Sums and Intersections of Subspaces, linear algebra II lesson

In $\mathbb{R}^3$ the plane $z = 0$ and the plane $y = 0$ each have dimension two, and adding vectors from the two planes reaches every vector of the space. Yet the result has dimension three, not four. The missing dimension is the line the planes share, which is counted once in each plane. This lesson defines sums and intersections of subspaces, proves the formula that corrects the double count, and gives a second proof through the product of two spaces.

Key terms

Sum of subspaces

Definition Let $U$ and $W$ be subspaces of a vector space $V$ over $\mathbb{F}$ , where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$ . Their sum is

U + W = \{u + w : u \in U,\ w \in W\}.

It is a subspace of $V$ . It contains $0 = 0 + 0$ , and a linear combination of two of its vectors regroups as

\begin{aligned} &a(u_1 + w_1) + b(u_2 + w_2) \\[0.4em] &\quad = (au_1 + bu_2) + (aw_1 + bw_2), \end{aligned}

a vector of $U$ plus a vector of $W$ . The sum is the smallest subspace of $V$ that contains both $U$ and $W$ . If $U$ and $W$ are spanned by two lists, then $U + W$ is spanned by the two lists together. When $U$ and $W$ share a nonzero vector $c$ , every vector of $U + W$ has more than one decomposition, since $u + w = (u + c) + (w - c)$ .

In plain words The sum collects every vector you can build by taking one piece from each subspace and adding the pieces. It combines directions, so it is usually larger than either subspace alone.

Example In $\mathbb{R}^3$ let $U = \{(x, y, z) : z = 0\}$ and $W = \{(x, y, z) : y = 0\}$ . Every vector splits as

(x, y, z) = (x, y, 0) + (0, 0, z),

so $U + W = \mathbb{R}^3$ . For instance $(0, 1, 1) = (0, 1, 0) + (0, 0, 1)$ lies in the sum but in neither plane. The split is not unique, since $(x + s, y, 0) + (-s, 0, z)$ gives the same vector for every real $s$ .

Intersection of subspaces

Definition For subspaces $U$ and $W$ of a vector space $V$ over $\mathbb{F}$ , the intersection $U \cap W$ is the set of vectors that lie in both $U$ and $W$ . It is a subspace of $V$ . It contains $0$ , and if $x$ and $y$ lie in both subspaces, then so does every combination $ax + by$ , because each subspace is closed under linear combinations. It is the largest subspace contained in both $U$ and $W$ . When the subspaces are described by equations, the intersection is found by imposing all the equations at once.

In plain words The intersection is the overlap, the directions the two subspaces have in common.

Example For the planes $U = \{z = 0\}$ and $W = \{y = 0\}$ in $\mathbb{R}^3$ , a common vector has $y = 0$ and $z = 0$ , so $U \cap W$ is the $x$ -axis, spanned by $(1, 0, 0)$ , with dimension $1$ . This shared line is what lets the split above slide, since the parameter $s$ moves the vector $(s, 0, 0)$ from one summand to the other.

Dimension formula for a sum

Definition Let $U$ and $W$ be finite-dimensional subspaces of a vector space $V$ over $\mathbb{F}$ . Then $U + W$ is finite-dimensional and

\begin{aligned} \dim(U + W) &= \dim U + \dim W \\[0.4em] &\quad - \dim(U \cap W). \end{aligned}

For the proof, choose a basis $c_1, \dots, c_r$ of $U \cap W$ . Extend it to a basis $c_1, \dots, c_r, u_1, \dots, u_p$ of $U$ , and separately to a basis $c_1, \dots, c_r, w_1, \dots, w_q$ of $W$ . The combined list of the $c_i$ , the $u_k$ and the $w_j$ spans $U + W$ , since any $u + w$ expands in it. To show independence, suppose

\begin{aligned} &\sum_{i} \alpha_i c_i + \sum_{k} \beta_k u_k \\[0.4em] &\quad + \sum_{j} \gamma_j w_j = 0. \end{aligned}

The vector $-\sum_j \gamma_j w_j$ lies in $W$ and equals $\sum_i \alpha_i c_i + \sum_k \beta_k u_k$ , which lies in $U$ . So it lies in $U \cap W$ and equals $\sum_i \delta_i c_i$ for some scalars $\delta_i$ . Then $\sum_i \delta_i c_i + \sum_j \gamma_j w_j = 0$ , and independence of the basis of $W$ gives every $\gamma_j = 0$ . The original relation now involves only the basis of $U$ , so every $\alpha_i$ and $\beta_k$ is $0$ too. Counting gives $r + p + q = (r + p) + (r + q) - r$ . The argument uses no angles or perpendicularity, so it holds over $\mathbb{R}$ and $\mathbb{C}$ and for spaces of functions or matrices alike.

