Intersection and sum of subspaces

Unit 6 Intersection and sum of subspaces

Key unit competence: Relate the sum and the intersection of subspaces of a vector space by the dimension formula.

Key ideas

• A subspace $W$ of a vector space $V$ is a non-empty subset closed under addition and scalar multiplication: $\alpha u+\beta v\in W$ for all $u,v\in W$, $\alpha ,\beta \in \mathbb{R}$. Every subspace contains $\vec{0}$.
• Typical subspaces of ${\mathbb{R}}^3$: $\{\vec{0}\}$, lines through $O$ (dimension 1), planes through $O$ (dimension 2), ${\mathbb{R}}^3$ itself. A line or plane not through $O$ is not a subspace.
• The intersection $U\cap W$ of two subspaces is always a subspace. The union $U\cup W$ is a subspace only when one contains the other.
• The sum $U+W=\{u+w:u\in U,w\in W\}$ is the smallest subspace containing $U\cup W$. If $U=\mathrm{span}(S)$ and $W=\mathrm{span}(T)$ then $U+W=\mathrm{span}(S\cup T)$.
• The sum is direct, written $U\oplus W$, when $U\cap W=\{\vec{0}\}$; then every vector of $U+W$ is written in exactly one way as $u+w$.
• Grassmann's formula: $\dim (U+W)=\dim U+\dim W-\dim (U\cap W)$. For a direct sum, $\dim (U\oplus W)=\dim U+\dim W$.
• To find $U\cap W$ when both are given by equations, solve all the equations together; when given by spanning vectors, write $a{\vec{u}}_1+b{\vec{u}}_2=c{\vec{w}}_1+\dots$ and solve.

Formulas to know

Common misconceptions

• "The union of two subspaces is a subspace." Usually not: the $x$-axis $\cup$ $y$-axis does not contain $(1,1)=(1,0)+(0,1)$.
• "$\{(x,y):y=2x+1\}$ is a subspace." No: it does not contain $(0,0)$.
• "Two planes of ${\mathbb{R}}^3$ can form a direct sum." No: by Grassmann $\dim (U\cap W)\ge 2+2-3=1$.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Show that $W=\{(x,y,z):x-2y+3z=0\}$ is a subspace of ${\mathbb{R}}^3$ and find a basis.

   Show answer

   $\vec{0}\in W$; if $u,v\in W$ then $\alpha u+\beta v$ satisfies the equation (it is linear). $x=2y-3z$: $(2y-3z,y,z)=y(2,1,0)+z(-3,0,1)$. Basis $\{(2,1,0),(-3,0,1)\}$, $\dim W=2$.

   Common mistake: Giving three vectors as a basis of a plane.

2. $U$: $x+y-z=0$ and $W=\mathrm{span}\{(1,0,1)\}$. Find $U\cap W$ and $\dim (U+W)$.

   Show answer

   $(1,0,1)$: $1+0-1=0$, so $W\subseteq U$. $U\cap W=W$ (dim 1) and $U+W=U$ (dim 2).

   Common mistake: Assuming the sum is always ${\mathbb{R}}^3$.

3. $U=\mathrm{span}\{(1,0,0),(0,1,0)\}$ and $W=\mathrm{span}\{(0,1,1)\}$. Is the sum direct? Find $U+W$.

   Show answer

   $(0,1,1)\notin U$ (its third coordinate is not 0), so $U\cap W=\{\vec{0}\}$: direct sum. $\dim (U\oplus W)=2+1=3$, so $U\oplus W={\mathbb{R}}^3$.

   Common mistake: Thinking the sum is not direct because the vectors share a component.

4. $U=\mathrm{span}\{(1,1,0),(0,1,1)\}$, $W=\mathrm{span}\{(1,0,-1),(1,2,1)\}$. Compare $U$ and $W$.

