v) = ⟨ v, w^{*} ⟩

for all

v \in V

. We define

T^{*} (w) = w^{*}

To show that

T^{*}

is linear, let

w_{1}, w_{2} \in V

and

a, b \in R

. For every

v \in V

⟨ v, T^{*} (a w_{1} + b w_{2})⟩ = ⟨ T v, a w_{1} + b w_{2} ⟩ = a ⟨ T v, w_{1} ⟩ + b ⟨ T v, w_{2} ⟩ = a ⟨ v, T^{*} w_{1} ⟩ + b ⟨ v, T^{*} w_{2} ⟩ = ⟨ v, a T^{*} w_{1} + b T^{*} w_{2} ⟩ .

Since this holds for every

v

, the positive definiteness of the inner product gives

T^{*} (a w_{1} + b w_{2}) = a T^{*} w_{1} + b T^{*} w_{2}

Uniqueness follows immediately: if

S

also satisfies

⟨ T v, w ⟩ = ⟨ v, Sw ⟩

for all

v, w

, then

⟨ v, T^{*} w - Sw ⟩ = 0

for all

v

, which forces

T^{*} w = Sw

for every

w

. □

Example 3.

Let

V = R^{n}

with the standard inner product, and let

T

be the linear operator with matrix

A

(so

T v = A v

). Then

⟨ A v, w ⟩ = (A v)^{T} w = v^{T} A^{T} w = ⟨ v, A^{T} w ⟩

. Hence the adjoint of

T

has matrix

A^{T}

. In the complex case with the standard Hermitian inner product

⟨ v, w ⟩ = w^{*} v

, the adjoint has matrix

A^{*} = \overline{A}^{T}

(the conjugate transpose).

Theorem 4 (Properties of the adjoint).

Let

S, T : V \to V

be linear operators and

c

a scalar. Then:

$(S + T)^{*} = S^{*} + T^{*}$ .
$(c T)^{*} = \overset{c}{ˉ} T^{*}$ .
$(T^{*})^{*} = T$ .
$(ST)^{*} = T^{*} S^{*}$ .
$ker T^{*} = (im T)^{⊥}$ .

Proof.

We verify each identity by checking the defining property of the adjoint.

(1) For all

v, w

⟨(S + T) v, w ⟩ = ⟨ S v, w ⟩ + ⟨ T v, w ⟩ = ⟨ v, S^{*} w ⟩ + ⟨ v, T^{*} w ⟩ = ⟨ v, (S^{*} + T^{*}) w ⟩

(2)

⟨(c T) v, w ⟩ = c ⟨ T v, w ⟩ = c ⟨ v, T^{*} w ⟩ = ⟨ v, \overset{c}{ˉ} T^{*} w ⟩

(using conjugate-linearity in the second argument in the complex case; in the real case

\overset{c}{ˉ} = c

(3)

⟨ v, (T^{*})^{*} w ⟩ = ⟨ T^{*} v, w ⟩ = \overline{⟨ w, T^{*} v ⟩} = \overline{⟨ Tw, v ⟩} = ⟨ v, Tw ⟩

. Hence

(T^{*})^{*} = T

(4)

⟨(ST) v, w ⟩ = ⟨ S (T v), w ⟩ = ⟨ T v, S^{*} w ⟩ = ⟨ v, T^{*} (S^{*} w)⟩ = ⟨ v, (T^{*} S^{*}) w ⟩

(5) We have

w \in ker T^{*}

if and only if

T^{*} w = 0

, which holds if and only if

⟨ v, T^{*} w ⟩ = 0

for all

v \in V

, i.e.,

⟨ T v, w ⟩ = 0

for all

v

. This is precisely the condition

w \in (im T)^{⊥}

. □

3. Self-Adjoint (Symmetric) Operators

Definition 5 (Self-adjoint operator).

A linear operator

T

on an inner product space

V

is called self-adjoint (or symmetric) if

T^{*} = T

, that is,

⟨ T v, w ⟩ = ⟨ v, Tw ⟩ for all v, w \in V .

In terms of matrices with respect to the standard inner product,

T

is self-adjoint if and only if its matrix satisfies

A = A^{T}

(real case) or

A = A^{*}

(complex case, where such operators are called Hermitian).

Example 6.

The matrix

A = (2 - 1 - 1 3)

is real symmetric, hence the corresponding operator on

R^{2}

is self-adjoint.

