Subspaces, Sums and Intersections, linear algebra I lesson

Picture the plane $z = x + y$ through the origin in space and a parallel copy lifted one unit higher. Both look flat and endless, yet they react differently to vector arithmetic. Add two vectors that end in the first plane and the sum ends there too, but double a vector that ends in the lifted plane and its tip leaves the plane. This lesson turns that difference into an exact test, shows which familiar sets pass it, and compares sums, intersections and unions of subspaces.

Key terms

Subspace

Definition Throughout, a tuple such as $(x, y, z)$ stands for a column vector. Let $W$ be a subset of $\mathbb{R}^n$ . The set $W$ is closed under addition if $\mathbf{u} + \mathbf{v}$ lies in $W$ for all $\mathbf{u}$ and $\mathbf{v}$ in $W$ . It is closed under scalar multiplication, or scaling, if $c\mathbf{u}$ lies in $W$ for every $\mathbf{u}$ in $W$ and every real number $c$ . A subspace of $\mathbb{R}^n$ is a subset $W$ that contains the zero vector $\mathbf{0}$ and is closed under addition and under scalar multiplication. The whole space $\mathbb{R}^n$ and the zero subspace $\{\mathbf{0}\}$ are subspaces of $\mathbb{R}^n$ .

In plain words A subspace is a set that contains the origin and that you cannot leave by adding its members or by stretching, shrinking or reversing them. Closure under addition speaks about every pair of members, and closure under scaling about every member and every real scalar, negative ones included, so a few examples that work prove nothing.

Example The line $L$ of all multiples $t(1, 2)$ in $\mathbb{R}^2$ is a subspace. It contains $\mathbf{0} = 0(1, 2)$ . For any members $s(1, 2)$ and $t(1, 2)$ and any real $c$ ,

\begin{aligned} s(1, 2) + t(1, 2) &= (s + t)(1, 2), \\[0.4em] c\,\big(t(1, 2)\big) &= (ct)(1, 2), \end{aligned}

and both results are again multiples of $(1, 2)$ . By contrast, the quadrant of vectors $(x, y)$ with $x \ge 0$ and $y \ge 0$ contains $\mathbf{0}$ and is closed under addition, but it is not a subspace, because $(-1)(1, 1) = (-1, -1)$ leaves it.

Subspace test

Definition For a subset $W$ of $\mathbb{R}^n$ the following three statements are equivalent.

The set $W$ is a subspace, that is, it contains $\mathbf{0}$ and is closed under addition and scalar multiplication.
The set $W$ is nonempty and closed under addition and scalar multiplication.
The set $W$ is a vector space in its own right, with the addition and scalar multiplication of $\mathbb{R}^n$ .

The reason is that a nonempty $W$ has some member $\mathbf{w}$ , and closure then gives $0\mathbf{w} = \mathbf{0}$ and $(-1)\mathbf{w} = -\mathbf{w}$ in $W$ . The other vector space rules, such as associativity and the distributive laws, already hold for all vectors of $\mathbb{R}^n$ , so they hold in $W$ . To prove that a set is a subspace, check $\mathbf{0}$ and then take arbitrary members and an arbitrary real scalar. To prove that it is not, one specific failure is enough. The empty set is never a subspace, since it has no zero vector.

In plain words Check three things, zero, sums and multiples, with letters rather than numbers. Everything else a vector space needs comes for free from $\mathbb{R}^n$ .

Example Let $W$ be the set of vectors $(x, y, z)$ in $\mathbb{R}^3$ with $z = x + y$ . The zero vector is in $W$ because $0 = 0 + 0$ . Take $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$ in $W$ , so $u_3 = u_1 + u_2$ and $v_3 = v_1 + v_2$ . The third entry of $\mathbf{u} + \mathbf{v}$ is

\begin{aligned} u_3 + v_3 &= (u_1 + u_2) + (v_1 + v_2) \\[0.4em] &= (u_1 + v_1) + (u_2 + v_2), \end{aligned}

the sum of its first two entries. For a real $c$ , the third entry of $c\mathbf{u}$ is $cu_3 = cu_1 + cu_2$ . So $W$ is a subspace of $\mathbb{R}^3$ . Specific vectors agree with the proof. The vectors $(1, 2, 3)$ and $(4, -1, 3)$ lie in $W$ , and so do their sum $(5, 1, 6)$ and the multiple $-2(4, -1, 3) = (-8, 2, -6)$ .

