Sum of subspaces and Grassmann's formula – Intersection and Sum of Subspaces (Senior 6 Mathematics) | Ganzaa

Intersection and Sum of Subspaces

Lessons

Senior 6 · Mathematics

Intersection and Sum of Subspaces

3 lessons

Lesson 3 of 3

Sum of subspaces and Grassmann's formula

35 min

Combining two subspaces

The union of two subspaces is too small to be a subspace (lesson 2): the $x$ -axis and the $y$ -axis together do not contain $(1, 1)$ . The repair is to allow all sums of a vector from one with a vector from the other. The result is the sum of the subspaces, and its dimension is tied to the dimension of the intersection by one of the neatest formulas in linear algebra.

Definition

Sum of subspaces: for subspaces $U$ and $W$ of $V$ ,

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

Example: in $\mathbb{R}^3$ , the $x$ -axis plus the $y$ -axis is the $x y$ -plane, since $(a, 0, 0) + (0, b, 0) = (a, b, 0)$ .

Important

Theorem. $U + W$ is a subspace of $V$ . It contains $U$ and $W$ , and it is the smallest subspace that does. If $U = \operatorname{span}(S)$ and $W = \operatorname{span}(T)$ , then $U + W = \operatorname{span}(S \cup T)$ .

Proof: $\mathbf{0} = \mathbf{0} + \mathbf{0} \in U + W$ . If $\mathbf{u}_1 + \mathbf{w}_1$ and $\mathbf{u}_2 + \mathbf{w}_2$ are in $U + W$ , then $\alpha(\mathbf{u}_1 + \mathbf{w}_1) + \beta(\mathbf{u}_2 + \mathbf{w}_2) = (\alpha\mathbf{u}_1 + \beta\mathbf{u}_2) + (\alpha\mathbf{w}_1 + \beta\mathbf{w}_2)$ , a vector of $U$ plus a vector of $W$ . Each $\mathbf{u} \in U$ is $\mathbf{u} + \mathbf{0}$ , so $U \subseteq U + W$ , and likewise $W \subseteq U + W$ . Any subspace containing $U$ and $W$ contains all their sums, so it contains $U + W$ . Last, a sum of a combination of $S$ and a combination of $T$ is a combination of $S \cup T$ , and conversely.

Finding a basis of $U + W$ . Put the spanning vectors of $U$ and of $W$ together as the rows of one matrix and row-reduce. The non-zero rows that remain form a basis of $U + W$ , and

\dim(U + W) = \text{rank of the combined matrix} .

Definition

Direct sum: the sum $U + W$ is direct, written $U \oplus W$ , when $U \cap W = \{\mathbf{0}\}$ . If moreover $U \oplus W = V$ , then $U$ and $W$ are complementary subspaces of $V$ .

Example: $\mathbb{R}^2 = (x\text{-axis}) \oplus (y\text{-axis})$ : the axes meet only at $O$ , and $(a, b) = (a, 0) + (0, b)$ .

Important

Theorem (unique decomposition). $U + W$ is direct if and only if every vector of $U + W$ can be written as $\mathbf{u} + \mathbf{w}$ in exactly one way.

Proof: suppose $U \cap W = \{\mathbf{0}\}$ and $\mathbf{u} + \mathbf{w} = \mathbf{u}' + \mathbf{w}'$ . Then $\mathbf{u} - \mathbf{u}' = \mathbf{w}' - \mathbf{w}$ ; the left side is in $U$ , the right side in $W$ , so both are in $U \cap W$ and equal $\mathbf{0}$ . Hence $\mathbf{u} = \mathbf{u}'$ and $\mathbf{w} = \mathbf{w}'$ . Conversely, if $\mathbf{v} \ne \mathbf{0}$ were in $U \cap W$ , then $\mathbf{0} = \mathbf{0} + \mathbf{0} = \mathbf{v} + (-\mathbf{v})$ would be two different decompositions of $\mathbf{0}$ .

ℝ³ = U ⊕ W: the plane U and the line W meet only at O, and every vector v splits in exactly one way into a part u in the plane and a part w along the line.

Important

Grassmann's dimension formula. For finite-dimensional subspaces $U$ and $W$ of $V$ ,

\dim(U + W) = \dim U + \dim W - \dim(U \cap W) .

In particular, $U + W$ is direct exactly when $\dim(U + W) = \dim U + \dim W$ .

