Linear-Algebra-Done-Right · Unit 30 · Video 2 · Interactive Practice

Defining V ⊗ W as Bilinear Functionals on V′ × W′

IKey Formulas

FormulaNameWhy
V⊗W=B(V′,W′)V \otimes W = \mathcal{B}(V', W')Tensor product of spacesBilinear functionals on the dual spaces
(v⊗w)(φ,τ)=φ(v) τ(w)(v \otimes w)(\varphi, \tau) = \varphi(v)\,\tau(w)Tensor product of vectorsFor all φ∈V′\varphi \in V', τ∈W′\tau \in W'; no basis used
dim⁡(V⊗W)=(dim⁡V′)(dim⁡W′)=(dim⁡V)(dim⁡W)\dim(V \otimes W) = (\dim V')(\dim W') = (\dim V)(\dim W)Dimensiondim⁡V′=dim⁡V\dim V' = \dim V, and dim⁡B(V′,W′)=(dim⁡V′)(dim⁡W′)\dim \mathcal{B}(V', W') = (\dim V')(\dim W')
(v1+v2)⊗w=v1⊗w+v2⊗w(v_1 + v_2) \otimes w = v_1 \otimes w + v_2 \otimes w,  λ(v⊗w)=(λv)⊗w=v⊗(λw)\ \lambda(v \otimes w) = (\lambda v) \otimes w = v \otimes (\lambda w)Bilinearity of ⊗\otimesEvaluate both sides at every (φ,τ)(\varphi, \tau)

Key Insight: v⊗wv \otimes w is not a pair: it is a function that takes a pair of functionals (φ,τ)(\varphi, \tau) and returns the number φ(v) τ(w)\varphi(v)\,\tau(w).

IIFeeding v ⊗ w a Pair of Functionals

With v=(2,3)v = (2, 3) and w=(1,0,4)w = (1, 0, 4), every pair (φ,τ)(\varphi, \tau) gets the value φ(v) τ(w)\varphi(v)\,\tau(w), wherever the scalar λ\lambda is placed.

IIIThe First Bilinearity Law, Line by Line

Two tensors are equal when they agree at every (φ,τ)(\varphi, \tau); here at φ(x)=x1+x2\varphi(x) = x_1 + x_2, τ(y)=y3\tau(y) = y_3.

v1=(1,2)v_1 = (1, 2),  v2=(1,1)\ v_2 = (1, 1),  w=(1,0,4)\ w = (1, 0, 4)
((v1+v2)⊗w)(φ,τ)=φ(v1+v2) τ(w)=φ(2,3)⋅4=5⋅4=20\big((v_1 + v_2) \otimes w\big)(\varphi, \tau) = \varphi(v_1 + v_2)\,\tau(w) = \varphi(2, 3) \cdot 4 = 5 \cdot 4 = 20

IVQuiz Questions

Problem 1 · Evaluate a Tensor

Given: v=(1,−2)∈F2v = (1, -2) \in \mathbf{F}^2, w=(3,0,1)∈F3w = (3, 0, 1) \in \mathbf{F}^3, φ(x)=3x1+x2\varphi(x) = 3x_1 + x_2 and τ(y)=y1−y3\tau(y) = y_1 - y_3. Find (v⊗w)(φ,τ)(v \otimes w)(\varphi, \tau).

✅ Correct! φ(v)=3−2=1\varphi(v) = 3 - 2 = 1 and τ(w)=3−1=2\tau(w) = 3 - 1 = 2, so the value is 1⋅2=21 \cdot 2 = 2.
❌ Check τ(w)\tau(w). τ(y)=y1−y3\tau(y) = y_1 - y_3, so τ(w)=3−1=2\tau(w) = 3 - 1 = 2, not 1−31 - 3.
❌ Close, but… the two numbers are multiplied, not added: (v⊗w)(φ,τ)=φ(v) τ(w)(v \otimes w)(\varphi, \tau) = \varphi(v)\,\tau(w).
❌ Not quite. Compute φ(v)=3⋅1+(−2)\varphi(v) = 3 \cdot 1 + (-2) and τ(w)\tau(w) separately, then multiply.
Show solution
φ(v)=3⋅1+(−2)=1,τ(w)=3−1=2,\varphi(v) = 3 \cdot 1 + (-2) = 1, \qquad \tau(w) = 3 - 1 = 2, (v⊗w)(φ,τ)=φ(v) τ(w)=1⋅2=2.(v \otimes w)(\varphi, \tau) = \varphi(v)\,\tau(w) = 1 \cdot 2 = 2.

