Linear-Algebra-Done-Right · Unit 27 · Video 3 · Interactive Practice

Symmetric Bilinear Forms Are Diagonal in Some Basis

IKey Formulas

Formula	Name	Hypotheses
$\rho(u, w) = \rho(w, u)$ for all $u, w \in V$ $\iff$ $\mathcal{M}(\rho, (e_1, \dots, e_n))$ is symmetric for some basis $\iff$ for every basis	Symmetric in every basis or in none	$\rho \in V^{(2)}$ , $V$ finite-dimensional
$\mathcal{M}(\rho, (e_1, \dots, e_n)) = C^{\mathrm{t}} B C$ , $\ (C^{\mathrm{t}} B C)^{\mathrm{t}} = C^{\mathrm{t}} B^{\mathrm{t}} C$	Why symmetry survives a change of basis	$B = \mathcal{M}(\rho, (f_1, \dots, f_n))$ , $C = \mathcal{M}(I, (e_1, \dots, e_n), (f_1, \dots, f_n))$
$2\rho(u, w) = \rho(u + w, u + w) - \rho(u, u) - \rho(w, w)$	Finding $v$ with $\rho(v, v) \neq 0$	$\rho$ symmetric, $\rho(u, w) \neq 0$
$U = \{u \in V : \rho(u, v) = 0\}$ , $\ \dim U = n - 1$ ; basis $e_1, \dots, e_{n-1}, v$	The induction step	$\rho$ symmetric, $\rho(v, v) \neq 0$ ; $e_1, \dots, e_{n-1}$ a basis of $U$ with $\rho(e_j, e_k) = 0$ for $j \neq k$

Key Insight: On a finite-dimensional real inner product space, let $\rho$ be symmetric and $B = \mathcal{M}(\rho, (f_1, \dots, f_n))$ for an orthonormal basis $f_1, \dots, f_n$ ; the operator $T$ with $\mathcal{M}(T, (f_1, \dots, f_n)) = B$ is then self-adjoint. The change-of-basis matrix $C$ between two orthonormal bases satisfies $C^{\mathrm{t}} = C^{-1}$ , so the orthonormal eigenbasis of $T$ that the real spectral theorem gives also diagonalizes $\rho$ , with the eigenvalues of $T$ on the diagonal.

IIOperator or Form: Which Matrix Stays Symmetric?

One symmetric matrix $B$ becomes $C^{-1}BC$ as an operator but $C^{\mathrm{t}}BC$ as a bilinear form.

Matrix B in the standard basis

IIIThe Induction Step: Which $v$ Works?

In $\mathbf{R}^2$ , appending $v$ to a basis of $U = \{u : \rho(u, v) = 0\}$ gives a basis of $\mathbf{R}^2$ that diagonalizes $\rho$ exactly when $\rho(v, v) \neq 0$ .

Form

💡 A nonzero symmetric form always has some $v$ with $\rho(v, v) \neq 0$ : if $\rho(u, w) \neq 0$ , one of $u + w$ , $u$ , $w$ is such a $v$ .

IVAn Orthonormal Basis That Diagonalizes $\rho$

Among orthonormal bases of $\mathbf{R}^2$ , the ones making $\mathcal{M}(\rho, (e_1, e_2))$ diagonal are eigenvector bases of $B$ .

Matrix B Angle θ of e₁ 20.0°

💡 Replacing a basis vector $v$ by $cv$ multiplies its diagonal entry by $c^2$ , since $\rho(cv, cv) = c^2 \rho(v, v)$ ; a non-orthonormal diagonalizing basis need not show the eigenvalues.

VQuiz Questions

Problem 1 · Symmetric in Every Basis or in None

Given: the symmetric bilinear form $\rho$ on $\mathbf{R}^3$ with $\mathcal{M}(\rho, (e_1, e_2, e_3)) = B = \begin{pmatrix} 1 & 0 & 2 \\ 0 & -1 & 1 \\ 2 & 1 & 0 \end{pmatrix}$ and the basis $f_1 = (1, 1, 0)$ , $f_2 = (0, 1, 1)$ , $f_3 = (1, 0, 1)$ — find $\mathcal{M}(\rho, (f_1, f_2, f_3))$ , and decide whether a bilinear form $\sigma$ on $\mathbf{R}^3$ with $\sigma(e_1, e_2) = 1$ and $\sigma(e_2, e_1) = -1$ has a diagonal matrix with respect to some basis.

$\mathcal{M}(\rho, (f_1, f_2, f_3))$

Is $\mathcal{M}(\sigma, \cdot)$ diagonal with respect to some basis of $\mathbf{R}^3$ ?

