Linear Algebra 7 | Examples for Subspaces [dark version]

Q: How does one quickly confirm that a constrained set in $\mathbb{R}^3$ contains the zero vector?

Substitute $(0,0,0)$ into the defining conditions. For $x_1=x_2$ and $x_3=-x_2$, both equalities hold immediately: $0=0$ and $0=-0$. That satisfies the “zero vector” requirement.

Q: Why does scalar multiplication preserve membership for the set $x_1=x_2$, $x_3=-x_2$?

Take any $u$ in the set, so $u_1=u_2$ and $u_3=-u_2$. For any scalar $\lambda$, the scaled vector $x=\lambda u$ has components $x_1=\lambda u_1$, $x_2=\lambda u_2$, $x_3=\lambda u_3$. Then $x_1=x_2$ because $\lambda u_1=\lambda u_2$, and $x_3=-x_2$ because $x_3=\lambda(-u_2)=-\lambda u_2$.

Q: How does closure under addition work for the same $\mathbb{R}^3$ subspace?

Let $u$ and $v$ both satisfy the constraints. Their sum has components $x_1=u_1+v_1$, $x_2=u_2+v_2$, $x_3=u_3+v_3$. Since $u_1=u_2$ and $v_1=v_2$, it follows that $x_1=x_2$. Since $u_3=-u_2$ and $v_3=-v_2$, then $x_3=u_3+v_3=-(u_2+v_2)=-x_2$, so the second constraint also holds.

Q: What counterexample shows that the set $\{(x_1,x_2): x_1^2=x_2\}$ is not a subspace?

Use $u=(1,1)$. It satisfies the condition because $1^2=1$. Scale by $\lambda=2$ to get $\lambda u=(2,2)$. Now $x_1^2=2^2=4$ but $x_2=2$, so $x_1^2=x_2$ fails. Since scalar multiplication can move the vector outside the set, it cannot be a subspace.

TL;DR

The subspace test uses three checks: zero vector inclusion, closure under scalar multiplication, and closure under addition.

Briefing Cornell Notes

Briefing

A set of vectors in $R^{3}$ defined by simple component constraints— $x_{1} = x_{2}$ and $x_{3} = - x_{2}$ —turns out to be a genuine linear subspace, and the proof comes down to checking the three standard closure conditions. First, the zero vector satisfies both equations: plugging in $(0, 0, 0)$ makes $x_{1} = x_{2}$ true and also forces $x_{3} = - x_{2}$ to hold. That establishes the “contains the zero vector” requirement.

Next, scaling cannot push the set out. Starting from an arbitrary vector $u = (u_{1}, u_{2}, u_{3})$ in the set, the defining relations guarantee $u_{1} = u_{2}$ and $u_{3} = - u_{2}$ . After multiplying by any scalar $λ$ , the new vector $x = λ u$ has components $(λ u_{1}, λ u_{2}, λ u_{3})$ . The same relations persist because $λ u_{1} = λ u_{2}$ and $λ u_{3} = λ (- u_{2}) = - λ u_{2}$ , which matches the required form $x_{3} = - x_{2}$ . The set is therefore closed under scalar multiplication.

Finally, the set is also closed under addition. Take two vectors $u = (u_{1}, u_{2}, u_{3})$ and $v = (v_{1}, v_{2}, v_{3})$ already inside the set, so $u_{1} = u_{2}$ , $u_{3} = - u_{2}$ , and $v_{1} = v_{2}$ , $v_{3} = - v_{2}$ . Their sum $u + v$ has components $(u_{1} + v_{1}, u_{2} + v_{2}, u_{3} + v_{3})$ . The first defining equation holds because $u_{1} + v_{1} = u_{2} + v_{2}$ . The second holds after substituting $u_{3} = - 2 \cdot (u_{2}) / ?$ in the transcript’s algebraic step (equivalently, using $u_{3} = - u_{2}$ and $v_{3} = - v_{2}$ ) and factoring out the common coefficient, yielding $(u_{3} + v_{3}) = - (u_{2} + v_{2})$ . With all three checks satisfied, the constrained set in $R^{3}$ is confirmed as a subspace.

The lesson then flips to a counterexample in $R^{2}$ : the set of all vectors $(x_{1}, x_{2})$ satisfying $x_{1}^{2} = x_{2}$ . The square makes the condition incompatible with linear structure. A single scaling test breaks it: $(1, 1)$ belongs because $1^{2} = 1$ . But scaling by $λ = 2$ gives $(2, 2)$ , and now $x_{1}^{2} = 2^{2} = 4$ while $x_{2} = 2$ , so the defining equality fails. Since scalar multiplication can take a vector out of the set, the set is not a linear subspace.

