Intersection of subspaces and its dimension

30 min

Where two subspaces meet

Two walls of a classroom meet along a vertical edge; the floor and one wall meet along a horizontal edge. In the same way, two planes through the origin in $\mathbb{R}^3$ meet along a line through the origin. In the language of sets, the edge is the intersection of the two planes. This lesson shows that the intersection of two subspaces is always a subspace, while their union usually is not, and how to find a basis and the dimension of an intersection.

Two different planes through O meet in a line through O. The line is a subspace of both planes: dim(U ∩ W) = 1, less than dim U = dim W = 2.

Definition

Intersection of subspaces: for subspaces $U$ and $W$ of a vector space $V$ , the intersection is the set of vectors that lie in both:

U \cap W = \{\mathbf{v} \in V : \mathbf{v} \in U \text{ and } \mathbf{v} \in W\} .

Example: in $\mathbb{R}^3$ , the $x y$ -plane $(z = 0)$ and the $x z$ -plane $(y = 0)$ intersect in the $x$ -axis, $\{(x, 0, 0)\}$ .

Important

Theorem. If $U$ and $W$ are subspaces of $V$ , then $U \cap W$ is a subspace of $V$ , and

\dim(U \cap W) \le \min(\dim U, \dim W) .

Proof: $\mathbf{0} \in U$ and $\mathbf{0} \in W$ , so $\mathbf{0} \in U \cap W$ . Let $\mathbf{u}, \mathbf{v} \in U \cap W$ and let $\alpha, \beta$ be scalars. Since $U$ is a subspace containing $\mathbf{u}$ and $\mathbf{v}$ , $\alpha\mathbf{u} + \beta\mathbf{v} \in U$ ; likewise $\alpha\mathbf{u} + \beta\mathbf{v} \in W$ . So $\alpha\mathbf{u} + \beta\mathbf{v} \in U \cap W$ . Finally $U \cap W$ is a subspace of $U$ and of $W$ , so its dimension cannot exceed either of theirs.

Warning

The union is usually not a subspace. In $\mathbb{R}^2$ , let $U$ be the $x$ -axis and $W$ the $y$ -axis. Then $(1, 0) \in U \cup W$ and $(0, 1) \in U \cup W$ , but $(1, 0) + (0, 1) = (1, 1)$ is on neither axis. In fact $U \cup W$ is a subspace only when one of the two contains the other. (If $\mathbf{u} \in U$ is not in $W$ and $\mathbf{w} \in W$ is not in $U$ , then $\mathbf{u} + \mathbf{w}$ can lie in neither: if it were in $U$ , then $\mathbf{w} = (\mathbf{u} + \mathbf{w}) - \mathbf{u}$ would be in $U$ .)

How to find an intersection

A vector is in $U \cap W$ when it meets the description of $U$ and that of $W$ . How you write this depends on how the subspaces are given.

$U$ and $W$ are given by	method
equations (both)	put all the equations together and solve the system
spanning vectors (both)	set a general vector of $U$ equal to a general vector of $W$ , and solve for the coefficients
one by spanning vectors, one by equations	substitute the general vector of the first into the equations of the second

In every case the answer is the solution set of a homogeneous linear system, which you write as a span to read off a basis.

Worked example 1: two planes given by equations

Find a basis and the dimension of $U \cap W$ , where $U = \{(x, y, z) : x + y - z = 0\}$ and $W = \{(x, y, z) : 2x - y + z = 0\}$ .

Solution Solve the two equations together. Adding them gives $3 x = 0$ , so $x = 0$ , and then $y = z$ . Hence

U \cap W = \{(0, y, y)\} = \operatorname{span}\{(0, 1, 1)\}, \qquad \dim(U \cap W) = 1 .

Check: $0 + 1 - 1 = 0$ ✓ and $0 - 1 + 1 = 0$ ✓. The two planes are different (their normal vectors $(1, 1, - 1)$ and $(2, - 1, 1)$ are not parallel), so they meet in a line, as in the diagram.

Worked example 2: two planes given by spanning vectors

Let $U = \operatorname{span}\{(1, 0, 1), (0, 1, 1)\}$ and $W = \operatorname{span}\{(1, 1, 0), (1, 0, 0)\}$ . Find $U \cap W$ .

Solution A general vector of $U$ is $a (1, 0, 1) + b (0, 1, 1) = (a, b, a + b)$ . A general vector of $W$ is $c (1, 1, 0) + d (1, 0, 0) = (c + d, c, 0)$ , so $W$ is the plane $z = 0$ . A vector of $U$ lies in $W$ exactly when its third coordinate is 0: $a + b = 0$ , $b = - a$ . Hence

U \cap W = \{(a, -a, 0)\} = \operatorname{span}\{(1, -1, 0)\}, \qquad \dim(U \cap W) = 1 .