Theorem 7 (Eigenvalues of self-adjoint operators are real).

Let

T

be a self-adjoint operator on a real or complex inner product space. Then every eigenvalue of

T

is real.

Proof.

Let

λ

be an eigenvalue with eigenvector

v \neq = 0

. Then

λ ⟨ v, v ⟩ = ⟨ λ v, v ⟩ = ⟨ T v, v ⟩ = ⟨ v, T v ⟩ = ⟨ v, λ v ⟩ = \overset{ˉ}{λ} ⟨ v, v ⟩ .

Since

⟨ v, v ⟩ > 0

, we conclude

λ = \overset{ˉ}{λ}

, so

λ \in R

. □

Theorem 8 (Orthogonality of eigenvectors).

Let

T

be a self-adjoint operator. If

v_{1}

and

v_{2}

are eigenvectors corresponding to distinct eigenvalues

λ_{1} \neq = λ_{2}

, then

v_{1} ⊥ v_{2}

Proof.

We compute:

λ_{1} ⟨ v_{1}, v_{2} ⟩ = ⟨ T v_{1}, v_{2} ⟩ = ⟨ v_{1}, T v_{2} ⟩ = λ_{2} ⟨ v_{1}, v_{2} ⟩,

where the last equality uses the fact that

λ_{2}

is real (by the previous theorem). Hence

(λ_{1} - λ_{2}) ⟨ v_{1}, v_{2} ⟩ = 0

. Since

λ_{1} \neq = λ_{2}

, we conclude

⟨ v_{1}, v_{2} ⟩ = 0

. □

Lemma 9 (Invariant subspaces and their complements).

Let

T

be a self-adjoint operator on an inner product space

V

, and let

W

be a

T

-invariant subspace (i.e.,

T (W) \subseteq W

). Then

W^{⊥}

is also

T

-invariant.

Proof.

Let

u \in W^{⊥}

. We must show

T u \in W^{⊥}

, i.e.,

⟨ T u, w ⟩ = 0

for all

w \in W

. Since

T

is self-adjoint and

W

T

-invariant, we have

Tw \in W

, and therefore

⟨ T u, w ⟩ = ⟨ u, Tw ⟩ = 0

because

u \in W^{⊥}

. □

4. The Real Spectral Theorem

We now arrive at the main result: every real symmetric matrix is orthogonally diagonalizable. We state this both in the language of operators and matrices.

Theorem 7 (Real spectral theorem).

Let

V

be a finite-dimensional real inner product space and

T : V \to V

a self-adjoint operator. Then

V

has an orthonormal basis consisting of eigenvectors of

T

. Equivalently, every real symmetric matrix

A

is orthogonally diagonalizable: there exists an orthogonal matrix

Q

such that

Q^{T} A Q = diag (λ_{1}, \dots, λ_{n})

Proof.

We proceed by induction on

n = dim V

Base case. If

n = 1

, every operator is multiplication by a scalar, and any unit vector is an orthonormal eigenbasis.

Inductive step. Assume the theorem holds for all real inner product spaces of dimension less than

n

. Let

T

be self-adjoint on the

n

-dimensional real inner product space

V

First, we must show that

T

has at least one real eigenvalue. Let

A

be the matrix of

T

with respect to any orthonormal basis. Then

A

is real symmetric. The characteristic polynomial

p (λ) = det (A - λ I)

has degree

n

with real coefficients. Viewing

A

as a complex matrix, its eigenvalues are all real by Theorem 7 (since

A = A^{T} = \overline{A}^{T} = A^{*}

, so

A

is Hermitian). In particular,

p (λ)

has a real root

λ_{1}

Let

v_{1}

be a unit eigenvector for

λ_{1}

, and set

W = span {v_{1}}

. Since

T (v_{1}) = λ_{1} v_{1} \in W

, the subspace

W

T

-invariant. By Lemma 9,

W^{⊥}

is also

T

-invariant.

Now

W^{⊥}

is an

(n - 1)

-dimensional inner product space (with the inner product inherited from

V

), and the restriction

T ∣_{W^{⊥}} : W^{⊥} \to W^{⊥}

is again self-adjoint: for

u, w \in W^{⊥}

⟨ T ∣_{W^{⊥}} u, w ⟩ = ⟨ T u, w ⟩ = ⟨ u, Tw ⟩ = ⟨ u, T ∣_{W^{⊥}} w ⟩ .

