Recipe

Apply the subspace dimension formula

Use when: you know, or can find, three of $\dim U$ , $\dim W$ , $\dim(U+W)$ , $\dim(U\cap W)$ , and need the fourth

Goal: the missing dimension, found algebraically without necessarily constructing a basis for the harder subspace

Taught in Sums of subspaces and the dimension formula

Quick recipe

List which of the four quantities are already known
If $\dim(U+W)$ is unknown, compute it from spanning sets
Solve the rearranged formula for whichever quantity is missing
Check for a degenerate containment case first
Verify the formula holds with every number plugged in

1
Write down $\dim U$ , $\dim W$ , $\dim(U+W)$ , and $\dim(U\cap W)$ , and mark which are given directly and which are still unknown.

Example

$U$ is $3$ -dimensional and $W$ is $2$ -dimensional inside a $5$ -dimensional $V$ , and $\dim(U+W)=4$ is given; only $\dim(U\cap W)$ is unknown.

Why: the dimension formula has exactly four quantities, so knowing any three of them always determines the fourth by algebra alone.
2
Combine both spanning sets into one list, then row-reduce to find its rank; that rank is $\dim(U+W)$ .

Example

For $U=\operatorname{span}\{(1,0,1,0),(0,1,0,1)\}$ and $W=\operatorname{span}\{(1,1,0,0),(0,0,1,1)\}$ , stacking all four vectors and reducing gives rank $4$ .
- an explicit basis or spanning set is available for both subspaces — row-reduce the combined list directly for the exact value.
- only the ambient space's own dimension is known — $\dim(U+W)$ can only be bounded above by it, without row reduction.
Why: $U+W$ is spanned by exactly the union of the two individual spanning sets, so its dimension is that union's rank.
3
Rearrange $\dim(U+W)=\dim U+\dim W-\dim(U\cap W)$ to isolate the one unknown, then substitute the other three numbers.

Example

$\dim(U\cap W) = \dim U + \dim W - \dim(U+W) = 3+2-4 = 1$ .

Why: this finds the missing dimension exactly, without ever exhibiting an actual basis for the harder subspace.
4
If one subspace's spanning vectors already look like they sit inside the other, confirm containment directly instead of computing ranks.

Example

The $x$ -axis lies entirely inside the $xy$ -plane, so their intersection is just the $x$ -axis, dimension $1$ , with no elimination needed.
- neither subspace obviously contains the other — proceed with the general rank computation from Step 2.
- one subspace's spanning vectors are visibly already inside the other — the intersection equals the smaller subspace directly, and the sum equals the larger one.
Why: recognizing containment on sight saves a full row reduction, and the formula still holds exactly in this simpler case.
5
Substitute every quantity, known and found, back into $\dim(U+W)=\dim U+\dim W-\dim(U\cap W)$ and confirm both sides match.

Example

$2 = 2+1-1$ , confirmed exactly for the degenerate $xy$ -plane and $x$ -axis case.

Why: this check catches an arithmetic slip or a wrong containment guess before the wrong dimension carries into later reasoning.

Common mistakes

Trying to find an explicit basis for $U\cap W$ instead of solving the formula algebraically for its dimension.
Forgetting $\dim(U+W)$ is capped by the ambient space's own dimension, especially when using that as a bound.
Assuming a given spanning set is automatically a basis, without checking independence first.
Missing a degenerate containment case and running a full computation that a one-line check would settle.