Linear-Algebra-Done-Right · Unit 27 · Video 5 · Interactive Practice

Quadratic Forms: Polarization and Sums of Squares

IKey Formulas

Formula	Name	Hypotheses
$q_\beta(v) = \beta(v, v)$ ; on $\mathbf{F}^n$ : $q(x_1, \dots, x_n) = \sum_{j,k} A_{j,k}\, x_j x_k$	Quadratic form	$\beta$ bilinear on $V$ ; $A_{j,k}$ the entries of $\mathcal{M}(\beta)$ in the standard basis
$\rho(u, w) = \tfrac{1}{2}\big(q(u + w) - q(u) - q(w)\big)$	Polarization formula: the unique symmetric $\rho$ with $q = q_\rho$	$q$ a quadratic form on $V$
$q(2v) = 4q(v)$ and $(u, w) \mapsto q(u + w) - q(u) - q(w)$ symmetric bilinear	Test for a quadratic form	any function $q \colon V \to \mathbf{F}$
$q(x_1e_1 + \cdots + x_ne_n) = \lambda_1x_1^2 + \cdots + \lambda_nx_n^2$	Diagonalization: a weighted sum of squares	$V$ finite-dimensional; some basis $e_1, \dots, e_n$ ; orthonormal if $\mathbf{F} = \mathbf{R}$ and $V$ is an inner product space

Key Insight: Only the sums $A_{j,k} + A_{k,j}$ reach $q$ , so the symmetric form behind $q$ has matrix $\tfrac{1}{2}\big(A + A^{\mathsf{t}}\big)$ ; the alternating part of $\beta$ is invisible to $q$ .

IIOne Symmetric Matrix per Quadratic Form

Matrices with the same $A_{1,2} + A_{2,1}$ give the same $q$ ; their common symmetric part is $\mathcal{M}(\rho)$ .

A₁₁ 1 A₁₂ −4 A₂₁ 0 A₂₂ 0

IIIBoth Halves of the Test

A quadratic form needs $q(2v) = 4q(v)$ and a polarization $q(u + w) - q(u) - q(w)$ linear in $w$ .

Function

💡 Neither half suffices alone: $\lvert xy\rvert$ passes the scaling test, while $x^2 + y$ has the bilinear polarization $2u_1w_1$ but fails $q(2v) = 4q(v)$ .

IVA Basis with No Cross Term

For $q(x, y) = x^2 - 4xy$ the $c_1c_2$ coefficient is $2\rho(e_1, e_2)$ , so it vanishes exactly when $\rho(e_1, e_2) = 0$ .

💡 Challenge: find two diagonalizing bases whose weights differ; only the existence of some basis without cross terms is guaranteed, not the weights.

VQuiz Questions

Problem 1 · The Symmetric Matrix of a Quadratic Form

Given: the quadratic form $q(x, y, z) = 3x^2 + 2xy - 6yz + 5z^2$ on $\mathbf{R}^3$ — find $\mathcal{M}(\rho)$ in the standard basis, where $\rho$ is the unique symmetric bilinear form with $q = q_\rho$ .

✅ Correct! The squares go on the diagonal, and each cross coefficient is split evenly between its two mirror entries.

❌ This matrix gives $4xy$ and $-12yz$ . The entries

A_{1,2}

and

A_{2,1}

both contribute to the

xy

term, so each must be half of

2

❌ This $\beta$ does give $q_\beta = q$ , but it is not symmetric. Its

(1,2)

entry is

2

while its

(2,1)

entry is

0

; symmetrize with

\tfrac12\big(A + A^{\mathsf{t}}\big)

❌ The $-6$ belongs to $yz$ , not $xz$ . Entry

(j, k)

multiplies

x_j x_k

; the

yz

term lives in positions

(2,3)

and

(3,2)

Show solution

Write $q = \sum_{j,k} A_{j,k}\, x_jx_k$ with $A$ symmetric. A square $c\,x_j^2$ gives $A_{j,j} = c$ ; a cross term $c\,x_jx_k$ ( $j \ne k$ ) equals $A_{j,k}x_jx_k + A_{k,j}x_kx_j$ , so $A_{j,k} = A_{k,j} = c/2$ .