Solution set of a homogeneous system

Definition Let $A$ be a real $m \times n$ matrix. The solution set of the homogeneous system $A\mathbf{x} = \mathbf{0}$ ,

\{\mathbf{x} \in \mathbb{R}^n \mid A\mathbf{x} = \mathbf{0}\},

is a subspace of $\mathbb{R}^n$ . Indeed $A\mathbf{0} = \mathbf{0}$ , and if $A\mathbf{u} = \mathbf{0}$ and $A\mathbf{v} = \mathbf{0}$ , then $A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v} = \mathbf{0}$ and $A(c\mathbf{u}) = cA\mathbf{u} = \mathbf{0}$ for every real $c$ . The argument uses one fixed matrix $A$ for both solutions, since solutions of two different systems say nothing about closure. In the same way, the span of any finite list $\mathbf{v}_1, \dots, \mathbf{v}_k$ in $\mathbb{R}^n$ is a subspace of $\mathbb{R}^n$ . All coefficients zero give $\mathbf{0}$ , adding two combinations adds their coefficients entry by entry, and scaling a combination scales every coefficient.

In plain words Everything a fixed matrix sends to zero forms a subspace, and so does everything you can build from a fixed list of vectors. These are the two standard ways to describe a subspace, by equations or by generators.

Example The system $x + y + z = 0$ , $x - z = 0$ has the augmented matrix

\left[\begin{array}{ccc|c} 1 & 1 & 1 & 0 \\ 1 & 0 & -1 & 0 \end{array}\right].

The step $R_2 \to R_2 - R_1$ turns the second row into $(0, -1, -2 \mid 0)$ , so $y = -2z$ and then $x = -y - z = z$ . With $z = t$ free, the solutions are $t(1, -2, 1)$ , the line $\operatorname{span}\{(1, -2, 1)\}$ through the origin. The plane $W$ from the subspace test is also a span, since every member equals $(x, y, x + y) = x(1, 0, 1) + y(0, 1, 1)$ .

Affine set

Definition Let $W$ be a subspace of $\mathbb{R}^n$ and $\mathbf{p}$ a vector in $\mathbb{R}^n$ . The translate

\mathbf{p} + W = \{\mathbf{p} + \mathbf{w} \mid \mathbf{w} \in W\}

is called an affine set. It is a subspace exactly when $\mathbf{p}$ lies in $W$ , and then $\mathbf{p} + W = W$ . If the system $A\mathbf{x} = \mathbf{b}$ has a particular solution $\mathbf{x}_p$ , its solution set is the affine set $\mathbf{x}_p + \{\mathbf{x} \mid A\mathbf{x} = \mathbf{0}\}$ . When $\mathbf{b} \ne \mathbf{0}$ the zero vector is not a solution, so that solution set is not a subspace. An inconsistent system has an empty solution set, which is not a subspace either.

In plain words An affine set is a subspace slid along a fixed vector. It keeps the shape and the directions of the subspace, but once it misses the origin, the sum of any two of its members lands off it.

Example Let $F$ be the set of vectors $(x, y, z)$ with $z = x + y + 1$ , the solution set of $x + y - z = -1$ . Every member can be written as

\begin{aligned} (x,\ y,\ x + y + 1) &= (0, 0, 1) \\[0.4em] &\quad + (x,\ y,\ x + y), \end{aligned}

so $F = (0, 0, 1) + W$ for the plane $W$ above. The zero vector is not in $F$ , because $0 = 0 + 0 + 1$ is false. Scaling fails too. The vector $(0, 0, 1)$ is in $F$ , but $2(0, 0, 1) = (0, 0, 2)$ is not, since $2 \ne 1$ .

Intersection and sum of subspaces

Definition Let $U$ and $V$ be subspaces of $\mathbb{R}^n$ . Their intersection $U \cap V$ is the set of vectors that lie in both $U$ and $V$ . Their sum is the set of all sums of a vector in $U$ and a vector in $V$ ,

U + V = \{\mathbf{u} + \mathbf{v} \mid \mathbf{u} \in U,\ \mathbf{v} \in V\}.

Both are subspaces of $\mathbb{R}^n$ . For the intersection, $\mathbf{0}$ lies in both, and sums and multiples of shared members stay in each subspace. For the sum, $\mathbf{0} = \mathbf{0} + \mathbf{0}$ is a member, and two members add as

\begin{aligned} &(\mathbf{u}_1 + \mathbf{v}_1) + (\mathbf{u}_2 + \mathbf{v}_2) \\[0.4em] &\quad = (\mathbf{u}_1 + \mathbf{u}_2) + (\mathbf{v}_1 + \mathbf{v}_2) \end{aligned}

with $\mathbf{u}_1 + \mathbf{u}_2$ in $U$ and $\mathbf{v}_1 + \mathbf{v}_2$ in $V$ . Likewise $c(\mathbf{u} + \mathbf{v}) = c\mathbf{u} + c\mathbf{v}$ with $c\mathbf{u}$ in $U$ and $c\mathbf{v}$ in $V$ . The sum contains $U$ and $V$ , because $\mathbf{u} = \mathbf{u} + \mathbf{0}$ , and it is the smallest subspace that contains both. If $U$ and $V$ are spanned by two lists, then $U + V$ is spanned by the two lists joined together.