In plain words Adding the two dimensions counts every shared direction twice, once in each subspace. Subtracting the dimension of the overlap corrects the count. Starting from a basis of the overlap is what makes the proof work, because two unrelated bases joined together hide the overlap inside a dependence.

Example Let $U = \{(x, y, z) : x + y + z = 0\}$ and $W = \{(x, y, z) : z = 0\}$ in $\mathbb{R}^3$ . Imposing both equations gives $z = 0$ and $y = -x$ , so $U \cap W = \operatorname{span}\{(1, -1, 0)\}$ . Extend this to the basis $(1, -1, 0), (1, 0, -1)$ of $U$ and to the basis $(1, -1, 0), (0, 1, 0)$ of $W$ . The combination

\begin{aligned} &a(1, -1, 0) + b(1, 0, -1) + d(0, 1, 0) \\[0.4em] &\quad = (a + b, -a + d, -b) \end{aligned}

is zero only when $b = 0$ , then $a = 0$ , then $d = 0$ . So the three vectors form a basis of $U + W = \mathbb{R}^3$ , and $3 = 2 + 2 - 1$ .

Union of subspaces

Definition The union $U \cup W$ of two subspaces of a vector space $V$ over $\mathbb{F}$ is the set of vectors that lie in at least one of them. It is a subspace exactly when one of the subspaces contains the other. If $U \subseteq W$ , the union is $W$ . If neither contains the other, pick $u \in U$ outside $W$ and $w \in W$ outside $U$ . Were $u + w$ in $U$ , then $w = (u + w) - u$ would be in $U$ , and were it in $W$ , then $u = (u + w) - w$ would be in $W$ . Both are false, so $u + w$ lies outside $U \cup W$ and the union is not closed under addition. The sum $U + W$ is the span of $U \cup W$ , the smallest subspace containing it.

In plain words The union only collects vectors that are already in one subspace or the other. It never adds them, so unless one subspace contains the other, some sums of its members escape it.

Example With $U = \{x + y + z = 0\}$ and $W = \{z = 0\}$ as above, the vector $(1, 0, -1)$ lies in $U$ but not in $W$ , and $(0, 1, 0)$ lies in $W$ but not in $U$ . Their sum $(1, 1, -1)$ has coordinate sum $1$ and third coordinate $-1$ , so it lies in neither plane. The union is not a subspace, although the sum $U + W = \mathbb{R}^3$ is.

Product of vector spaces

Definition Let $U$ and $W$ be vector spaces over the same field $\mathbb{F}$ . Their product $U \times W$ , also called their external product, is the set of ordered pairs $(u, w)$ with $u \in U$ and $w \in W$ , with the componentwise operations

\begin{aligned} &(u_1, w_1) + (u_2, w_2) \\[0.4em] &\quad = (u_1 + u_2, w_1 + w_2), \\[0.4em] &\lambda (u, w) = (\lambda u, \lambda w). \end{aligned}

It is a vector space over $\mathbb{F}$ with zero vector $(0, 0)$ . If $U$ and $W$ are finite-dimensional with bases $u_1, \dots, u_p$ and $w_1, \dots, w_q$ , the pairs $(u_1, 0), \dots, (u_p, 0), (0, w_1), \dots, (0, w_q)$ form a basis. They span because each component expands in its own basis, and a zero combination forces both lists of coefficients to vanish. So

\dim(U \times W) = \dim U + \dim W,

even when $U$ and $W$ are overlapping subspaces of one space. For finite-dimensional subspaces $U$ and $W$ of $V$ , the addition map $\sigma(u, w) = u + w$ from $U \times W$ onto $U + W$ is linear and surjective. If $u + w = 0$ , then $u = -w$ lies in both subspaces, so the kernel of $\sigma$ is $\{(c, -c) : c \in U \cap W\}$ , and $c \mapsto (c, -c)$ is an isomorphism from $U \cap W$ onto it. Rank-nullity for $\sigma$ then gives the dimension formula a second time.

