How Do You Find the Dimensions of Subspace Sums and Intersections?

When people ask about the “operations” on subspaces, they usually mean the sum, intersection, and sometimes the union of two subspaces. The key dimension rule concerns the sum and intersection: it counts the dimensions of both subspaces, then subtracts the dimension they share.

Assume throughout that the vector spaces are finite-dimensional. The union needs separate treatment because it is usually not a subspace at all.

The dimension formula

For subspaces $U$ and $V$ of a finite-dimensional vector space, their sum is

$$U+V=\{u+v : u\in U,\ v\in V\}.$$

It is the smallest subspace containing both $U$ and $V$, and it is also the span of $U\cup V$. Their intersection $U\cap V$ consists of vectors that lie in both. The dimensions satisfy

$$\dim(U+V)=\dim U+\dim V-\dim(U\cap V).$$

The subtraction matters because a direction in the intersection is counted once in $\dim U$ and again in $\dim V$. Subtracting $\dim(U\cap V)$ removes the duplicate count.

A worked example in $\mathbb{R}^3$

Let

$$U=\operatorname{span}\{(1,0,0),(0,1,0)\}, \qquad V=\operatorname{span}\{(0,1,0),(0,0,1)\}.$$

Each subspace has dimension $2$. Both contain $(0,1,0)$, and every vector in $U$ has third coordinate $0$, while every vector in $V$ has first coordinate $0$. Thus their intersection is exactly

$$U\cap V=\operatorname{span}\{(0,1,0)\},$$

which has dimension $1$. Therefore,

$$\dim(U+V)=2+2-1=3.$$

Indeed, the vectors $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ all lie in $U+V$ and form a basis for $\mathbb{R}^3$. So $U+V=\mathbb{R}^3$.

Why the formula works

Take a basis for $U\cap V$ and extend it to a basis for each of $U$ and $V$. Write the shared basis as $c_1,\ldots,c_k$; add vectors $u_1,\ldots,u_r$ to complete a basis of $U$, and vectors $v_1,\ldots,v_s$ to complete a basis of $V$.

The combined list

$$c_1,\ldots,c_k,u_1,\ldots,u_r,v_1,\ldots,v_s$$

is a basis for $U+V$. It spans the sum because it includes bases for both subspaces. It is linearly independent because any combination of the added $u$-vectors that equals a combination of the added $v$-vectors must lie in $U\cap V$; the basis extensions ensure that this can happen only when those added-vector coefficients are zero. Hence

$$\dim(U+V)=k+r+s,$$

while $\dim U=k+r$, $\dim V=k+s$, and $\dim(U\cap V)=k$. Substitution gives the formula.

Direct sums and useful bounds

The sum is called a direct sum, written $U\oplus V$, when $U\cap V=\{0\}$. Since the zero subspace has dimension $0$, the formula becomes

$$\dim(U\oplus V)=\dim U+\dim V.$$

For example, the $xy$-plane and the $z$-axis in $\mathbb{R}^3$ intersect only at the zero vector, so their direct sum is all of $\mathbb{R}^3$ and has dimension $2+1=3$.

The formula also gives bounds on the intersection. Since $U+V$ is a subspace of the ambient space $W$,

$$\max(0,\dim U+\dim V-\dim W) \leq \dim(U\cap V) \leq \min(\dim U,\dim V).$$

For instance, two planes in $\mathbb{R}^3$ must intersect in at least a line: the lower bound is $2+2-3=1$.

What about the union?

The union $U\cup V$ is usually not a subspace. A subspace must be closed under addition, but a vector from $U$ added to a vector from $V$ need not belong to either one.

In $\mathbb{R}^2$, let $U$ be the $x$-axis and $V$ the $y$-axis. The vectors $(1,0)\in U$ and $(0,1)\in V$ have sum $(1,1)$, which belongs to neither axis. Thus $U\cup V$ is not a subspace and does not have a vector-space dimension. For two subspaces, the union is a subspace exactly when one is contained in the other; then the union is simply the larger subspace.

Common mistakes

• Adding dimensions without subtracting the overlap: this is correct only when $U\cap V=\{0\}$.
• Treating the union as the sum: $U\cup V$ contains vectors from either subspace; $U+V$ also contains sums of vectors from the two subspaces.
• Calling a zero intersection “empty”: every pair of subspaces contains the zero vector in its intersection. A trivial intersection means $U\cap V=\{0\}$.
• Using the formula without finite dimensions: the stated arithmetic formula is intended for finite-dimensional subspaces; infinite-dimensional settings require more care with dimension conventions.