3x^2 \to A_{1,1} = 3, \qquad 2xy \to A_{1,2} = A_{2,1} = 1, \qquad -6yz \to A_{2,3} = A_{3,2} = -3, \qquad 5z^2 \to A_{3,3} = 5.

\mathcal{M}(\rho) = \begin{pmatrix} 3 & 1 & 0 \\ 1 & 0 & -3 \\ 0 & -3 & 5 \end{pmatrix}.

The missing $y^2$ and $xz$ terms give zeros in positions $(2,2)$ , $(1,3)$ and $(3,1)$ .

Problem 2 · Which One Is a Quadratic Form?

Given: four functions $\mathbf{R}^2 \to \mathbf{R}$ — find the one that is a quadratic form.

✅ Correct!

(2x - y)(x + 3y) = 2x^2 + 5xy - 3y^2

, a sum of degree-two terms.

❌ The scaling test fails.

q(0, 1) = 3

but

q(0, 2) = 6 \ne 4 \cdot 3

❌ It passes $q(2v) = 4q(v)$ , but the polarization is not linear. With

u = (1, 0)

w = (0, 1)

gives

q(1, 1) - 1 - 1 = -1

, and

-w

gives

q(1, -1) - 1 - 1 = -1

again, not

+1

❌ It passes $q(2v) = 4q(v)$ , but a quadratic form also has $q(-v) = q(v)$ . Here

q(1, 1) = 2

while

q(-1, -1) = -1 + 1 = 0

Show solution

(b) Expanding, $q(x, y) = 2x^2 + 6xy - xy - 3y^2 = 2x^2 + 5xy - 3y^2$ , which has the form $\sum_{j,k} A_{j,k}x_jx_k$ (for instance $A_{1,1} = 2$ , $A_{1,2} = A_{2,1} = \tfrac52$ , $A_{2,2} = -3$ ). So it is a quadratic form.

(a) $q(2v) = 4x^2 + 6y$ differs from $4q(v) = 4x^2 + 12y$ whenever $y \ne 0$ .

(c) $\max\{4x^2, 4y^2\} = 4\max\{x^2, y^2\}$ , so the scaling test holds. But for $u = (1, 0)$ the polarization $P(u, w) = q(u + w) - q(u) - q(w)$ takes the value $-1$ at both $w = (0, 1)$ and $w = (0, -1)$ ; a function linear in $w$ would change sign.

(d) $q(2v) = 2x \cdot 2\lvert y\rvert + 4y^2 = 4q(v)$ , so the test with $\lambda = 2$ alone cannot rule it out. A quadratic form satisfies $q(\lambda v) = \lambda^2 q(v)$ for every $\lambda$ , in particular $q(-v) = q(v)$ ; here $q(-1, -1) = 0 \ne 2 = q(1, 1)$ . (So its polarization cannot be bilinear either, since $q(2v) = 4q(v)$ plus a symmetric bilinear polarization would make $q$ a quadratic form.)

Problem 3 · Polarization on $\mathbf{R}^3$

Given: $q(x, y, z) = x^2 + 3y^2 - 2z^2 + 4xy - 6yz$ on $\mathbf{R}^3$ , $u = (1, 0, 2)$ and $w = (0, 1, 1)$ — find $q(u + w)$ and $\rho(u, w)$ , where $\rho$ is the symmetric bilinear form with $q = q_\rho$ .

$q(u + w)$

$\rho(u, w)$

✅ Correct!

\rho(u, w) = \tfrac12\big(-28 - (-7) - (-5)\big) = -8

❌ That is $q(u) + q(w) = -7 + (-5)$ . A quadratic form is not additive; the difference

q(u + w) - q(u) - q(w)

is exactly the polarization.

❌ The cross terms are missing. At

u + w = (1, 1, 3)

4xy = 4

and

-6yz = -18

also count.

❌ Check the signs.

u + w = (1, 1, 3)

, and

q(1, 1, 3) = 1 + 3 - 18 + 4 - 18

❌ That is the polarization, $2\rho(u, w)$ . The symmetric form is half of it.

❌ $0$ is $\beta(u, w)$ for the upper triangular $\beta$ with $q_\beta = q$ . That

\beta

is not symmetric:

\beta(w, u) = -16

. The symmetric

\rho

averages the two.

❌ Subtract the negative values with care.

