Linear-Algebra-Done-Right · Unit 2 · Video 2 · Interactive Practice

Unique Zero and Inverses, 0v = 0, and the Subspace Test

IKey Formulas

Formula	Name	Meaning
$0' = 0' + 0 = 0 + 0' = 0$	Unique additive identity	Two additive identities of $V$ must be equal
$w = w + (v + w') = (w + v) + w' = 0 + w' = w'$	Unique additive inverse	So $-v$ is well defined, and $w - v$ means $w + (-v)$
$0v = 0$ , $a0 = 0$ , $(-1)v = -v$	Scalars and zero	Each proof uses a distributive property, the only axiom linking addition with scalar multiplication
$0 \in U$ ; $u, w \in U \Rightarrow u + w \in U$ ; $a \in \mathbf{F},\ u \in U \Rightarrow au \in U$	Subspace test	A subset $U \subseteq V$ is a subspace of $V$ if and only if all three hold

Key Insight: The subspace test never asks for additive inverses: if $u \in U$ , then $-u = (-1)u$ , which lies in $U$ by closure under scalar multiplication.

IIWhich Axioms Does Each Proof Use?

Every proof that links scalar multiplication with vector addition needs a distributive property, the only axiom connecting the two.

Step 0

IIIThe Subspace Test in $\mathbf{R}^2$

A subset $U$ of $\mathbf{R}^2$ is a subspace exactly when it contains $0$ and is closed under addition and scalar multiplication.

Subset U Scalar a −1

💡 Challenge: for each set that is not a subspace, find $u$ , $w$ and $a$ that break one of the three conditions. A single failure proves $U$ is not a subspace; no number of successes proves that it is.

IVQuiz Questions

Problem 1 · Which Zero Is Which?

Given: $V$ is a vector space over $\mathbf{F}$ , $v \in V$ and $a \in \mathbf{F}$ . Identify the two zeros in each identity.

In 0v = 0

In a0 = 0

✅ Correct! In

0v = 0

a scalar multiplies a vector, so the left zero is in

\mathbf{F}

; in

a0 = 0

the scalar is

a

, so the zero it multiplies is a vector. A scalar times a vector is always a vector, so both right-hand zeros are in

V

❌ Check the right zero. It is the value of the left-hand side, a scalar times a vector, and scalar multiplication produces an element of

V

, never of

\mathbf{F}

❌ Check the left zero of $0v$ . It multiplies the vector

v

, so it must be a scalar; the product is a vector.

❌ Check the left zero of $a0$ . The scalar is

a

, so the zero after it is the thing being multiplied: a vector.

Show solution

Scalar multiplication is a function $\mathbf{F} \times V \to V$ : the first factor is a scalar, the second a vector, and the result a vector.

In $0v$ , the second factor is $v \in V$ , so the first factor $0$ is the scalar $0 \in \mathbf{F}$ . The result $0v$ is in $V$ , so the right side is the additive identity of $V$ .

In $a0$ , the first factor is the scalar $a \in \mathbf{F}$ , so the second factor $0$ is the additive identity of $V$ , and again the result is the additive identity of $V$ .

Problem 2 · The Proof of a0 = 0

Given: the proof of $a0 = 0$ for $a \in \mathbf{F}$ begins $a0 = a(0 + 0) = a0 + a0$ . Justify each of the two equalities.

Why is a0 = a(0 + 0)?

Why is a(0 + 0) = a0 + a0?

✅ Correct! Here the zero being split is the vector

0

, so the distributive property that applies is the one over a sum of vectors,

a(u + v) = au + av

, with

u = v = 0

❌ That is the proof of $0v = 0$ . In

a0

the zero is a vector, so the splitting

0 = 0 + 0

happens in

V

and is justified by the additive identity.

❌ Not quite. The expression

a(0 + 0)

comes from replacing the vector

0

0 + 0

. Which axiom says

0 + 0 = 0

V

❌ That is the distributive property used for $0v = 0$ . It splits a sum of scalars; here the sum

0 + 0

is a sum of vectors.

❌ Not quite. That axiom concerns a product of two scalars. The step turns a scalar times a sum into a sum of scalar multiples.

❌ Check the axiom. The step turns a scalar times a sum of vectors into a sum of scalar multiples.

Show solution

Since $0$ is the additive identity of $V$ , $0 + 0 = 0$ in $V$ , so $a0 = a(0 + 0)$ . The distributive property $a(u + v) = au + av$ with $u = v = 0$ gives $a(0 + 0) = a0 + a0$ .

Adding $-(a0)$ to both sides of $a0 = a0 + a0$ and using associativity, the additive inverse and the additive identity:

0 = a0 + \big(a0 + (-(a0))\big) = a0 + 0 = a0.