✅ Correct! The entries

\rho(f_j, f_k)

form a symmetric matrix because

\rho

is symmetric, and a form whose matrix is not symmetric in one basis is symmetric in no basis, so it is diagonal in none.

❌ That is the operator formula $C^{-1}BC$ . A bilinear form changes by

C^{\mathrm{t}}BC

; the result here is not even symmetric, which already rules it out, since

\rho

is symmetric.

❌ That is only $BC$ . Its

(j, k)

entry is a coordinate of

Bf_k

, not

\rho(f_j, f_k)

; multiply by

C^{\mathrm{t}}

on the left. A matrix of

\rho

must be symmetric, and this one is not.

❌ The transposes are on the wrong side. This is

CBC^{\mathrm{t}}

; it is symmetric, but compare its corner entry

5

with

\rho(f_1, f_1) = f_1^{\mathrm{t}} B f_1

❌ Not quite. The

(j, k)

entry is

\rho(f_j, f_k) = f_j^{\mathrm{t}} B f_k

; start with

\rho(f_1, f_1)

❌ The induction uses symmetry. The identity

2\rho(u, w) = \rho(u + w, u + w) - \rho(u, u) - \rho(w, w)

and the zero last row both need

\rho(u, w) = \rho(w, u)

❌ That operator is not self-adjoint. Its standard matrix is not symmetric, so the real spectral theorem gives no orthonormal eigenbasis, and no basis could make

\mathcal{M}(\sigma, \cdot)

diagonal anyway.

❌ A vector with $\sigma(v, v) \neq 0$ is not enough. Diagonal implies symmetric, and

\sigma

fails symmetry at

(e_1, e_2)

whatever

\sigma(v, v)

is.

❌ Not quite. Ask what a diagonal

\mathcal{M}(\sigma, \cdot)

would say about

\sigma

itself, then use the theorem's equivalences.

Show solution

Let $C$ have columns $f_1, f_2, f_3$ , so $C = \mathcal{M}(I, (f_1, f_2, f_3), (e_1, e_2, e_3))$ and the change-of-basis formula gives $\mathcal{M}(\rho, (f_1, f_2, f_3)) = C^{\mathrm{t}}BC$ :

BC = \begin{pmatrix} 1 & 2 & 3 \\ -1 & 0 & 1 \\ 3 & 1 & 2 \end{pmatrix}, \qquad C^{\mathrm{t}}BC = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ -1 & 0 & 1 \\ 3 & 1 & 2 \end{pmatrix} = \begin{pmatrix} 0 & 2 & 4 \\ 2 & 1 & 3 \\ 4 & 3 & 5 \end{pmatrix}.

Check one entry directly: $\rho(f_1, f_3) = f_1^{\mathrm{t}} (Bf_3) = (1, 1, 0) \cdot (3, 1, 2) = 4$ . The result is symmetric, as it must be: $\rho$ is symmetric, so its matrix is symmetric with respect to every basis.

For $\sigma$ : if $\mathcal{M}(\sigma, (g_1, g_2, g_3))$ were diagonal for some basis, it would be symmetric for that basis, so $\sigma$ would be symmetric, and then $\mathcal{M}(\sigma, (e_1, e_2, e_3))$ would be symmetric too. Its $(1, 2)$ and $(2, 1)$ entries are $1$ and $-1$ , so no such basis exists.

Problem 2 · Finding $v$ When the Diagonal Is Zero

Given: the symmetric bilinear form $\rho(x, y) = x_1y_2 + x_2y_1 + 2x_2y_3 + 2x_3y_2$ on $\mathbf{R}^3$ , for which $\rho(e_1, e_1) = \rho(e_2, e_2) = \rho(e_3, e_3) = 0$ — find a vector $v$ with $\rho(v, v) \neq 0$ to start the induction, and the value $\rho(v, v)$ .

Which vector has $\rho(v, v) \neq 0$ ?

The value $\rho(v, v)$ for that vector

✅ Correct!

\rho(e_2, e_3) = 2 \neq 0

, so the identity guarantees that one of

e_2 + e_3

e_2

e_3

works; the last two give

0

, so

v = e_2 + e_3

with

\rho(v, v) = 2\rho(e_2, e_3) = 4

❌ Every standard basis vector gives $0$ here. The stem says so:

\rho(e_j, e_j) = 0

for each

j

❌ The identity needs $\rho(u, w) \neq 0$ . Here

\rho(e_1, e_3) = 0

, so

\rho(e_1 + e_3, e_1 + e_3) = 0 + 2 \cdot 0 + 0 = 0

; pair

e_2

with

e_3

instead.

❌ Not quite. Look for a pair

u, w

with

\rho(u, w) \neq 0

and use

2\rho(u, w) = \rho(u + w, u + w) - \rho(u, u) - \rho(w, w)

❌ Both middle terms count.