   Show answer

   $(1,0,-1)=(1,1,0)-(0,1,1)$ and $(1,2,1)=(1,1,0)+(0,1,1)$, so $W\subseteq U$; both have dimension 2, so $W=U$. Then $U\cap W=U+W=U$ (dimension 2).

   Common mistake: Assuming different spanning sets give different subspaces.

5. Verify Grassmann's formula for $U$ = the $xy$-plane and $W$ = the $yz$-plane.

   Show answer

   $U\cap W$ = the $y$-axis (dimension 1); $U+W={\mathbb{R}}^3$. $3=2+2-1$. ✓

   Common mistake: Saying the intersection is $\{\vec{0}\}$.

6. Let $U$ be the plane $z=0$. Find a subspace $W$ with ${\mathbb{R}}^3=U\oplus W$.

   Show answer

   Any line through $O$ not in $U$, e.g. $W=\mathrm{span}\{(0,0,1)\}$: $U\cap W=\{\vec{0}\}$ and $2+1=3$.

   Common mistake: Choosing a line inside $U$.

7. In ${\mathbb{R}}^4$: $U=\mathrm{span}\{e_1,e_2\}$, $W=\mathrm{span}\{e_2,e_3\}$. Find $U\cap W$, $U+W$ and check the formula.

   Show answer

   $U\cap W=\mathrm{span}\{e_2\}$ (dim 1); $U+W=\mathrm{span}\{e_1,e_2,e_3\}$ (dim 3). $3=2+2-1$. ✓

   Common mistake: Counting $e_2$ twice in the basis of $U+W$.

8. Show that the union of the $x$-axis and the $y$-axis is not a subspace of ${\mathbb{R}}^2$.

   Show answer

   $(1,0)$ and $(0,1)$ are in the union, but $(1,0)+(0,1)=(1,1)$ is on neither axis: not closed under addition.

   Common mistake: Only checking that $\vec{0}$ is in the union.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. $U=\mathrm{span}\{(1,0,0)\}$ and $W=\mathrm{span}\{(0,1,0)\}$. Describe $U+W$.	a. the plane $z=0$
   2. $U=\mathrm{span}\{(1,0,0),(0,1,0)\}$ and $W=\mathrm{span}\{(0,1,0),(0,0,1)\}$. Find $U\cap W$.	b. $2$
   3. Find $\dim \mathrm{span}\{(1,2,1),(2,1,0),(3,3,1)\}$.	c. $\mathrm{span}\{(0,1,0)\}$
   4. In $M_2(\mathbb{R})$, $S$ is the set of symmetric matrices and $A$ the set of skew-symmetric matrices. Find $\dim S$ and $\dim A$.	d. $3$ and $1$

   Show answer

   1 → a (the plane $z=0$) · 2 → c ($\mathrm{span}\{(0,1,0)\}$) · 3 → b ($2$) · 4 → d ($3$ and $1$)

   • Step 1. $U+W=\{(a,b,0)\}$.
     Step 2. That is the $xy$-plane, $z=0$, of dimension 2.
   • Step 1. $U$ is the plane $z=0$ and $W$ is the plane $x=0$.
     Step 2. Both conditions: $x=z=0$, which is the $y$-axis $\mathrm{span}\{(0,1,0)\}$.
   • Step 1. $(3,3,1)=(1,2,1)+(2,1,0)$.
     Step 2. The first two are not multiples of each other, so the dimension is 2.
   • Step 1. Symmetric: $\left(\begin{array}{cc}a& b\\ b& d\end{array}\right)$, 3 free entries.
     Step 2. Skew-symmetric: $\left(\begin{array}{cc}0& b\\ -b& 0\end{array}\right)$, 1 free entry. $3+1=4=\dim M_2$, and $M_2=S\oplus A$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   1 and 4  ·  $\dim E=2$, $\dim O=1$ and $P_2=E\oplus O$  ·  $1$  ·  1 and 3
   1. In $P_2$, $E$ = even polynomials and $O$ = odd polynomials. Then .
   2. Two different planes through the origin in ${\mathbb{R}}^3$ meet in a subspace of dimension .
   3. $U=\{(x,y,z):x=0\}$, $W=\{(x,y,z):y=0\}$. $\dim (U\cap W)$ and $\dim (U+W)$ are .
   Show answer