Proof (sketch): take a basis $\{\mathbf{e}_1, \dots, \mathbf{e}_k\}$ of $U \cap W$ . Extend it to a basis $\{\mathbf{e}_1, \dots, \mathbf{e}_k, \mathbf{u}_1, \dots, \mathbf{u}_p\}$ of $U$ and to a basis $\{\mathbf{e}_1, \dots, \mathbf{e}_k, \mathbf{w}_1, \dots, \mathbf{w}_q\}$ of $W$ . Together the $k + p + q$ vectors span $U + W$ , and one checks that they are independent: in a relation, the $\mathbf{w}$ -part would be a vector of $W$ equal to a vector of $U$ , so it lies in $U \cap W$ , which forces every coefficient to be 0. So

\dim(U + W) = k + p + q = (k + p) + (k + q) - k = \dim U + \dim W - \dim(U \cap W) .

Tip

The formula looks like the counting rule for sets, $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ : the shared part is counted twice, so it is subtracted once. But notice that it is the sum $U + W$ , not the union, that appears on the left.

Worked example 1: dimensions in $\mathbb{R}^3$ and $\mathbb{R}^4$

(a) For the planes $U : x + y - z = 0$ and $W : 2 x - y + z = 0$ of lesson 2, find $U + W$ . (b) In $\mathbb{R}^4$ , let $U = \operatorname{span}\{(1, 0, 1, 0), (0, 1, 0, 1)\}$ and $W = \{\mathbf{x} : x_1 = x_2\}$ . Find $\dim(U + W)$ .

Solution

$\dim U = \dim W = 2$ and $\dim(U \cap W) = 1$ . So $\dim(U + W) = 2 + 2 - 1 = 3$ , and $U + W = \mathbb{R}^3$ . Every vector of $\mathbb{R}^3$ is a sum of one from each plane, but not in a unique way, since the sum is not direct.
$\dim U = 2$ , $\dim W = 4 - 1 = 3$ , and lesson 2 found $U \cap W = \operatorname{span}\{(1, 1, 1, 1)\}$ . So $\dim(U + W) = 2 + 3 - 1 = 4$ , and $U + W = \mathbb{R}^4$ .

Worked example 2: from the rank to the intersection

Let $U = \operatorname{span}\{(1, 2, 0, 1), (0, 1, 1, 0)\}$ and $W = \operatorname{span}\{(1, 3, 1, 1), (1, 0, 0, 0)\}$ in $\mathbb{R}^4$ . Find $\dim(U + W)$ and $\dim(U \cap W)$ , and a basis of $U \cap W$ .

Solution Row-reduce the four spanning vectors:

\begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 3 & 1 & 1 \\ 1 & 0 & 0 & 0 \end{pmatrix} \xrightarrow{\substack{R_3 - R_1 - R_2 \\ R_4 - R_1 + 2R_2}} \begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & -1 \end{pmatrix} .

Three non-zero rows remain, so $\dim(U + W) = 3$ . Each pair of spanning vectors is independent, so $\dim U = \dim W = 2$ , and by Grassmann

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

The zero row came from $(1, 3, 1, 1) = (1, 2, 0, 1) + (0, 1, 1, 0)$ : this vector of $W$ lies in $U$ . So $U \cap W = \operatorname{span}\{(1, 3, 1, 1)\}$ .

Worked example 3: a direct sum and a decomposition

Let $P = \{(x, y, z) : x - y + z = 0\}$ and $L = \operatorname{span}\{(1, 1, 1)\}$ . Show that $\mathbb{R}^3 = P \oplus L$ , and write $\mathbf{v} = (2, 3, 4)$ as $\mathbf{u} + \mathbf{w}$ with $\mathbf{u} \in P$ and $\mathbf{w} \in L$ .

Solution In lesson 2 we found $P \cap L = \{\mathbf{0}\}$ . Then $\dim(P + L) = 2 + 1 - 0 = 3$ , so $P + L = \mathbb{R}^3$ , and the sum is direct. Write $\mathbf{w} = t(1, 1, 1)$ ; then $\mathbf{u} = (2 - t, 3 - t, 4 - t)$ must satisfy the equation of $P$ :

(2 - t) - (3 - t) + (4 - t) = 3 - t = 0, \qquad t = 3 .

