Linear-Algebra-Done-Right · Unit 27 · Video 1 · Interactive Practice

Bilinear Forms: Linear in Each Slot, a Matrix per Basis

IKey Formulas

Formula	Name	Hypotheses
$\beta(v + w, u) = \beta(v, u) + \beta(w, u)$ , $\ \beta(\lambda v, u) = \lambda\,\beta(v, u)$ , and the same in the second slot	Bilinear form on $V$	$\beta\colon V \times V \to \mathbf{F}$ ; $u, v, w \in V$ , $\lambda \in \mathbf{F}$
$\beta_A(x, y) = \sum_{k=1}^{n}\sum_{j=1}^{n} A_{j,k}\, x_j\, y_k$	The bilinear form of a matrix	$A \in \mathbf{F}^{n,n}$ , $x, y \in \mathbf{F}^n$
$\mathcal{M}(\beta)_{j,k} = \beta(e_j, e_k)$	Matrix of a bilinear form	$\beta \in V^{(2)}$ , the space of bilinear forms on $V$ (the $(2)$ counts slots, not a power); $e_1, \dots, e_n$ a basis of $V$
$\beta \mapsto \mathcal{M}(\beta)$ is an isomorphism $V^{(2)} \to \mathbf{F}^{n,n}$ , so $\dim V^{(2)} = (\dim V)^2$	The space of bilinear forms	$\dim V = n$ , with a basis $e_1, \dots, e_n$ of $V$ fixed

Key Insight: A bilinear form is pinned down by its $n^2$ values $\beta(e_j, e_k)$ on pairs of basis vectors, and every choice of those $n^2$ numbers comes from exactly one bilinear form.

IIFix One Slot, Get a Linear Functional

For $\beta(u, v) = u_1 v_1 + 2u_1 v_2 - u_2 v_1$ on $\mathbf{R}^2$ , fixing either vector leaves a linear functional of the other.

Fixed slot

IIIInner Products Ask for More

A real inner product is a bilinear form that is also symmetric and positive definite; a bilinear form need be neither.

β(e₁, e₁) 2 β(e₁, e₂) 1 β(e₂, e₁) −1 β(e₂, e₂) 1

IVThe Matrix of a Bilinear Form

Entry $(j, k)$ of $\mathcal{M}(\beta)$ is $\beta(e_j, e_k)$ : only a term containing $x_j y_k$ survives.

Form

VQuiz Questions

Problem 1 · Read Off the Matrix

Given: the bilinear form on $\mathbf{F}^3$ defined by $\beta(x, y) = 3x_1 y_3 - x_2 y_1 + 4x_2 y_2 - 2x_3 y_2$ — find $\mathcal{M}(\beta)$ with respect to the standard basis.

✅ Correct! Entry

(j, k)

\beta(e_j, e_k)

, the coefficient of

x_j y_k

: for instance

\beta(e_2, e_1) = -1

sits in row

2

, column

1

❌ Rows and columns are swapped. The row index belongs to the first slot:

\mathcal{M}(\beta)_{2,1} = \beta(e_2, e_1)

, the coefficient of

x_2 y_1

, which is

-1

. This matrix is

\mathcal{M}(\beta)^{\mathrm{t}}

, the matrix of

(x, y) \mapsto \beta(y, x)

❌ The signs belong to the entries.

\beta(e_2, e_1) = -1

and

\beta(e_3, e_2) = -2

, not

1

and

2

❌ Two slips. The entries are transposed (the row index belongs to the first slot) and the signs of

-x_2 y_1

and

-2x_3 y_2

were dropped.

Show solution

At $x = e_j$ , $y = e_k$ every coordinate is $0$ except $x_j = 1$ and $y_k = 1$ , so each term $c\,x_a y_b$ vanishes unless $(a, b) = (j, k)$ . Hence $\beta(e_j, e_k)$ is the coefficient of $x_j y_k$ :

\beta(e_1, e_3) = 3, \quad \beta(e_2, e_1) = -1, \quad \beta(e_2, e_2) = 4, \quad \beta(e_3, e_2) = -2,

and the other five values are $0$ . Placing $\beta(e_j, e_k)$ in row $j$ , column $k$ :

\mathcal{M}(\beta) = \begin{pmatrix} 0 & 0 & 3 \\ -1 & 4 & 0 \\ 0 & -2 & 0 \end{pmatrix}.

The matrix is not symmetric, matching $\beta(e_1, e_3) = 3 \neq 0 = \beta(e_3, e_1)$ .

Problem 2 · Which Is Bilinear?

Given: four functions $\beta\colon \mathcal{P}_2(\mathbf{R}) \times \mathcal{P}_2(\mathbf{R}) \to \mathbf{R}$ — find the one that is a bilinear form.

✅ Correct! It is

\varphi(p)\,\tau(q)

with the linear functionals

\varphi(p) = p'(0)

and

\tau(q) = \int_0^1 q

, and a product of a functional of

p

with a functional of

q

is bilinear.

❌ That map is linear on $V \times V$ , not bilinear. With

q

fixed,

p \mapsto p(1) + q(1)

is not linear: doubling

p

gives

2p(1) + q(1) \neq 2\bigl(p(1) + q(1)\bigr)

when

q(1) \neq 0

. Only the zero form is both.