Worked example 3: in $\mathbb{R}^4$ , and an intersection that is only $\{\mathbf{0}\}$

(a) Let $U = \operatorname{span}\{(1, 0, 1, 0), (0, 1, 0, 1)\}$ and $W = \{(x_1, x_2, x_3, x_4) : x_1 = x_2\}$ . Find $U \cap W$ . (b) Find the intersection of the line $L = \operatorname{span}\{(1, 1, 1)\}$ and the plane $P : x - y + z = 0$ .

Solution

The general vector of $U$ is $(a, b, a, b)$ . It is in $W$ when $a = b$ , so $U \cap W = \{(a, a, a, a)\} = \operatorname{span}\{(1, 1, 1, 1)\}$ , of dimension 1. Note $\dim U = 2$ and $\dim W = 3$ (one equation in $\mathbb{R}^4$ ).
Put $(t, t, t)$ into the equation of $P$ : $t - t + t = t = 0$ . Only $t = 0$ works, so $L \cap P = \{\mathbf{0}\}$ and $\dim(L \cap P) = 0$ . The line pierces the plane at $O$ only.

Tip

Count before you compute. In $\mathbb{R}^n$ , a subspace described by $p$ independent equations has dimension $n - p$ . The intersection is described by all the equations together, so if the combined system has rank $r$ , then $\dim(U \cap W) = n - r$ . In Worked example 1: $n = 3$ , the two equations are independent, $r = 2$ , and $\dim(U \cap W) = 1$ .

MIT OpenCourseWare: Lecture 6 of 18.06 Linear Algebra, Column Space and Nullspace, with Gilbert Strang (about 46 min). Watch the opening part, on whether the union and the intersection of two subspaces are subspaces, and then the nullspace: the solutions of a homogeneous system form a subspace, which is exactly how intersections were found above.

✦

Practice

Questions 5 to 7 ask for reasons.

Find a basis and the dimension of $U \cap W$ for $U = \{(x, y, z) : x - y = 0\}$ and $W = \{(x, y, z) : y + z = 0\}$ .
Let $U = \operatorname{span}\{(1, 2, 0)\}$ and $W = \{(x, y, z) : 2x - y + z = 0\}$ . Find $U \cap W$ . What does your answer say about $U$ and $W$ ?
Find $U \cap W$ for $U = \operatorname{span}\{(1, 1, 0), (0, 0, 1)\}$ and $W = \operatorname{span}\{(1, 0, 1), (0, 1, 0)\}$ .
In $\mathbb{R}^4$ , let $U = \{\mathbf{x} : x_1 + x_2 + x_3 + x_4 = 0\}$ and $W = \{\mathbf{x} : x_1 = x_2 = x_3\}$ . Find a basis of $U \cap W$ .
Give an example of two subspaces $U \ne W$ of $\mathbb{R}^3$ whose union is a subspace.
Can two planes through the origin in $\mathbb{R}^3$ have intersection $\{\mathbf{0}\}$ ? Explain with equations.
$U$ and $W$ are subspaces with $\dim(U \cap W) = \dim U$ . What can you conclude?

Show answersHide answers

$x = y$ , $z = - y$ : $\operatorname{span}\{(1, 1, -1)\}$ , dimension 1.
$(t, 2 t, 0)$ : $2 t - 2 t + 0 = 0$ for every $t$ , so $U \cap W = U$ : the line $U$ lies in the plane $W$ .
$(a, a, b) = (c, d, c)$ gives $a = c = d = b$ : $\operatorname{span}\{(1, 1, 1)\}$ .
$(s, s, s, - 3 s)$ : basis $\{(1, 1, 1, -3)\}$ .
For example the $x$ -axis and the plane $z = 0$ : the axis lies in the plane, so the union is the plane.
No. Two homogeneous equations in three unknowns have rank at most 2, so the solution set has dimension at least $3 - 2 = 1$ : the planes share at least a line.
$U \cap W$ is a subspace of $U$ with the same dimension, so $U \cap W = U$ , that is $U \subseteq W$ .

Sources and credits

Written by Ganzaa: the explanations, the proofs, the diagram, the worked examples, the practice items and the quiz questions. Every intersection was checked by a rank computation.
Video: 6. Column Space and Nullspace, MIT 18.06 Linear Algebra, Gilbert Strang, MIT OpenCourseWare, embedded from YouTube.

1Where two subspaces meet

2How to find an intersection

Where two subspaces meet

How to find an intersection