Linear-Algebra-Done-Right · Unit 30 · Video 4 · Interactive Practice

The Universal Property: Bilinear Maps Become Linear

IKey Formulas

Formula	Name	Meaning
$\gamma(2v, 2w) = 4\,\gamma(v, w)$	Bilinear map $\gamma \colon V \times W \to U$	Linear in each slot, not on the pair
$\hat{\gamma}(v \otimes w) = \gamma(v, w)$	Universal property (part A)	A unique linear $\hat{\gamma} \colon V \otimes W \to U$ for each bilinear $\gamma$
$\hat{\gamma}(e_j \otimes f_k) := \gamma(e_j, f_k)$	Construction of $\hat{\gamma}$	Prescribe a linear map on the basis $e_j \otimes f_k$
$T^\sharp(v, w) = T(v \otimes w)$	Converse (part B)	Every linear $T$ on $V \otimes W$ comes from a bilinear map

Key Insight: $\gamma \mapsto \hat{\gamma}$ and $T \mapsto T^\sharp$ undo each other, so bilinear maps on $V \times W$ and linear maps on $V \otimes W$ correspond one to one.

IILinear in Each Slot Is Not Linear on the Pair

For $\gamma(v, w) = \langle v, w \rangle$ on $\mathbf{R}^2 \times \mathbf{R}^2$ with $v = (1, 1)$ , $w = (1, 2)$ : scaling one slot scales by $c$ , scaling both by $c^2$ .

c 2

IIIBuilding γ̂ From Its Values on a Basis

For $\gamma(v, w) = 2v_1 w_1 - v_1 w_2 + 3 v_2 w_2$ , the values $\gamma(e_j, f_k)$ define $\hat{\gamma}$ , and $\hat{\gamma}(v \otimes w) = \gamma(v, w)$ .

v₁ 1 v₂ 2 w₁ 1 w₂ 1

IVQuiz Questions

Problem 1 · Scalars in Both Slots

Given: a bilinear map $\gamma \colon V \times W \to U$ with $\gamma(v, w) = u$ . Find $\gamma(3v, -2w)$ .

✅ Correct! Each scalar comes out of its own slot:

\gamma(3v, -2w) = 3 \cdot (-2)\,\gamma(v, w)

❌ Not quite. The scalars multiply,

3 \cdot (-2)

; they do not add.

❌ Check the sign. The second factor is

-2

❌ Not quite. Use linearity in the first slot, then in the second.

Show solution

\gamma(3v, -2w) = 3\,\gamma(v, -2w) = 3 \cdot (-2)\,\gamma(v, w) = -6u.

Problem 2 · Not Bilinear

Given: $V$ and $W$ nonzero, and four maps. Which one is not bilinear?

✅ Correct! For fixed

T \neq 0

S \mapsto S + T

sends

0

T

, so it is not linear in

S

. (It is linear on the pair, a different condition.)

❌ That one is bilinear.

(S_1 + S_2)T = S_1 T + S_2 T

, and

S(T_1 + T_2) = ST_1 + ST_2

because

S

is linear.

❌ That one is bilinear.

T

is linear in

v

, and

(T_1 + T_2)v = T_1 v + T_2 v

by the definition of addition in

\mathcal{L}(V, W)

❌ That one is bilinear. It is the bilinearity of the tensor product of vectors.

Show solution

A bilinear map must be linear in each slot, so in particular it sends $(0, T)$ to $0$ . For $\gamma(S, T) = S + T$ , $\gamma(0, T) = T \neq 0$ when $T \neq 0$ , so $\gamma$ is not bilinear. Maps (a), (b), (c) are the composition, evaluation and tensor-product examples, each linear in each slot.

Problem 3 · Evaluate γ̂

Given: $\gamma \colon \mathbf{F}^2 \times \mathbf{F}^2 \to \mathbf{F}$ , $\gamma(v, w) = v_1 w_2 - v_2 w_1$ , with standard bases $e_1, e_2$ and $f_1, f_2$ . Find $\hat{\gamma}(e_1 \otimes f_2 + 3\, e_2 \otimes f_1)$ .

✅ Correct!

\hat{\gamma}

is linear, so the value is

\gamma(e_1, f_2) + 3\gamma(e_2, f_1) = 1 - 3 = -2

❌ Check $\gamma(e_2, f_1)$ . With

v = e_2

and

w = f_1

v_1 w_2 - v_2 w_1 = 0 - 1 = -1

❌ Not quite. Use linearity of

\hat{\gamma}

and

\hat{\gamma}(e_j \otimes f_k) = \gamma(e_j, f_k)

Show solution

\gamma(e_1, f_2) = 1 \cdot 1 - 0 \cdot 0 = 1, \qquad \gamma(e_2, f_1) = 0 \cdot 0 - 1 \cdot 1 = -1,

\hat{\gamma}(e_1 \otimes f_2 + 3\, e_2 \otimes f_1) = \gamma(e_1, f_2) + 3\,\gamma(e_2, f_1) = 1 - 3 = -2.

Problem 4 · From T Back to T♯

Given: the linear map $T \colon \mathbf{F}^2 \otimes \mathbf{F}^2 \to \mathbf{F}$ with $T(e_j \otimes f_k) = jk$ . Find $T^\sharp\big((1, 1),\ (1, -1)\big)$ .

✅ Correct!

T^\sharp(v, w) = T(v \otimes w) = 1 - 2 + 2 - 4 = -3

❌ Check the signs.

w_2 = -1

, so the terms with

k = 2

are negative.

❌ Not quite. Expand

v \otimes w

in the basis

e_j \otimes f_k

first: its coefficients are

v_j w_k

❌ Not quite.

T^\sharp(v, w) = \sum_{j,k} v_j w_k\, T(e_j \otimes f_k)

Show solution

With $v = (1, 1)$ and $w = (1, -1)$ , $v \otimes w = \sum_{j,k} v_j w_k\, e_j \otimes f_k$ has coefficients $1, -1, 1, -1$ for $(j, k) = (1,1), (1,2), (2,1), (2,2)$ . So

T^\sharp(v, w) = T(v \otimes w) = 1 \cdot 1 + (-1) \cdot 2 + 1 \cdot 2 + (-1) \cdot 4 = -3.

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