By the inductive hypothesis,

W^{⊥}

has an orthonormal basis

{v_{2}, \dots, v_{n}}

of eigenvectors of

T ∣_{W^{⊥}}

(hence of

T

Since

v_{1} ⊥ W^{⊥}

, the set

{v_{1}, v_{2}, \dots, v_{n}}

is an orthonormal basis of

V

consisting of eigenvectors of

T

. □

Remark 8.

The converse also holds: if a real matrix

A

is orthogonally diagonalizable, then

A

is symmetric. Indeed, if

Q^{T} A Q = D

with

Q

orthogonal and

D

diagonal, then

A = Q D Q^{T}

, so

A^{T} = (Q D Q^{T})^{T} = Q D^{T} Q^{T} = Q D Q^{T} = A

5. Normal Operators and the Complex Spectral Theorem

Over $C$ , the appropriate generalization of self-adjoint operators is the class of normal operators.

Definition 9 (Normal operator).

A linear operator

T

on an inner product space

V

is normal if it commutes with its adjoint:

T^{*} T = T T^{*} .

A complex matrix

A

is normal if

A^{*} A = A A^{*}

Example 10.

Self-adjoint operators are normal (since

T^{*} = T

implies

T^{*} T = T^{2} = T T^{*}

). Unitary operators (

T^{*} T = I

) are also normal. The rotation matrix

(cos θ sin θ - sin θ cos θ)

is a real normal matrix that is not symmetric (for

θ \in / {0, π}

), but as a complex matrix it is unitarily diagonalizable.

Lemma 14.

Let

T

be a normal operator. Then

∥ T v ∥ = ∥ T^{*} v ∥

for every

v \in V

Proof.

∥ T v ∥^{2} = ⟨ T v, T v ⟩ = ⟨ T^{*} T v, v ⟩ = ⟨ T T^{*} v, v ⟩ = ⟨ T^{*} v, T^{*} v ⟩ = ∥ T^{*} v ∥^{2}

. □

Lemma 15.

T

is normal and

T v = λ v

, then

T^{*} v = \overset{ˉ}{λ} v

Proof.

The operator

T - λ I

is also normal (one checks that

(T - λ I)^{*} = T^{*} - \overset{ˉ}{λ} I

and that these commute when

T

is normal). By Lemma 14 applied to

T - λ I

, we have

∥ (T - λ I) v ∥ = ∥ (T^{*} - \overset{ˉ}{λ} I) v ∥

. The left side is

0

since

T v = λ v

, so

T^{*} v = \overset{ˉ}{λ} v

. □

Theorem 11 (Complex spectral theorem).

Let

V

be a finite-dimensional complex inner product space and

T : V \to V

a normal operator. Then

V

has an orthonormal basis of eigenvectors of

T

. Equivalently, every normal matrix

A \in M_{n} (C)

is unitarily diagonalizable: there exists a unitary matrix

U

(with

U^{*} U = I

) such that

U^{*} A U = diag (λ_{1}, \dots, λ_{n})

Proof.

We again use induction on

n = dim V

Base case. For

n = 1

the result is trivial.

Inductive step. Since

C

is algebraically closed,

T

has at least one eigenvalue

λ_{1}

. Let

v_{1}

be a unit eigenvector, and set

W = span {v_{1}}

We claim

W^{⊥}

T

-invariant. Let

u \in W^{⊥}

. For any

w \in W

, write

w = c v_{1}

. By Lemma 15,

T^{*} v_{1} = \overset{ˉ}{λ}_{1} v_{1}

, so

⟨ T u, w ⟩ = ⟨ T u, c v_{1} ⟩ = \overset{c}{ˉ} ⟨ T u, v_{1} ⟩ = \overset{c}{ˉ} ⟨ u, T^{*} v_{1} ⟩ = \overset{c}{ˉ} ⟨ u, \overset{ˉ}{λ}_{1} v_{1} ⟩ = \overset{c}{ˉ} λ_{1} ⟨ u, v_{1} ⟩ = 0,

since

u \in W^{⊥}

. Hence

T u \in W^{⊥}

The restriction

T ∣_{W^{⊥}}

is again normal on the

(n - 1)

-dimensional inner product space

W^{⊥}

. By the inductive hypothesis,

W^{⊥}

has an orthonormal eigenbasis

{v_{2}, \dots, v_{n}}

, and

{v_{1}, v_{2}, \dots, v_{n}}

is the desired orthonormal eigenbasis of

V

. □

Remark 12.

