Linear-Algebra-Done-Right · Unit 30 · Video 3 · Interactive Practice

A Basis of V ⊗ W, and v ⊗ w as a Grid of Products

IKey Formulas

Formula	Name	Why
$\{e_j \otimes f_k\}$ independent (resp. a basis) when both lists are	Basis of $V \otimes W$	Part A: evaluate at $(\varphi_M, \tau_N)$ ; part B: length $mn = \dim(V \otimes W)$
$(e_j \otimes f_k)(\varphi_M, \tau_N) = \begin{cases} 1 & j = M,\ k = N \\ 0 & \text{otherwise} \end{cases}$	Picking out one coefficient	$\varphi_M(e_j)\,\tau_N(f_k)$ with dual-basis functionals
$v \otimes w = \displaystyle\sum_{j=1}^{m} \sum_{k=1}^{n} v_j w_k\, e_j \otimes f_k$	Coordinates in $\mathbf{F}^m \otimes \mathbf{F}^n$	Bilinearity; the grid has $v_j w_k$ in row $j$ , column $k$
$e_1 \otimes f_1 + e_2 \otimes f_2 \neq v \otimes w$	Not every tensor is simple	Identity grid: its rows are not multiples of one row

Key Insight: Row $j$ of the grid of $v \otimes w$ is $v_j$ times the row $w$ , so a simple tensor's grid has all its rows proportional.

IIv ⊗ w as a Grid of Products

The coefficient of $e_j \otimes f_k$ in $v \otimes w$ is $v_j w_k$ , the value of $v \otimes w$ at $(\varphi_j, \tau_k)$ .

v₁ 1 v₂ 2 w₁ 3 w₂ −1 w₃ 4

IIISimple or Not?

A $2 \times 2$ coefficient grid comes from a single $v \otimes w$ exactly when its rows are proportional.

💡 Challenge: starting from the identity grid, change one entry to make the element a simple tensor.

IVQuiz Questions

Problem 1 · One Coefficient

Given: $v = (2, -1) \in \mathbf{F}^2$ and $w = (1, 3, 0) \in \mathbf{F}^3$ , with the standard bases. Find the coefficient of $e_2 \otimes f_2$ in $v \otimes w$ .

✅ Correct! The coefficient is

v_2 w_2 = (-1) \cdot 3 = -3

❌ Check the sign.

v_2 = -1

❌ Check the indices.

6 = v_1 w_2

, the coefficient of

e_1 \otimes f_2

; the first index picks the entry of

v

❌ Not quite. The coefficient of

e_j \otimes f_k

v_j w_k

Show solution

Expanding by bilinearity, $v \otimes w = \sum_{j,k} v_j w_k\, e_j \otimes f_k$ , so the coefficient of $e_2 \otimes f_2$ is

v_2 w_2 = (-1)(3) = -3.

The full grid is $\begin{pmatrix} 2 & 6 & 0 \\ -1 & -3 & 0 \end{pmatrix}$ .

Problem 2 · Counting the Basis

Given: a basis $e_1, e_2$ of $V$ and a basis $f_1, \dots, f_4$ of $W$ . How many vectors are in the basis $\{e_j \otimes f_k\}$ of $V \otimes W$ ?

✅ Correct! One tensor

e_j \otimes f_k

for each of the

2 \cdot 4

index pairs.

❌ Not quite.

6 = 2 + 4

is the size of a basis of

V \times W

❌ Not quite.

16 = 4^2

does not count index pairs

(j, k)

with

j \in \{1, 2\}

and

k \in \{1, 2, 3, 4\}

❌ Not quite. Count the pairs

(j, k)

Show solution

The list $e_j \otimes f_k$ has one entry for each $j \in \{1, 2\}$ and $k \in \{1, 2, 3, 4\}$ : $2 \cdot 4 = 8$ vectors, matching $\dim(V \otimes W) = (\dim V)(\dim W) = 8$ .

Problem 3 · Spot the Simple Tensor

Given: coefficient grids of elements of $\mathbf{F}^2 \otimes \mathbf{F}^3$ in the basis $e_j \otimes f_k$ . Which one is a simple tensor $v \otimes w$ ?

✅ Correct! Row 2 is

-\tfrac{1}{2}

times row 1, so the grid is

(2, -1) \otimes (1, -2, 3)

❌ Close, but… two of the three columns are proportional, but the last entry breaks it: the rows are not multiples of one row.

❌ Not quite. In a simple tensor every row is a multiple of the same row

w

Show solution

Row $j$ of the grid of $v \otimes w$ is $v_j\, w$ . In (b), row 1 is $2\,(1, -2, 3)$ and row 2 is $-1\,(1, -2, 3)$ , so the element is

(2, -1) \otimes (1, -2, 3).

In (a), $(2, 4, 5)$ is not a multiple of $(1, 2, 3)$ ; in (c) the rows $(1, 0, 0)$ and $(0, 1, 0)$ are not multiples of one row; in (d), $(1, 1, 2)$ is not a multiple of $(1, 1, 1)$ .

Problem 4 · Factor a Tensor

Given: $u = e_1 \otimes f_1 + 2\, e_1 \otimes f_2 + 3\, e_2 \otimes f_1 + 6\, e_2 \otimes f_2$ in $\mathbf{F}^2 \otimes \mathbf{F}^2$ . Write $u$ as a simple tensor.

✅ Correct! The grid is

\begin{pmatrix} 1 & 2 \\ 3 & 6 \end{pmatrix}

: rows are

1 \cdot (1, 2)

and

3 \cdot (1, 2)

❌ Close, but…

(1, 2) \otimes (1, 3)

has grid

\begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix}

, the transpose. The first factor runs down the rows.

❌ Not quite. Write the grid, with the coefficient of

e_j \otimes f_k

in row

j

, column

k

, and read off

v

and

w

Show solution

The coefficient grid of $u$ is $\begin{pmatrix} 1 & 2 \\ 3 & 6 \end{pmatrix}$ . Its rows are $1 \cdot (1, 2)$ and $3 \cdot (1, 2)$ , so with $v = (1, 3)$ and $w = (1, 2)$ the grid of $v \otimes w$ , entries $v_j w_k$ , is exactly this one. Coefficients in a basis are unique, so

u = (1, 3) \otimes (1, 2).

Solved: 0 / 4