Problem 2 · The Dimension of V ⊗ W

Given: dim⁡V=3\dim V = 3 and dim⁡W=5\dim W = 5. Find dim⁡(V⊗W)\dim(V \otimes W), where V⊗W=B(V′,W′)V \otimes W = \mathcal{B}(V', W').

✅ Correct! dim⁡V′=3\dim V' = 3 and dim⁡W′=5\dim W' = 5, so dim⁡B(V′,W′)=3⋅5=15\dim \mathcal{B}(V', W') = 3 \cdot 5 = 15.
❌ Not quite. 8=3+58 = 3 + 5 is dim⁡(V×W)\dim(V \times W); bilinear functionals multiply the dimensions.
❌ Not quite. Passing to the duals does not change dimensions: dim⁡V′=dim⁡V\dim V' = \dim V and dim⁡W′=dim⁡W\dim W' = \dim W.
❌ Not quite. Use dim⁡B(V′,W′)=(dim⁡V′)(dim⁡W′)\dim \mathcal{B}(V', W') = (\dim V')(\dim W').
Show solution

A space and its dual have the same dimension, so dim⁡V′=3\dim V' = 3 and dim⁡W′=5\dim W' = 5. By the dimension formula for bilinear functionals,

dim⁡(V⊗W)=dim⁡B(V′,W′)=(dim⁡V′)(dim⁡W′)=3⋅5=15.\dim(V \otimes W) = \dim \mathcal{B}(V', W') = (\dim V')(\dim W') = 3 \cdot 5 = 15.

Problem 3 · Moving Scalars

Given: v∈Vv \in V and w∈Ww \in W. Which element equals (2v)⊗(3w)(2v) \otimes (3w)?

✅ Correct! Each scalar comes out of its own slot: (2v)⊗(3w)=2⋅3 (v⊗w)(2v) \otimes (3w) = 2 \cdot 3\,(v \otimes w).
❌ Not quite. That element is 5(v⊗w)5(v \otimes w). Scalars in different slots multiply, because ⊗\otimes is linear in each slot separately.
❌ Not quite. (6v)⊗(6w)=36(v⊗w)(6v) \otimes (6w) = 36(v \otimes w): the whole factor 66 may move into one slot, not into both.
Show solution

Evaluate at any (φ,τ)∈V′×W′(\varphi, \tau) \in V' \times W':

((2v)⊗(3w))(φ,τ)=φ(2v) τ(3w)=2φ(v)⋅3τ(w)=6 (v⊗w)(φ,τ).\big((2v) \otimes (3w)\big)(\varphi, \tau) = \varphi(2v)\,\tau(3w) = 2\varphi(v) \cdot 3\tau(w) = 6\,(v \otimes w)(\varphi, \tau).

The two functions agree at every pair, so (2v)⊗(3w)=6(v⊗w)(2v) \otimes (3w) = 6(v \otimes w).

Problem 4 · A Tensor With a Zero Factor

Given: v∈Vv \in V and 0∈W0 \in W. What is v⊗0v \otimes 0?

✅ Correct! (v⊗0)(φ,τ)=φ(v) τ(0)=0(v \otimes 0)(\varphi, \tau) = \varphi(v)\,\tau(0) = 0 for every pair, so v⊗0v \otimes 0 is the zero functional.
❌ Not quite. (v,0)(v, 0) lives in V×WV \times W; v⊗0v \otimes 0 is an element of V⊗W=B(V′,W′)V \otimes W = \mathcal{B}(V', W'), a function on pairs of functionals.
❌ Not quite. τ(0)=0\tau(0) = 0 for every linear τ\tau, so the value is 00 whatever vv is.
❌ Not quite. Evaluate v⊗0v \otimes 0 at a pair (φ,τ)(\varphi, \tau).
Show solution

For every φ∈V′\varphi \in V' and τ∈W′\tau \in W', linearity gives τ(0)=0\tau(0) = 0, so

(v⊗0)(φ,τ)=φ(v) τ(0)=0.(v \otimes 0)(\varphi, \tau) = \varphi(v)\,\tau(0) = 0.

A function that is 00 at every input is the zero element of V⊗WV \otimes W. (The same follows from bilinearity: v⊗0=v⊗(0⋅0)=0 (v⊗0)=0v \otimes 0 = v \otimes (0 \cdot 0) = 0\,(v \otimes 0) = 0.)

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