Linear-Algebra-Done-Right · Unit 8 · Video 1 · Interactive Practice

Products of Vector Spaces: When Dimensions Add

IKey Formulas

Formula	Name	What it says
$V_1 \times \cdots \times V_m = \{(v_1, \dots, v_m) : v_k \in V_k\}$	Product of vector spaces	Lists with one entry from each space; addition and scalar multiplication slot by slot
$\dim(V_1 \times \cdots \times V_m) = \dim V_1 + \cdots + \dim V_m$	Dimension of a product	For finite-dimensional $V_k$ : each basis vector of each $V_k$ , placed in slot $k$ , gives a basis of the product
$\Gamma(v_1, \dots, v_m) = v_1 + \cdots + v_m$	The map $\Gamma$	Linear and onto $V_1 + \cdots + V_m$ ; the sum is direct $\iff \Gamma$ injective
$\dim(V_1 + \cdots + V_m) = \dim V_1 + \cdots + \dim V_m - \dim \text{null}\,\Gamma$	Dimension test	For finite-dimensional subspaces $V_k$ : the sum is direct $\iff$ the dimensions add

Key Insight: For three or more subspaces, checking that every pair meets only at $0$ is not enough to make a sum direct; for finite-dimensional subspaces, comparing $\dim(V_1 + \cdots + V_m)$ with $\dim V_1 + \cdots + \dim V_m$ always is.

IIVisualization 1 — Arithmetic Slot by Slot

In $\mathcal{P}_5(\mathbf{R}) \times \mathbf{R}^3$ every operation acts on each slot separately, inside that slot's own space.

Operation λ 2

IIIVisualization 2 — Each Basis Vector in Its Own Slot

Placing each basis vector of $\mathcal{P}_2(\mathbf{R})$ and of $\mathbf{R}^2$ in its own slot gives a basis of the product, so dimensions add.

Slot 1 Slot 2

IVVisualization 3 — Gamma and the Dimension Test

$\dim(V_1 + V_2 + V_3)$ falls short of $\dim V_1 + \dim V_2 + \dim V_3$ by exactly $\dim \text{null}\,\Gamma$ .

V₁ V₂ V₃

VQuiz Questions

Problem 1 · Compute in a Product

Given: in $\mathcal{P}_3(\mathbf{R}) \times \mathbf{R}^2$ , let $u = (1 - 2x + x^3,\ (4, -1))$ and $w = (x - x^2,\ (-2, 5))$ — find $3u + w$ .

✅ Correct! The scalar

3

multiplies both slots of

u

, then each slot is added in its own space.

❌ Not quite. The polynomial slot is right, but

3u

scales the list slot too:

3(4, -1) = (12, -3)

❌ Not quite. That is

3(u + w)

: the scalar was applied to

w

as well.

❌ Not quite. The slots never interact: the entry

10

of the list slot has been added into the constant term, and the list was dropped. The answer must be a list of length two: a polynomial, then a list in

\mathbf{R}^2

❌ Not quite. Compute

3u

slot by slot, then add

w

slot by slot.

Show solution

Scalar multiplication acts on each slot:

3u = \big(3 - 6x + 3x^3,\ (12, -3)\big).

Addition acts on each slot, inside its own space:

\text{slot 1: } (3 - 6x + 3x^3) + (x - x^2) = 3 - 5x - x^2 + 3x^3,

\text{slot 2: } (12, -3) + (-2, 5) = (10, 2).

So $3u + w = \big(3 - 5x - x^2 + 3x^3,\ (10, 2)\big)$ .

Problem 2 · Equal or Isomorphic?

Given: the spaces $\mathbf{R}^3 \times \mathbf{R}^2$ and $\mathbf{R}^5$ — which statement is correct?

✅ Correct! Different kinds of object, identified by a relabeling that only removes the inner brackets.

❌ Close, but… An element of

\mathbf{R}^3 \times \mathbf{R}^2

is a list of two lists, not a single list of five numbers. The two spaces are isomorphic, not equal.

❌ Not quite. The difference in list length rules out equality, not isomorphism: both spaces have dimension

5

, and removing the inner brackets is an invertible linear map.

❌ Not quite. Dimensions of a product add, they do not multiply:

\dim(\mathbf{R}^3 \times \mathbf{R}^2) = 3 + 2 = 5

❌ Not quite. Compare the kind of object in each space, then the dimensions.

Show solution

An element of $\mathbf{R}^3 \times \mathbf{R}^2$ is $\big((a, b, c), (d, e)\big)$ , a list of length $2$ whose entries are lists of lengths $3$ and $2$ . An element of $\mathbf{R}^5$ is a single list $(a, b, c, d, e)$ of length $5$ . So the sets are different: $\mathbf{R}^3 \times \mathbf{R}^2 \neq \mathbf{R}^5$ .