Over

R

, a normal matrix need not be orthogonally diagonalizable (its eigenvalues may be non-real). For example, the rotation matrix above has eigenvalues

e^{\pm i θ} \in / R

when

θ \in / {0, π}

. The real spectral theorem handles the real case by restricting to symmetric (self-adjoint) operators, which guarantees real eigenvalues.

6. The Spectral Decomposition

The spectral theorem allows us to express any self-adjoint operator as a weighted sum of orthogonal projections onto its eigenspaces.

Definition 13 (Orthogonal projection onto an eigenspace).

Let

λ

be an eigenvalue of a self-adjoint operator

T

, and let

E_{λ} = ker (T - λ I)

be the corresponding eigenspace. The orthogonal projection

P_{λ} : V \to V

is the linear map that projects every vector onto

E_{λ}

along

E_{λ}^{⊥}

. In coordinates, if

{u_{1}, \dots, u_{k}}

is an orthonormal basis of

E_{λ}

, then

P_{λ} (v) = j = 1 \sum k ⟨ v, u_{j} ⟩ u_{j} .

Theorem 14 (Spectral decomposition).

Let

T

be a self-adjoint operator on a finite-dimensional real inner product space

V

with distinct eigenvalues

λ_{1}, \dots, λ_{s}

. Let

P_{i} = P_{λ_{i}}

be the orthogonal projection onto the eigenspace

E_{λ_{i}}

. Then:

$P_{i}^{2} = P_{i}$ and $P_{i}^{*} = P_{i}$ for each $i$ (each $P_{i}$ is an orthogonal projection).
$P_{i} P_{j} = 0$ for $i \neq = j$ (the projections are mutually orthogonal).
$P_{1} + P_{2} + \dots + P_{s} = I$ (resolution of the identity).
$T = λ_{1} P_{1} + λ_{2} P_{2} + \dots + λ_{s} P_{s}$ (spectral decomposition).

Proof.

By the spectral theorem,

V = E_{λ_{1}} \oplus E_{λ_{2}} \oplus \dots \oplus E_{λ_{s}}

, where the summands are mutually orthogonal. Every vector

v \in V

can be written uniquely as

v = v_{1} + v_{2} + \dots + v_{s}

with

v_{i} \in E_{λ_{i}}

(1)

P_{i}

sends

v

v_{i}

, so

P_{i}^{2} (v) = P_{i} (v_{i}) = v_{i} = P_{i} (v)

. For self-adjointness,

⟨ P_{i} v, w ⟩ = ⟨ v_{i}, w ⟩ = ⟨ v_{i}, w_{i} ⟩

(since

v_{i} ⊥ w_{j}

for

j \neq = i

), and

⟨ v, P_{i} w ⟩ = ⟨ v, w_{i} ⟩ = ⟨ v_{i}, w_{i} ⟩

by the same reasoning. Hence

P_{i}^{*} = P_{i}

(2)

P_{i} P_{j} (v) = P_{i} (v_{j})

. Since

v_{j} \in E_{λ_{j}}

and

j \neq = i

, the component of

v_{j}

E_{λ_{i}}

0

, so

P_{i} (v_{j}) = 0

(3)

(P_{1} + \dots + P_{s}) (v) = v_{1} + \dots + v_{s} = v

(4)

T v = T (v_{1} + \dots + v_{s}) = λ_{1} v_{1} + \dots + λ_{s} v_{s} = λ_{1} P_{1} v + \dots + λ_{s} P_{s} v

. □

Example 15.

Let

A = (10 0 - 1)

. The eigenvalues are

λ_{1} = 1

and

λ_{2} = - 1

with eigenspaces spanned by

e_{1}

and

e_{2}

, respectively. The orthogonal projections are

P_{1} = (1000), P_{2} = (0001),

and indeed

A = 1 \cdot P_{1} + (- 1) \cdot P_{2} = P_{1} - P_{2}

Remark 16.

The spectral decomposition

T = \sum λ_{i} P_{i}

makes it easy to compute functions of

T

. For any polynomial (or more generally any function defined on the spectrum),

f (T) = \sum f (λ_{i}) P_{i}

. For instance, if all eigenvalues are positive,

T^{1/2} = \sum λ_{i} P_{i}

7. Worked Example: A $3 \times 3$ Symmetric Matrix

We carry out the complete orthogonal diagonalization of the symmetric matrix

A = 211121112 .