Cornell Notes

A subspace in $R^{3}$ can be verified by checking three closure properties against its defining equations. For the set defined by $x_{1} = x_{2}$ and $x_{3} = - x_{2}$ , the zero vector satisfies both relations. If $u$ satisfies the constraints, then $λ u$ also satisfies them for any scalar $λ$ , because scaling preserves equalities between components and preserves the sign relation in $x_{3} = - x_{2}$ . If both $u$ and $v$ satisfy the constraints, then $u + v$ also satisfies them since component-wise addition preserves $x_{1} = x_{2}$ and the corresponding relation for $x_{3}$ . A contrasting $R^{2}$ example with $x_{1}^{2} = x_{2}$ fails under scaling, so it cannot be a subspace.

How does one quickly confirm that a constrained set in $R^{3}$ contains the zero vector?

Substitute

(0, 0, 0)

into the defining conditions. For

x_{1} = x_{2}

and

x_{3} = - x_{2}

, both equalities hold immediately:

0 = 0

and

0 = - 0

. That satisfies the “zero vector” requirement.

Why does scalar multiplication preserve membership for the set $x_{1} = x_{2}$ , $x_{3} = - x_{2}$ ?

Take any

u

in the set, so

u_{1} = u_{2}

and

u_{3} = - u_{2}

. For any scalar

λ

, the scaled vector

x = λ u

has components

x_{1} = λ u_{1}

x_{2} = λ u_{2}

x_{3} = λ u_{3}

. Then

x_{1} = x_{2}

because

λ u_{1} = λ u_{2}

, and

x_{3} = - x_{2}

because

x_{3} = λ (- u_{2}) = - λ u_{2}

How does closure under addition work for the same $R^{3}$ subspace?

Let

u

and

v

both satisfy the constraints. Their sum has components

x_{1} = u_{1} + v_{1}

x_{2} = u_{2} + v_{2}

x_{3} = u_{3} + v_{3}

. Since

u_{1} = u_{2}

and

v_{1} = v_{2}

, it follows that

x_{1} = x_{2}

. Since

u_{3} = - u_{2}

and

v_{3} = - v_{2}

, then

x_{3} = u_{3} + v_{3} = - (u_{2} + v_{2}) = - x_{2}

, so the second constraint also holds.

What counterexample shows that the set ${(x_{1}, x_{2}) : x_{1}^{2} = x_{2}}$ is not a subspace?

Use

u = (1, 1)

. It satisfies the condition because

1^{2} = 1

. Scale by

λ = 2

to get

λ u = (2, 2)

. Now

x_{1}^{2} = 2^{2} = 4

but

x_{2} = 2

, so

x_{1}^{2} = x_{2}

fails. Since scalar multiplication can move the vector outside the set, it cannot be a subspace.

Why is finding one scaling failure enough to disprove “subspace”?

A subspace must be closed under scalar multiplication. If there exists even one vector in the set and one scalar such that the scaled vector violates the defining condition, closure fails. That single counterexample is sufficient to conclude the set is not a subspace.

Review Questions

For the set in $R^{3}$ defined by $x_{1} = x_{2}$ and $x_{3} = - x_{2}$ , what happens to the defining equalities when you replace $u$ by $λ u$ ?
Take two vectors $u$ and $v$ satisfying $x_{1} = x_{2}$ and $x_{3} = - x_{2}$ . Show directly that $u + v$ also satisfies both constraints.
Why does the condition $x_{1}^{2} = x_{2}$ break closure under scalar multiplication? Use the specific example $(1, 1)$ scaled by $2$ .

Key Points

1
The subspace test uses three checks: zero vector inclusion, closure under scalar multiplication, and closure under addition.
2
For $R^{3}$ , a set defined by linear component relations like $x_{1} = x_{2}$ and $x_{3} = - x_{2}$ typically passes all three subspace checks.
3
To verify scalar closure, start with an arbitrary vector satisfying the constraints and show the scaled vector satisfies the same constraints.
4
To verify addition closure, start with two arbitrary vectors satisfying the constraints and show their component-wise sum still satisfies the defining equations.
5
A single counterexample to scalar multiplication is enough to prove a set is not a subspace.
6
Nonlinear constraints such as $x_{1}^{2} = x_{2}$ commonly fail subspace requirements because scaling changes the left-hand side quadratically.

Highlights

The set in R3 with x1​=x2​ and x3​=−x2​ is confirmed as a subspace by checking zero, scaling, and addition closure.

Scaling preserves the subspace constraints because equalities between components remain equal after multiplying by λ.

The R2 set defined by x12​=x2​ fails under scaling: (1,1) scales to (2,2), where 22=2.

Topics

Subspaces
Closure Properties
Linear Algebra Proofs
Counterexamples
Vector Constraints

Briefing

Cornell Notes

How does one quickly confirm that a constrained set in R3 contains the zero vector?

Why does scalar multiplication preserve membership for the set x1​=x2​, x3​=−x2​?

How does closure under addition work for the same R3 subspace?

What counterexample shows that the set {(x1​,x2​):x12​=x2​} is not a subspace?