The map $\big((a, b, c), (d, e)\big) \mapsto (a, b, c, d, e)$ is linear, because both spaces add and scale number by number and the map keeps each number in its position. It is invertible: splitting a list of length $5$ after its third entry undoes it. Hence it is an isomorphism, consistent with $\dim(\mathbf{R}^3 \times \mathbf{R}^2) = 3 + 2 = 5 = \dim \mathbf{R}^5$ .

Problem 3 · Three Subspaces of R⁴

Given: subspaces of $\mathbf{R}^4$ : $V_1 = \{(a, b, 0, 0)\}$ , $V_2 = \{(0, 0, c, d)\}$ , $V_3 = \{(t, t, -t, 0)\}$ , and $\Gamma : V_1 \times V_2 \times V_3 \to V_1 + V_2 + V_3$ . Every pair of them meets only at $0$ . Find $\dim \text{null}\,\Gamma$ and a nonzero element of $\text{null}\,\Gamma$ .

What is dim null Γ?

Which list lies in null Γ?

✅ Correct!

\dim \text{null}\,\Gamma = 5 - 4 = 1

, so the sum is not direct even though every pair meets only at

0

❌ Not quite. Pairwise intersections

\{0\}

do not make

\Gamma

injective for three subspaces. Compare

\dim V_1 + \dim V_2 + \dim V_3

with

\dim(V_1 + V_2 + V_3)

❌ Check the dimension.

\dim \text{null}\,\Gamma = \dim(V_1 \times V_2 \times V_3) - \dim(V_1 + V_2 + V_3)

, and the sum lies in

\mathbf{R}^4

❌ Check the sum. Each entry is in the right subspace, but the three vectors do not add to

(0, 0, 0, 0)

❌ Not quite. The entries do add to

0

, but

(-1, 0, 1, 0)

is not in

V_3

: its first two coordinates must be equal.

❌ Check the list. Need

v_1 \in V_1

v_2 \in V_2

v_3 \in V_3

, not all

0

, with

v_1 + v_2 + v_3 = 0

Show solution

Dimensions. $\dim V_1 = 2$ , $\dim V_2 = 2$ , $\dim V_3 = 1$ , so $\dim(V_1 \times V_2 \times V_3) = 2 + 2 + 1 = 5$ . Already $V_1 + V_2 = \mathbf{R}^4$ , so $\dim(V_1 + V_2 + V_3) = 4$ . Since $\Gamma$ is onto the sum, the fundamental theorem of linear maps gives

\dim \text{null}\,\Gamma = 5 - 4 = 1.

An element. Take $v_3 = (-1, -1, 1, 0) \in V_3$ (that is $t = -1$ ). Then $v_1 + v_2 = (1, 1, -1, 0)$ is solved by $v_1 = (1, 1, 0, 0) \in V_1$ and $v_2 = (0, 0, -1, 0) \in V_2$ :

\Gamma\big((1, 1, 0, 0),\ (0, 0, -1, 0),\ (-1, -1, 1, 0)\big) = (0, 0, 0, 0).

So $\Gamma$ is not injective and $V_1 + V_2 + V_3$ is not a direct sum, although every pair meets only at $0$ .

Problem 4 · Too Many Dimensions

Given: subspaces $V_1, V_2, V_3$ of $\mathbf{R}^6$ with $\dim V_1 = 3$ and $\dim V_2 = \dim V_3 = 2$ — which statement must be true?

✅ Correct!

\dim(V_1 + V_2 + V_3) \le 6 \lt 7 = \dim V_1 + \dim V_2 + \dim V_3

, so the dimensions cannot add.

❌ Close, but…

\dim \text{null}\,\Gamma = 7 - \dim(V_1 + V_2 + V_3)

, and the sum can have dimension less than

6

— for example if

V_3 \subseteq V_1

, then

\dim \text{null}\,\Gamma \ge 2

❌ Not quite. No condition can make this sum direct: the dimensions

3 + 2 + 2 = 7

exceed

\dim \mathbf{R}^6

. And for three or more subspaces, pairwise intersections

\{0\}

do not by themselves guarantee a direct sum.

❌ Not quite. The product is a new space of lists, not a subspace of

\mathbf{R}^6

\dim(V_1 \times V_2 \times V_3) = 3 + 2 + 2 = 7

❌ Not quite. Compare

\dim V_1 + \dim V_2 + \dim V_3

with the largest possible

\dim(V_1 + V_2 + V_3)

Show solution

The product has dimension $\dim V_1 + \dim V_2 + \dim V_3 = 3 + 2 + 2 = 7$ . The sum $V_1 + V_2 + V_3$ is a subspace of $\mathbf{R}^6$ , so its dimension is at most $6$ . By the fundamental theorem of linear maps applied to $\Gamma$ , which is onto the sum,

\dim \text{null}\,\Gamma = 7 - \dim(V_1 + V_2 + V_3) \ge 7 - 6 = 1.

So $\Gamma$ is not injective and the sum is not direct, whatever the subspaces are. The exact value of $\dim \text{null}\,\Gamma$ depends on them: it is $1$ when the sum is all of $\mathbf{R}^6$ , and larger when it is not.

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