Step 1: Find the eigenvalues. The characteristic polynomial is

det (A - λ I) = det

Adding all three rows to the first row gives $(4 - λ) (111)$ . Factoring out $(4 - λ)$ and subtracting the first row from the others, we obtain

det (A - λ I) = (4 - λ) (1 - λ)^{2} .

The eigenvalues are

λ_{1} = 4

(multiplicity

1

) and

λ_{2} = 1

(multiplicity

2

Step 2: Find eigenvectors. For $λ_{1} = 4$ : solving $(A - 4 I) v = 0$ gives $v_{1} = \frac{1}{3} 111$ .

For $λ_{2} = 1$ : solving $(A - I) v = 0$ yields the system $x_{1} + x_{2} + x_{3} = 0$ , a two-dimensional eigenspace. Two linearly independent solutions are $w_{1} = 1 - 1 0$ and $w_{2} = 10 - 1$ .

Step 3: Orthonormalize. The vector $v_{1}$ is already a unit vector and is automatically orthogonal to the $λ_{2}$ -eigenspace (by Theorem 8). We apply Gram–Schmidt to ${w_{1}, w_{2}}$ :

u_{1} = \frac{w _{1}}{∥ w _{1} ∥} = \frac{1}{2} 1 - 1 0 .

w_{2}^{'} = w_{2} - ⟨ w_{2}, u_{1} ⟩ u_{1} = 10 - 1 - \frac{1}{2} 1 - 1 0 = 1/2 1/2 - 1, u_{2} = \frac{w _{2}^{'}}{∥ w _{2}^{'} ∥} = \frac{1}{6} 11 - 2 .

Step 4: Form the orthogonal matrix.

Q = \frac{1}{3} \frac{1}{3} \frac{1}{3} \frac{1}{2} - \frac{1}{2} 0 \frac{1}{6} \frac{1}{6} - \frac{2}{6}, Q^{T} A Q = 400010001 .

The spectral decomposition is $A = 4 P_{1} + 1 \cdot P_{2}$ , where

P_{1} = v_{1} v_{1}^{T} = \frac{1}{3} 111111111, P_{2} = I - P_{1} = \frac{1}{3} 2 - 1 - 1 - 1 2 - 1 - 1 - 1 2 .

One can verify:

4 P_{1} + P_{2} = \frac{1}{3} 4 + 2 4 - 1 4 - 1 4 - 1 4 + 2 4 - 1 4 - 1 4 - 1 4 + 2 = 211121112 = A

8. Application to Quadratic Forms

A quadratic form on $R^{n}$ is a function $q : R^{n} \to R$ of the form

q (x) = x^{T} A x = i, j \sum a_{ij} x_{i} x_{j},

where

A

is a symmetric matrix (we may always assume symmetry by replacing

A

with

\frac{1}{2} (A + A^{T})

). The matrix

A

is called the matrix of the quadratic form.

Example 17.

The quadratic form

q (x_{1}, x_{2}) = 2 x_{1}^{2} + 6 x_{1} x_{2} + 5 x_{2}^{2}

has matrix

A = (2335)

. To classify this form, we orthogonally diagonalize

A

. The eigenvalues are

λ = \frac{7 \pm 3 5}{2}

, both positive, so

q

is positive definite.

Definition 18 (Classification of quadratic forms).

A quadratic form

q (x) = x^{T} A x

(equivalently, its symmetric matrix

A

) is called:

positive definite if $q (x) > 0$ for all $x \neq = 0$ ;
positive semidefinite if $q (x) \geq 0$ for all $x$ ;
negative definite if $q (x) < 0$ for all $x \neq = 0$ ;
negative semidefinite if $q (x) \leq 0$ for all $x$ ;
indefinite if $q$ takes both positive and negative values.

Theorem 19.

A real symmetric matrix

A

is positive definite if and only if all its eigenvalues are positive.

Proof.

By the spectral theorem, write

A = Q diag (λ_{1}, \dots, λ_{n}) Q^{T}

with

Q

orthogonal. Setting

y = Q^{T} x

, we have

q (x) = y^{T} diag (λ_{1}, \dots, λ_{n}) y = \sum_{i = 1}^{n} λ_{i} y_{i}^{2}

If all

λ_{i} > 0

, then

q (x) = \sum λ_{i} y_{i}^{2} > 0

whenever

y \neq = 0

(equivalently,

x \neq = 0

). Conversely, if some

λ_{k} \leq 0

, choose

x = Q e_{k}

(so

y = e_{k}

), giving

q (x) = λ_{k} \leq 0

. □

Theorem 20 (Sylvester's law of inertia).