\rho(v, v) = \rho(e_2, e_2) + \rho(e_2, e_3) + \rho(e_3, e_2) + \rho(e_3, e_3)

, and the two middle terms are equal.

❌ Each cross term appears once. The coefficient

2

2x_2y_3

is already

\rho(e_2, e_3)

; it is not doubled again.

❌ Not quite. Expand

\rho(e_2 + e_3, e_2 + e_3)

by bilinearity into four terms.

Show solution

The quadratic expression is $\rho(x, x) = 2x_1x_2 + 4x_2x_3$ , which vanishes at $e_1$ , $e_2$ , $e_3$ and at $e_1 + e_3 = (1, 0, 1)$ . Since $\rho(e_2, e_3) = 2 \neq 0$ , the identity

2\rho(e_2, e_3) = \rho(e_2 + e_3, e_2 + e_3) - \rho(e_2, e_2) - \rho(e_3, e_3)

gives $\rho(e_2 + e_3, e_2 + e_3) = 4 + 0 + 0 = 4 \neq 0$ . Directly, $x = (0, 1, 1)$ gives $2 \cdot 0 \cdot 1 + 4 \cdot 1 \cdot 1 = 4$ . The pair $e_1, e_3$ fails because $\rho(e_1, e_3) = 0$ : the identity only promises a good vector among $u + w$ , $u$ , $w$ when $\rho(u, w) \neq 0$ .

Problem 3 · Run the Induction on $\mathbf{R}^3$

Given: $\rho(x, y) = x^{\mathrm{t}} B y$ on $\mathbf{R}^3$ with $B = \begin{pmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & -1 \end{pmatrix}$ . Take $v = e_1$ , so $\rho(v, v) = 1$ ; inside $U = \{u : \rho(u, v) = 0\}$ take $v' = (1, -1, 0)$ , and let $u'$ span $\{x \in U : \rho(x, v') = 0\}$ — find $U$ and the matrix of $\rho$ with respect to the basis $u', v', v$ .

$U = \{u \in \mathbf{R}^3 : \rho(u, e_1) = 0\}$

$\mathcal{M}(\rho, (u', v', v))$

✅ Correct!

U = \{x_1 + x_2 = 0\}

, the second step gives

u' = (1, -1, 1)

with

\rho(u', u') = -2

, and the basis

(1, -1, 1)

(1, -1, 0)

(1, 0, 0)

makes

\rho

diagonal with entries

-2

1

1

❌ That is the orthogonal complement of $e_1$ for the dot product.

U

is defined by

\rho

\rho(x, e_1) = x^{\mathrm{t}} B e_1 = x_1 + x_2

❌ That is $\rho(x, e_2) = 0$ .

\rho(x, e_1)

uses the first column of

B

(1, 1, 0)

❌ $U$ has dimension $n - 1 = 2$ . It is the null space of a nonzero functional on

\mathbf{R}^3

, a plane, not a line.

❌ Not quite. Compute the functional

x \mapsto \rho(x, e_1) = x^{\mathrm{t}} (Be_1)

and take its null space.

❌ $e_3$ is in $U$ , but it is not $\rho$ -orthogonal to $v'$ .

\rho(e_3, v') = -1 \neq 0

, so the basis

e_3, v', v

gives a matrix with nonzero off-diagonal entries; the second step must produce

u'

❌ Check the sign of the cross term. With

a = (1, -1, 0)

and

b = e_3

u' = a + b

and

\rho(u', u') = \rho(a, a) + 2\rho(a, b) + \rho(b, b)

, where

\rho(a, b) = -1

❌ The induction does not produce eigenvalues. Only an orthonormal diagonalizing basis is forced to show them, and

u', v', v

is not orthonormal; compute

\rho(u', u')

\rho(v', v')

and

\rho(v, v)

directly.

❌ Not quite. Find

u'

from

u' \in U

and

\rho(u', v') = 0

, then compute

\rho(u', u')

\rho(v', v')

and

\rho(v, v)

Show solution

First step. $Be_1 = (1, 1, 0)$ , so $\rho(x, e_1) = x_1 + x_2$ and $U = \{x : x_1 + x_2 = 0\} = \operatorname{span}\bigl((1, -1, 0), (0, 0, 1)\bigr)$ , of dimension $2$ .

Second step, inside $U$ . $v' = (1, -1, 0) \in U$ has $\rho(v', v') = 1 - 1 - 1 + 2 = 1 \neq 0$ . Since $Bv' = (0, -1, -1)$ , $\rho(x, v') = -x_2 - x_3$ . A vector of $U$ is $x = (s, -s, t)$ , and $\rho(x, v') = s - t = 0$ gives $u' = (1, -1, 1)$ .