   1: $\dim E=2$, $\dim O=1$ and $P_2=E\oplus O$ · 2: $1$ · 3: 1 and 3

   • Step 1. Even: $a+c{x}^2$, basis $\{1,x^2\}$. Odd: $bx$, basis $\{x\}$.
     Step 2. Only 0 is both even and odd, and $2+1=3=\dim P_2$.
   • Step 1. $U+W={\mathbb{R}}^3$ (two different planes span the space).
     Step 2. $\dim (U\cap W)=2+2-3=1$: a line.
   • Step 1. $U\cap W$: $x=y=0$, the $z$-axis, dimension 1.
     Step 2. $\dim (U+W)=2+2-1=3$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. In ${\mathbb{R}}^4$, $\dim U=3$ and $\dim W=3$. Find the smallest possible $\dim (U\cap W)$.	a. $u=(2,2)$, $w=(1,-1)$
   2. $U$: $x-y+z=0$ and $W$: $x+y=0$. Find a basis of $U\cap W$.	b. $k=2$
   3. Find $k$ so that $(2,3,7)\in \mathrm{span}\{(1,0,k),(0,1,1)\}$.	c. $\{(1,-1,-2)\}$
   4. ${\mathbb{R}}^2=U\oplus W$ with $U=\mathrm{span}\{(1,1)\}$ and $W=\mathrm{span}\{(1,-1)\}$. Split $(3,1)=u+w$.	d. $2$

   Show answer

   1 → d ($2$) · 2 → c ($\{(1,-1,-2)\}$) · 3 → b ($k=2$) · 4 → a ($u=(2,2)$, $w=(1,-1)$)

   • Step 1. $\dim (U\cap W)=6-\dim (U+W)$.
     Step 2. $\dim (U+W)\le 4$, so $\dim (U\cap W)\ge 6-4=2$.
   • Step 1. $W$ gives $y=-x$.
     Step 2. In $U$: $x+x+z=0\Rightarrow z=-2x$. With $x=1$: $(1,-1,-2)$.
   • Step 1. $a(1,0,k)+b(0,1,1)=(a,b,ak+b)$; so $a=2$, $b=3$.
     Step 2. $2k+3=7\Rightarrow k=2$.
   • Step 1. $a(1,1)+b(1,-1)=(3,1)$: $a+b=3$, $a-b=1$.
     Step 2. $a=2$, $b=1$: $u=(2,2)$, $w=(1,-1)$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   3  ·  2  ·  $V=U+W$ and $U\cap W=\{\vec{0}\}$  ·  4
   1. $V=U\oplus W$ (direct sum) means .
   2. In ${\mathbb{R}}^5$, $\dim U=3$ and $\dim W=4$. The smallest possible $\dim (U\cap W)$ is .
   3. In ${\mathbb{R}}^4$, $\dim U=\dim W=2$ and $U\cap W=\{\vec{0}\}$. $\dim (U+W)=$ →
   Show answer

   1: $V=U+W$ and $U\cap W=\{\vec{0}\}$ · 2: 2 · 3: 4

   • Step 1. Every vector of $V$ must be a sum $u+w$ ($V=U+W$).
     Step 2. That sum must be unique, which happens exactly when $U\cap W=\{\vec{0}\}$.
   • Step 1. $\dim (U\cap W)=3+4-\dim (U+W)$.
     Step 2. $\dim (U+W)\le 5$, so $\dim (U\cap W)\ge 7-5=2$.
   • Step 1. $\dim (U+W)=\dim U+\dim W-\dim (U\cap W)$.
     Step 2. $2+2-0=4$ (a direct sum).

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