Let

A

be a real symmetric

n \times n

matrix with

p

positive eigenvalues,

z

zero eigenvalues, and

q

negative eigenvalues (so

p + z + q = n

). If

B = M^{T} A M

for some invertible matrix

M

(not necessarily orthogonal), then

B

has the same numbers

p

z

, and

q

of positive, zero, and negative eigenvalues as

A

. The triple

(p, z, q)

is called the signature (or inertia) of

A

Proof.

By the spectral theorem, there exists an orthogonal

Q

such that

Q^{T} A Q = D = diag (λ_{1}, \dots, λ_{n})

, where

λ_{1}, \dots, λ_{p} > 0

λ_{p + 1} = \dots = λ_{p + z} = 0

, and

λ_{p + z + 1}, \dots, λ_{n} < 0

Suppose for contradiction that

B = M^{T} A M

has

p^{'}

positive eigenvalues with

p^{'} > p

(the case

p^{'} < p

is symmetric). Let

W_{+}

be the span of eigenvectors of

B

corresponding to positive eigenvalues, so

dim W_{+} = p^{'}

. Let

W_{0} = span {Q e_{p + 1}, \dots, Q e_{n}}

, the eigenspace of

A

corresponding to non-positive eigenvalues;

dim W_{0} = n - p

Consider the map

x \mapsto M x

from

W_{+}

into

R^{n}

. Since

p^{'} + (n - p) > n

, the subspaces

M (W_{+})

and

W_{0}

must have nontrivial intersection. Choose a nonzero

v \in W_{+}

with

M v \in W_{0}

. Then

v^{T} B v > 0

(since

v

lies in the positive-definite subspace of

B

), but

v^{T} B v = (M v)^{T} A (M v) \leq 0

(since

M v \in W_{0}

where

A

is negative semidefinite). This is a contradiction. □

Remark 21.

Sylvester's law tells us that the signature is a complete invariant for quadratic forms under change of basis. Two real quadratic forms are equivalent (related by an invertible linear substitution) if and only if they have the same signature.

Remark 22.

The spectral theorem stands at the crossroads of algebra and analysis. In quantum mechanics, self-adjoint operators represent physical observables, and the spectral theorem guarantees that every measurement yields a real eigenvalue. In statistics, the spectral decomposition of covariance matrices underlies principal component analysis. The infinite-dimensional generalization of the spectral theorem, due to von Neumann and Stone, is a cornerstone of functional analysis.

Linear Algebra Algebra Textbook Spectral Theorem Symmetric Matrices Orthogonal Diagonalization

Folio Official

Mathematics "between the lines" — exploring the intuition textbooks leave out, written in LaTeX on Folio.

1 followers·107 articles

The Spectral Theorem: Orthogonal Diagonalization of Symmetric Matrices

1. Motivation: From Diagonalization to Orthogonal Diagonalization

2. The Adjoint of a Linear Map

3. Self-Adjoint (Symmetric) Operators

4. The Real Spectral Theorem

5. Normal Operators and the Complex Spectral Theorem

6. The Spectral Decomposition

7. Worked Example: A $3 \times 3$ Symmetric Matrix

8. Application to Quadratic Forms

Share your expertise with the world

More from Folio Official

Matrices and Representation of Linear Maps

The Determinant: A Scalar Invariant of Square Matrices

Jordan Normal Form: Beyond Diagonalization

The Singular Value Decomposition: Structure of Arbitrary Matrices

1. Motivation: From Diagonalization to Orthogonal Diagonalization

2. The Adjoint of a Linear Map

3. Self-Adjoint (Symmetric) Operators

4. The Real Spectral Theorem

5. Normal Operators and the Complex Spectral Theorem

6. The Spectral Decomposition

7. Worked Example: A3×3Symmetric Matrix

8. Application to Quadratic Forms

Share your expertise with the world

More from Folio Official

Matrices and Representation of Linear Maps

The Determinant: A Scalar Invariant of Square Matrices

Jordan Normal Form: Beyond Diagonalization

The Singular Value Decomposition: Structure of Arbitrary Matrices

7. Worked Example: A $3 \times 3$ Symmetric Matrix