❌ Linear in $p$ only. For fixed

q

it is evaluation of

p

at the number

q(0)

, which is linear; but with

p(x) = x^2

it is

q \mapsto q(0)^2

, which is not linear in

q

❌ The constant breaks linearity. Every bilinear form has

\beta(0, q) = 0

, but here

\beta(0, q) = 1

Show solution

(b) Fix $q$ . Then $c = \int_0^1 q(x)\,dx$ is a fixed number and $p \mapsto c\,p'(0)$ is linear, since differentiation and evaluation at $0$ are linear. Fix $p$ . Then $p'(0)$ is a fixed number and $q \mapsto p'(0)\int_0^1 q$ is linear, since integration is linear. So (b) is bilinear.

(a) fails: $\beta(2p, q) = 2p(1) + q(1)$ , while $2\beta(p, q) = 2p(1) + 2q(1)$ . (c) fails in the second slot: with $p(x) = x^2$ , $\beta(p, 2q) = 4q(0)^2 \neq 2q(0)^2 = 2\beta(p, q)$ whenever $q(0) \neq 0$ . (d) fails because $\beta(0, q) = 1 \neq 0$ .

Problem 3 · An Integral Form on $\mathcal{P}_2(\mathbf{R})$

Given: $\beta(p, q) = \displaystyle\int_{-1}^{1} p(x)\, q'(x)\,dx$ on $\mathcal{P}_2(\mathbf{R})$ and the basis $1, x, x^2$ — find $\mathcal{M}(\beta)$ .

✅ Correct! The first column is

0

because the constant

1

has derivative

0

, and entries such as

\int_{-1}^1 x\,dx

vanish because an odd function integrates to

0

[-1, 1]

❌ That is the transpose. Row

j

goes with the first slot, so entry

(1, 2)

\beta(1, x) = \int_{-1}^1 1 \cdot 1\,dx = 2

, and the first column (second slot

= 1

, derivative

0

) is the zero column.

❌ The derivative is missing. These are the integrals

\int_{-1}^1 e_j e_k

; the form differentiates its second argument, so entry

(j, k)

\int_{-1}^1 e_j\, e_k'

❌ Check the derivative of $x^2$ . It is

2x

, so

\beta(x, x^2) = \int_{-1}^1 x \cdot 2x\,dx = \tfrac43

, not

\tfrac23

Show solution

Write $e_1 = 1$ , $e_2 = x$ , $e_3 = x^2$ , so $e_1' = 0$ , $e_2' = 1$ , $e_3' = 2x$ . Entry $(j, k)$ is $\beta(e_j, e_k) = \int_{-1}^1 e_j\, e_k'\,dx$ .

Column $1$ ( $e_1' = 0$ ): all entries $0$ . Column $2$ ( $e_2' = 1$ ): $\int_{-1}^1 1\,dx = 2$ , $\ \int_{-1}^1 x\,dx = 0$ , $\ \int_{-1}^1 x^2\,dx = \tfrac23$ . Column $3$ ( $e_3' = 2x$ ): $\int_{-1}^1 2x\,dx = 0$ , $\ \int_{-1}^1 2x^2\,dx = \tfrac43$ , $\ \int_{-1}^1 2x^3\,dx = 0$ .

\mathcal{M}\bigl(\beta, (1, x, x^2)\bigr) = \begin{pmatrix} 0 & 2 & 0 \\ 0 & 0 & \tfrac43 \\ 0 & \tfrac23 & 0 \end{pmatrix}.

Problem 4 · Forms That Vanish on the Constant

Given: $V = \mathcal{P}_3(\mathbf{R})$ and the subspace $U = \{\beta \in V^{(2)} : \beta(p, 1) = 0 \text{ for every } p \in V\}$ — find $\dim U$ .

✅ Correct! In the basis

1, x, x^2, x^3

the condition says exactly that the first column of

\mathcal{M}(\beta)

is zero, which leaves

16 - 4 = 12

free entries.

❌ That is $\dim V^{(2)} = 4^2$ . The condition

\beta(p, 1) = 0

is a real restriction: it forces four entries of

\mathcal{M}(\beta)

to be

0

❌ It is more than one condition.

\beta(p, 1) = 0

must hold for every

p

, which by linearity in the first slot means

\beta(e_j, 1) = 0

for each of the four basis vectors

e_j

❌ Only the second slot is restricted. Nothing forces

\beta(1, q) = 0

, so the first row of

\mathcal{M}(\beta)

stays free; only the first column is zero.

Show solution

Use the basis $e_1 = 1$ , $e_2 = x$ , $e_3 = x^2$ , $e_4 = x^3$ , so $\dim V = 4$ and $\beta \mapsto \mathcal{M}(\beta)$ is an isomorphism $V^{(2)} \to \mathbf{R}^{4,4}$ .

For $p = a_1 e_1 + \cdots + a_4 e_4$ , linearity in the first slot gives $\beta(p, 1) = \sum_{j=1}^4 a_j\, \beta(e_j, e_1)$ . This is $0$ for every $p$ exactly when $\beta(e_j, e_1) = 0$ for $j = 1, \dots, 4$ , that is, when column $1$ of $\mathcal{M}(\beta)$ is zero.

So the isomorphism carries $U$ onto the matrices in $\mathbf{R}^{4,4}$ with zero first column, a space with $12$ free entries:

\dim U = 16 - 4 = 12.

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