In plain words A pair remembers which subspace supplied each piece, while a sum forgets it. Along the overlap a piece can move from one slot to the other without changing the sum, and that freedom is the dimension the formula subtracts.

Example Take the planes $U = \{z = 0\}$ and $W = \{y = 0\}$ in $\mathbb{R}^3$ . The product $U \times W$ has dimension $2 + 2 = 4$ , even though both planes sit inside $\mathbb{R}^3$ . The addition map onto $U + W = \mathbb{R}^3$ has as kernel the pairs $((s, 0, 0), (-s, 0, 0))$ with $s \in \mathbb{R}$ , a space of dimension $1$ , and rank-nullity gives $\dim(U + W) = 4 - 1 = 3$ .

Check yourself

Two subspaces $U$ and $W$ of $\mathbb{R}^5$ both have dimension $3$ . Can their intersection be $\{0\}$ ?

Show the answer

No. The sum $U + W$ is a subspace of $\mathbb{R}^5$ , so its dimension is at most $5$ , and the dimension formula gives

\begin{aligned} \dim(U \cap W) &= 3 + 3 - \dim(U + W) \\[0.4em] &\ge 6 - 5 = 1. \end{aligned}

The bound is attained. For the standard basis $e_1, \dots, e_5$ of $\mathbb{R}^5$ , the subspaces $\operatorname{span}(e_1, e_2, e_3)$ and $\operatorname{span}(e_3, e_4, e_5)$ meet in the line spanned by $e_3$ .

Over $\mathbb{R}$ , let $U = \{p \in \mathcal{P}_2 : p(1) = 0\}$ and $W = \{p \in \mathcal{P}_2 : p(-1) = 0\}$ , where $\mathcal{P}_2$ is the space of polynomials of degree at most two, including the zero polynomial. Find $U \cap W$ and $\dim(U + W)$ . Is $U \cup W$ a subspace?

Show the answer

Evaluation at $1$ is a surjective linear map from $\mathcal{P}_2$ to $\mathbb{R}$ with kernel $U$ , so rank-nullity gives $\dim U = 3 - 1 = 2$ , and likewise $\dim W = 2$ . A polynomial of degree at most two that vanishes at both $1$ and $-1$ is a multiple of $(t - 1)(t + 1) = t^2 - 1$ , so $U \cap W = \operatorname{span}(t^2 - 1)$ has dimension $1$ . The formula gives $\dim(U + W) = 2 + 2 - 1 = 3$ , so $U + W = \mathcal{P}_2$ .

The union is not a subspace. The polynomial $t - 1$ lies in $U$ but not in $W$ , and $t + 1$ lies in $W$ but not in $U$ . Their sum $2t$ takes the value $2$ at $1$ and $-2$ at $-1$ , so it lies in neither.

Let $L$ be the line in $\mathbb{R}^2$ spanned by $(1, 2)$ . Find $\dim(L \times L)$ , $\dim(L + L)$ and the kernel of the addition map from $L \times L$ onto $L + L$ .

Show the answer

The product has dimension $1 + 1 = 2$ , with basis $((1, 2), 0)$ and $(0, (1, 2))$ . The sum $L + L$ is $L$ itself, of dimension $1$ , because $L$ is closed under addition. The addition map sends $(u, w)$ to $u + w$ , so its kernel is $\{(c, -c) : c \in L\}$ , the line spanned by $((1, 2), (-1, -2))$ , of dimension $1$ . Rank-nullity checks out as $2 = 1 + 1$ , matching $\dim L + \dim L - \dim(L \cap L) = 1$ . The product is not the sum. The pairs $((2, 4), 0)$ and $((1, 2), (1, 2))$ are different elements of $L \times L$ with the same sum $(2, 4)$ .

Sums and Intersections of Subspaces

Key terms

Sum of subspaces

Intersection of subspaces

Dimension formula for a sum

Union of subspaces

Product of vector spaces

Check yourself

How this lesson connects

Builds on

Used later in