In plain words The intersection keeps only what the two subspaces share. The sum collects everything you can reach by taking one step in $U$ and one step in $V$ .

Example In $\mathbb{R}^3$ let $U$ be the $xy$ -plane of vectors $(x, y, 0)$ and $V$ the $xz$ -plane of vectors $(x, 0, z)$ . A vector in both has third entry $0$ and second entry $0$ , so $U \cap V$ is the $x$ -axis $\operatorname{span}\{\mathbf{e}_1\}$ . The sum is all of $\mathbb{R}^3$ , since every vector splits as

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

For instance $(1, 2, 3) = (1, 2, 0) + (0, 0, 3)$ lies in $U + V$ although it lies in neither plane. The split is not unique here. The same vector also equals $(0, 2, 0) + (1, 0, 3)$ , because a piece of the shared $x$ -axis can move from one part to the other.

Union of subspaces

Definition The union $U \cup V$ of two subspaces of $\mathbb{R}^n$ is the set of vectors that lie in $U$ or in $V$ . It contains $\mathbf{0}$ and is closed under scalar multiplication, but it is usually not closed under addition. Precisely, $U \cup V$ is a subspace if and only if $U \subseteq V$ or $V \subseteq U$ . If one contains the other, the union is simply the larger subspace. If neither contains the other, pick $\mathbf{u}$ in $U$ but not in $V$ and $\mathbf{v}$ in $V$ but not in $U$ . If $\mathbf{u} + \mathbf{v}$ were in $U$ , then $\mathbf{v} = (\mathbf{u} + \mathbf{v}) - \mathbf{u}$ would be in $U$ , and similarly for $V$ , so $\mathbf{u} + \mathbf{v}$ lies outside $U \cup V$ .

In plain words Pooling the members of two subspaces usually does not give a subspace, because adding a vector from each can land outside both. The sum $U + V$ repairs this, since it includes every such vector.

Example In $\mathbb{R}^2$ take the $x$ -axis and the $y$ -axis. Their union contains $\mathbf{0}$ , and multiples of a horizontal or vertical vector stay horizontal or vertical. But $(1, 0) + (0, 1) = (1, 1)$ lies on neither axis, so the union is not a subspace. Their intersection is $\{\mathbf{0}\}$ , which is a subspace, the smallest one there is. The same failure happens for the two coordinate planes above, where $\mathbf{e}_2 + \mathbf{e}_3 = (0, 1, 1)$ lies in neither plane.

Check yourself

Which of these subsets of $\mathbb{R}^2$ are subspaces? (a) The line $y = 3x$ . (b) The line $y = 3x + 1$ .

Show the answer

(a) Yes. It is the solution set of the homogeneous equation $3x - y = 0$ , and it equals $\operatorname{span}\{(1, 3)\}$ .

(b) No. The zero vector fails, because $0 = 3 \cdot 0 + 1$ is false. This line is the affine set $(0, 1) + \operatorname{span}\{(1, 3)\}$ .

The set $S$ of vectors $(x, y)$ with $xy \ge 0$ contains $\mathbf{0}$ , and it is closed under scalar multiplication because $(cx)(cy) = c^2 xy \ge 0$ . Is $S$ a subspace of $\mathbb{R}^2$ ?

Show the answer

No. Passing the zero test and the scaling test is not enough. The vectors $(1, 2)$ and $(-2, -1)$ lie in $S$ , since both products equal $2$ . Their sum $(-1, 1)$ has product $-1 < 0$ , so $S$ is not closed under addition. In fact $S$ is the union of the closed first and third quadrants, and as with the union of the two axes, addition is the condition that fails.

In $\mathbb{R}^3$ let $U$ be the plane $x + y + z = 0$ and $V$ the plane $x = y$ . Find $U \cap V$ and show that $U + V = \mathbb{R}^3$ .

Show the answer

A vector in both satisfies $x = y$ and $x + y + z = 0$ , so $z = -2x$ and the vector is $x(1, 1, -2)$ . Hence

U \cap V = \operatorname{span}\{(1, 1, -2)\}.

For the sum, write any $(a, b, c)$ as

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

The first vector lies in $U$ , because $a + b + (-a - b) = 0$ , and the second lies in $V$ , because its first two entries are equal. So every vector of $\mathbb{R}^3$ is in $U + V$ . For example $(1, 2, 3) = (1, 2, -3) + (0, 0, 6)$ .

Subspaces, Sums and Intersections

Key terms

Subspace

Subspace test

Solution set of a homogeneous system

Affine set

Intersection and sum of subspaces

Union of subspaces

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Linear Algebra II