-28 - (-7) - (-5) = -28 + 7 + 5 = -16

, and

\rho(u, w)

is half of that.

❌ Not quite. Use

\rho(u, w) = \tfrac12\big(q(u + w) - q(u) - q(w)\big)

Show solution

q(u) = q(1, 0, 2) = 1 + 0 - 8 + 0 - 0 = -7, \qquad q(w) = q(0, 1, 1) = 0 + 3 - 2 + 0 - 6 = -5,

q(u + w) = q(1, 1, 3) = 1 + 3 - 18 + 4 - 18 = -28.

By the polarization formula,

\rho(u, w) = \tfrac12\big(q(u + w) - q(u) - q(w)\big) = \tfrac12(-28 + 7 + 5) = -8.

Check with the matrix: $\mathcal{M}(\rho) = \begin{pmatrix} 1 & 2 & 0 \\ 2 & 3 & -3 \\ 0 & -3 & -2 \end{pmatrix}$ and $\mathcal{M}(\rho)\,w = (2, 0, -5)$ , so $\rho(u, w) = 1 \cdot 2 + 0 \cdot 0 + 2 \cdot (-5) = -8$ .

Problem 4 · A Weighted Sum of Squares on $\mathbf{R}^3$

Given: $q(x, y, z) = x^2 + 6xy + 2xz + 7y^2 + 10yz + 2z^2$ — complete the square in $x$ first, then in $y$ , and call the bracketed quantities $c_1, c_2, c_3$ (with $c_3 = z$ ). Find $q$ in these coordinates and the basis vector $e_2$ for which $(x, y, z) = c_1e_1 + c_2e_2 + c_3e_3$ .

$q(c_1e_1 + c_2e_2 + c_3e_3)$

$e_2$

✅ Correct! In the basis

(1, 0, 0)

(-3, 1, 0)

(-4, 1, 1)

the weights are

1

-2

3

❌ Those are the diagonal entries of $\mathcal{M}(\rho)$ . The standard basis leaves the cross terms

6xy

2xz

10yz

; the weights come from completing the square.

❌ The second square brings its own $z^2$ .

-2(y - z)^2 = -2y^2 + 4yz - 2z^2

, so

-2y^2 + 4yz + z^2 = -2(y - z)^2 + 3z^2

❌ Check the $y^2$ remainder.

(x + 3y + z)^2

already contains

9y^2

, so

7y^2 - 9y^2 = -2y^2

❌ Not quite. Start from

(x + 3y + z)^2 = x^2 + 6xy + 2xz + 9y^2 + 6yz + z^2

and subtract it from

q

❌ That lists the coefficients of $c_2 = y - z$ . The basis vectors come from solving back for

(x, y, z)

in terms of the

c_j

❌ Watch the sign in the solve-back.

x = c_1 - 3y - z

, so the coefficient of

c_2

x

-3

❌ That is $e_3$ . It is the coefficient vector of

c_3

;

e_2

is the coefficient vector of

c_2

❌ Not quite. Express

x

y

z

in terms of

c_1, c_2, c_3

and read off the coefficient of

c_2

Show solution

Square in $x$ . The $x$ -terms $x^2 + 6xy + 2xz$ start $(x + 3y + z)^2 = x^2 + 6xy + 2xz + 9y^2 + 6yz + z^2$ , so

q = (x + 3y + z)^2 - 2y^2 + 4yz + z^2.

Square in $y$ . $-2y^2 + 4yz = -2(y - z)^2 + 2z^2$ , hence

q = (x + 3y + z)^2 - 2(y - z)^2 + 3z^2.

With $c_1 = x + 3y + z$ , $c_2 = y - z$ , $c_3 = z$ this is $c_1^2 - 2c_2^2 + 3c_3^2$ .

Solve back. $z = c_3$ , $y = c_2 + c_3$ , $x = c_1 - 3y - z = c_1 - 3c_2 - 4c_3$ , so

(x, y, z) = c_1(1, 0, 0) + c_2(-3, 1, 0) + c_3(-4, 1, 1).

Thus $e_2 = (-3, 1, 0)$ . Check: $q(e_1) = 1$ , $q(e_2) = 9 - 18 + 7 = -2$ , $q(e_3) = 16 - 24 - 8 + 7 + 10 + 2 = 3$ .

Solved: 0 / 4