So $\mathbf{w} = (3, 3, 3)$ and $\mathbf{u} = (-1, 0, 1)$ . Check: $- 1 - 0 + 1 = 0$ ✓ and $(- 1, 0, 1) + (3, 3, 3) = (2, 3, 4)$ ✓.

Warning

$\dim(U + W)$ can never exceed $\dim V$ . This gives a lower bound for every intersection: $\dim(U \cap W) \ge \dim U + \dim W - \dim V$ . For example, two subspaces of dimensions 3 and 4 in $\mathbb{R}^5$ must share at least a 2-dimensional subspace. Forgetting this bound leads to impossible answers such as " $U \cap W = \{\mathbf{0}\}$ ".

MIT OpenCourseWare: Lecture 9 of 18.06 Linear Algebra, Independence, Basis, and Dimension, with Gilbert Strang (about 50 min). Watch how independence and spanning together define a basis, and why every basis of a space has the same number of vectors: that number is the dimension counted by Grassmann's formula.

✦

Practice

Questions 6 to 8 ask for proofs or reasons.

In $\mathbb{R}^3$ , let $U = \operatorname{span}\{(1, 0, 0), (0, 1, 0)\}$ and $W = \operatorname{span}\{(0, 1, 0), (0, 0, 1)\}$ . Find $U + W$ and $U \cap W$ , and check Grassmann's formula.
Is $\mathbb{R}^3 = U \oplus W$ for $U = \operatorname{span}\{(1, 0, 1)\}$ and $W = \{(x, y, z) : x + y + z = 0\}$ ?
With $U$ and $W$ as in question 2, write $(4, 1, 3)$ as $\mathbf{u} + \mathbf{w}$ with $\mathbf{u} \in U$ , $\mathbf{w} \in W$ .
$U$ and $W$ are subspaces of $\mathbb{R}^5$ with $\dim U = 3$ and $\dim W = 4$ . What are the possible values of $\dim(U \cap W)$ ?
Let $U = \operatorname{span}\{(1, 1, 0, 0), (0, 1, 1, 0)\}$ and $W = \operatorname{span}\{(0, 0, 1, 1), (1, 2, 1, 0)\}$ . Find $\dim(U + W)$ , $\dim(U \cap W)$ and a basis of $U \cap W$ .
Show that $\mathbb{R}^4 = U \oplus W$ for $U = \operatorname{span}\{(1, 1, 0, 0), (0, 1, 1, 0)\}$ and $W = \operatorname{span}\{(0, 0, 1, 1), (1, 0, 0, -1)\}$ .
$U$ and $W$ are two different 2-dimensional subspaces of $\mathbb{R}^3$ . Prove that $\dim(U \cap W) = 1$ .
Explain why, in $\mathbb{R}^n$ , the sum of two subspaces of dimension greater than $\frac{n}{2}$ can never be direct.

Show answersHide answers

$U + W = \mathbb{R}^3$ ; $U \cap W = \operatorname{span}\{(0, 1, 0)\}$ ; $3 = 2 + 2 - 1$ ✓
Yes: $(1, 0, 1)$ gives $1 + 0 + 1 = 2 \ne 0$ , so $U \cap W = \{\mathbf{0}\}$ , and $1 + 2 = 3$ .
$t (1, 0, 1)$ with $(4 - t) + 1 + (3 - t) = 0$ : $t = 4$ , so $(4, 1, 3) = (4, 0, 4) + (0, 1, - 1)$ .
$\dim(U + W) \le 5$ gives $\dim(U \cap W) \ge 2$ , and $\dim(U \cap W) \le 3$ : so 2 or 3.
$(1, 2, 1, 0) = (1, 1, 0, 0) + (0, 1, 1, 0) \in U$ ; the rank of all four is 3, so $\dim(U + W) = 3$ , $\dim(U \cap W) = 1$ , basis $\{(1, 2, 1, 0)\}$ .
The four vectors have rank 4, so $\dim(U + W) = 4$ , $U + W = \mathbb{R}^4$ , and $\dim(U \cap W) = 2 + 2 - 4 = 0$ .
$U \ne W$ , so $U + W$ is strictly bigger than $U$ : $\dim(U + W) = 3$ . Then $\dim(U \cap W) = 2 + 2 - 3 = 1$ .
If $\dim U + \dim W \gt n$ , then $\dim(U \cap W) \ge \dim U + \dim W - n \gt 0$ , so $U \cap W \ne \{\mathbf{0}\}$ .

1Combining two subspaces

Combining two subspaces