Linear-Algebra-Done-Right · Unit 30 · Video 6 · Interactive Practice

Tensor Products of m Spaces and m-Linear Maps

IKey Formulas

Formula	Name	Meaning
$\dim \mathcal{B}(V_1, \dots, V_m) = (\dim V_1) \cdots (\dim V_m)$	$m$ -linear functionals	Linear in each slot with the others fixed
$V_1 \otimes \cdots \otimes V_m = \mathcal{B}(V_1', \dots, V_m')$	Tensor product of $m$ spaces	$m$ -linear functionals on the duals
$(v_1 \otimes \cdots \otimes v_m)(\varphi_1, \dots, \varphi_m) = \varphi_1(v_1) \cdots \varphi_m(v_m)$	Tensor product of $m$ vectors	With $m = 2$ , the definition for $V \otimes W$
$e^1_{j_1} \otimes \cdots \otimes e^m_{j_m}$ , $\ 1 \le j_k \le n_k$	Basis	$n_1 n_2 \cdots n_m$ tensors; coefficients form an array with $m$ indices

Key Insight: An $m$ -linear map scales by $c^m$ when every slot is scaled by $c$ , and every $m$ -linear map on $V_1 \times \cdots \times V_m$ becomes a single linear map on $V_1 \otimes \cdots \otimes V_m$ .

IIA Product of Dimensions, Not a Sum

A basis tensor picks one basis vector from each space, so the coefficients of an element fill an $n_1 \times n_2 \times n_3$ block.

n₁ 2 n₂ 3 n₃ 4

IIIThe Array of a Simple Tensor u ⊗ v ⊗ w

With $u = (1, 2)$ and $v = (3, -1, 4)$ , the entry at $(j_1, j_2, j_3)$ of $u \otimes v \otimes w$ is $u_{j_1} v_{j_2} w_{j_3}$ .

w₁ 1 w₂ −2

IVQuiz Questions

Problem 1 · Dimension of a Triple Tensor Product

Given: $\dim V_1 = 2$ , $\dim V_2 = 2$ , $\dim V_3 = 3$ . Find $\dim(V_1 \otimes V_2 \otimes V_3)$ .

✅ Correct! One basis tensor for each choice of indices:

2 \cdot 2 \cdot 3 = 12

❌ Not quite.

7 = 2 + 2 + 3

\dim(V_1 \times V_2 \times V_3)

❌ Not quite. Multiply the dimensions:

n_1 n_2 n_3

Show solution

The basis tensors $e^1_{j_1} \otimes e^2_{j_2} \otimes e^3_{j_3}$ have $j_1 \in \{1, 2\}$ , $j_2 \in \{1, 2\}$ , $j_3 \in \{1, 2, 3\}$ , so there are

\dim(V_1 \otimes V_2 \otimes V_3) = 2 \cdot 2 \cdot 3 = 12.

Problem 2 · Scaling Every Slot

Given: a $3$ -linear functional $\beta$ with $\beta(v_1, v_2, v_3) = 3$ . Find $\beta(2v_1, 2v_2, 2v_3)$ .

✅ Correct! Each of the three slots gives a factor

2

2^3 \cdot 3 = 24

❌ Not quite.

6 = 2 \cdot 3

is what a linear map on the product would give;

\beta

is linear in each slot separately.

❌ Close, but… that uses two slots; all three slots are scaled.

❌ Not quite. Pull the

2

out of each slot in turn.

Show solution

\beta(2v_1, 2v_2, 2v_3) = 2\,\beta(v_1, 2v_2, 2v_3) = 2^2\,\beta(v_1, v_2, 2v_3) = 2^3\,\beta(v_1, v_2, v_3) = 8 \cdot 3 = 24.

Problem 3 · Evaluate a Tensor of Three Vectors

Given: $v_1 = (1, 2)$ , $v_2 = (0, 3)$ , $v_3 = (2, 2)$ in $\mathbf{F}^2$ , and $\varphi_1(x) = x_1 + x_2$ , $\varphi_2(x) = x_2$ , $\varphi_3(x) = x_1$ . Find $(v_1 \otimes v_2 \otimes v_3)(\varphi_1, \varphi_2, \varphi_3)$ .

✅ Correct!

\varphi_1(v_1)\,\varphi_2(v_2)\,\varphi_3(v_3) = 3 \cdot 3 \cdot 2 = 18

❌ Not quite.

8 = 3 + 3 + 2

: the three values multiply.

❌ Not quite. Compute

\varphi_1(v_1)

\varphi_2(v_2)

and

\varphi_3(v_3)

separately, then multiply.

Show solution

\varphi_1(v_1) = 1 + 2 = 3, \qquad \varphi_2(v_2) = 3, \qquad \varphi_3(v_3) = 2,

(v_1 \otimes v_2 \otimes v_3)(\varphi_1, \varphi_2, \varphi_3) = 3 \cdot 3 \cdot 2 = 18.

Problem 4 · One Entry of the Array

Given: $v_1 = (1, -2) \in \mathbf{F}^2$ , $v_2 = (3, 1) \in \mathbf{F}^2$ , $v_3 = (1, 0, 5) \in \mathbf{F}^3$ , with standard bases $e^1_j$ , $e^2_j$ , $e^3_j$ . Find the coefficient of $e^1_2 \otimes e^2_1 \otimes e^3_3$ in $v_1 \otimes v_2 \otimes v_3$ .

✅ Correct! The coefficient is

(v_1)_2\,(v_2)_1\,(v_3)_3 = (-2) \cdot 3 \cdot 5 = -30

❌ Check the sign. The second entry of

v_1

-2

❌ Check the indices. The lower index of

e^2_1

picks the first entry of

v_2

, which is

3

❌ Not quite. Multiply the

j_1

-th entry of

v_1

, the

j_2

-th of

v_2

and the

j_3

-th of

v_3

Show solution

Expanding each vector in its basis and using linearity in each slot,

v_1 \otimes v_2 \otimes v_3 = \sum_{j_1, j_2, j_3} (v_1)_{j_1} (v_2)_{j_2} (v_3)_{j_3}\; e^1_{j_1} \otimes e^2_{j_2} \otimes e^3_{j_3},

so the coefficient at $(j_1, j_2, j_3) = (2, 1, 3)$ is $(-2) \cdot 3 \cdot 5 = -30$ .

Solved: 0 / 4