\rho(u', u') = u'^{\mathrm{t}} B u' = (1, -1, 1) \cdot (0, 0, -2) = -2, \qquad \rho(v, v) = 1.

The off-diagonal entries vanish: $\rho(u', v') = 0$ by the choice of $u'$ , and $\rho(u', v) = \rho(v', v) = 0$ because $u', v' \in U$ . So

\mathcal{M}\bigl(\rho, (u', v', v)\bigr) = \begin{pmatrix} -2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}.

Problem 4 · An Orthonormal Diagonalizing Basis

Given: $\mathbf{R}^3$ with the usual inner product and $\rho(x, y) = x^{\mathrm{t}} B y$ , $B = \begin{pmatrix} 2 & 0 & 2 \\ 0 & 3 & 0 \\ 2 & 0 & -1 \end{pmatrix}$ — find the diagonal entries of $\mathcal{M}(\rho, (q_1, q_2, q_3))$ for an orthonormal basis $q_1, q_2, q_3$ that diagonalizes $\rho$ , and one such basis.

The diagonal entries, in some order

An orthonormal basis that diagonalizes $\rho$

✅ Correct!

B

is symmetric, so the operator

T

with matrix

B

is self-adjoint; its orthonormal eigenbasis

\tfrac{1}{\sqrt{5}}(2, 0, 1)

e_2

\tfrac{1}{\sqrt{5}}(1, 0, -2)

has

C^{\mathrm{t}} = C^{-1}

, so

\mathcal{M}(\rho, \cdot) = C^{-1}BC

is diagonal with the eigenvalues

3

3

-2

❌ Those come from the unnormalized eigenvectors.

(2, 0, 1)

and

(1, 0, -2)

have length

\sqrt{5}

, and

\rho(cv, cv) = c^2 \rho(v, v)

multiplies the entries by

5

❌ Those are the diagonal entries of $B$ itself. They have the right sum,

\operatorname{tr} B = 4

, but

B

is not diagonal:

\rho(e_1, e_3) = 2

❌ Rescaling multiplies an entry by $c^2$ , not $c$ . And an orthonormal basis has unit vectors, so no rescaling factor appears at all.

❌ Not quite. With an orthonormal diagonalizing basis, the diagonal entries are the eigenvalues of the operator with matrix

B

❌ This basis diagonalizes $\rho$ but is not orthonormal. Each of

(2, 0, 1)

and

(1, 0, -2)

has length

\sqrt{5}

, giving diagonal entries

15

and

-10

❌ Orthonormal, but not eigenvectors.

B(2, 0, -1) = (2, 0, 5)

, so

\rho\bigl((1, 0, 2), (2, 0, -1)\bigr) = 2 + 10 = 12 \neq 0

and the matrix has off-diagonal entries

\tfrac{12}{5}

❌ The standard basis gives $B$ itself. It is orthonormal, but

B

has off-diagonal entries

2

❌ Not quite. You need unit vectors, pairwise orthogonal, each an eigenvector of

B

Show solution

$e_2$ is an eigenvector with eigenvalue $3$ . On $\operatorname{span}(e_1, e_3)$ , $B$ acts by $\begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}$ :

B(2, 0, 1) = (6, 0, 3) = 3(2, 0, 1), \qquad B(1, 0, -2) = (-2, 0, 4) = -2(1, 0, -2).

These three eigenvectors are pairwise orthogonal; dividing $(2, 0, 1)$ and $(1, 0, -2)$ by $\sqrt{5}$ gives an orthonormal basis $q_1, q_2, q_3$ . With $C$ the matrix whose columns are $q_1, q_2, q_3$ , $C^{\mathrm{t}} = C^{-1}$ , so

\mathcal{M}\bigl(\rho, (q_1, q_2, q_3)\bigr) = C^{\mathrm{t}}BC = C^{-1}BC = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -2 \end{pmatrix}.

Check: $\rho(q_1, q_1) = \tfrac{1}{5}(2, 0, 1) \cdot (6, 0, 3) = 3$ , and the entries add up to $\operatorname{tr} B = 4$ . Without normalizing, the same directions give $15$ , $3$ , $-10$ .

Solved: 0 / 4

Symmetric Bilinear Forms Are Diagonal in Some Basis

IKey Formulas

IIOperator or Form: Which Matrix Stays Symmetric?

IIIThe Induction Step: Which vvv Works?

IVAn Orthonormal Basis That Diagonalizes ρ\rhoρ

VQuiz Questions

IIIThe Induction Step: Which $v$ Works?

IVAn Orthonormal Basis That Diagonalizes $\rho$