The proof of $0v = 0$ runs the same way but splits the scalar, $0 = 0 + 0$ in $\mathbf{F}$ , and uses the other distributive property $(a + b)v = av + bv$ .

Problem 3 · Apply the Subspace Test

Given: $U = \{(x_1, x_2, x_3) \in \mathbf{R}^3 : x_1 = 2x_3 \text{ and } x_2 = x_3^2\}$ . Which conditions of the subspace test fail?

✅ Correct!

0 \in U

(take

x_3 = 0

), but

(2, 1, 1) + (2, 1, 1) = (4, 2, 2)

and

2(2, 1, 1) = (4, 2, 2)

both leave

U

, since

2 \ne 2^2

❌ Not quite. The condition

x_2 = x_3^2

is not linear. Test it on

(2, 1, 1) + (2, 1, 1)

❌ Not quite.

(0, 0, 0)

satisfies

0 = 2 \cdot 0

and

0 = 0^2

, so

0 \in U

. The failures are elsewhere.

❌ Close, but… one counterexample was found, but the other closure condition fails too. Check both

(2, 1, 1) + (2, 1, 1)

and

2(2, 1, 1)

❌ Not quite. Every element of

U

has the form

(2t, t^2, t)

; test each condition on that form.

Show solution

Every element of $U$ has the form $(2t, t^2, t)$ with $t \in \mathbf{R}$ .

Additive identity: $t = 0$ gives $(0, 0, 0) \in U$ . ✓

Addition: $(2t, t^2, t) + (2s, s^2, s) = \big(2(t + s),\ t^2 + s^2,\ t + s\big)$ , which lies in $U$ only if $t^2 + s^2 = (t + s)^2$ , that is $2ts = 0$ . With $t = s = 1$ :

(2, 1, 1) + (2, 1, 1) = (4, 2, 2), \qquad 2 \ne 2^2 = 4,

so $U$ is not closed under addition. ✗

Scalar multiplication: $a(2t, t^2, t) = (2at, at^2, at)$ lies in $U$ only if $at^2 = a^2t^2$ . With $a = 2$ , $t = 1$ : $2(2, 1, 1) = (4, 2, 2) \notin U$ . ✗

Both closure conditions fail, so $U$ is not a subspace.

Problem 4 · An Unusual Vector Space

Given: $V = \mathbf{R}$ is a vector space over $\mathbf{R}$ with "addition" $u \oplus v = u + v + 3$ and "scalar multiplication" $\lambda \odot v = \lambda v + 3(\lambda - 1)$ . Use $0v = 0$ and $(-1)v = -v$ inside this space to find its zero vector and the vector $(-1) \odot v$ .

The zero vector of V

The vector (−1) ⊙ v

✅ Correct!

0 \odot 7 = 0 + 3(0 - 1) = -3

is the zero vector, and

(-1) \odot v = -v + 3(-2) = -v - 6

, which is the additive inverse of

v

here:

v \oplus (-v - 6) = -3

❌ That is the ordinary zero. Here

v \oplus 0 = v + 3 \ne v

. By

0v = 0

, the zero vector is

0 \odot v

for any

v

❌ Not quite.

v \oplus 3 = v + 6 \ne v

. Compute

0 \odot 7 = 0 \cdot 7 + 3(0 - 1)

❌ Not quite.

v \oplus (-6) = v - 3 \ne v

. Compute

0 \odot 7 = 0 \cdot 7 + 3(0 - 1)

❌ Check the zero vector. By

0v = 0

, it equals

0 \odot v = 3(0 - 1)

for every

v

❌ That is the ordinary negative. Here

v \oplus (-v) = 3

, which is not the zero vector of this space. Compute

(-1) \odot v

from the definition.

❌ Close, but… with

\lambda = -1

the constant term is

3(\lambda - 1) = 3(-2) = -6

, not

-3

❌ Check the sign.

(-1) \odot v = (-1)v + 3(-1 - 1)

: the

v

term is

-v

❌ Check the computation. Substitute

\lambda = -1

into

\lambda v + 3(\lambda - 1)

Show solution

Zero vector. By $0v = 0$ , the zero vector of $V$ is $0 \odot v$ for any $v \in V$ ; with $v = 7$ ,

0 \odot 7 = 0 \cdot 7 + 3(0 - 1) = -3.

Check: $v \oplus (-3) = v + (-3) + 3 = v$ for every $v$ . By uniqueness of the additive identity, $-3$ is the only zero vector. (Also $\lambda \odot (-3) = -3\lambda + 3\lambda - 3 = -3$ , which is $a0 = 0$ in this space.)

The vector $(-1) \odot v$ . From the definition,

(-1) \odot v = (-1)v + 3(-1 - 1) = -v - 6.

By $(-1)v = -v$ , this is the additive inverse of $v$ in $V$ . Indeed

v \oplus (-v - 6) = v + (-v - 6) + 3 = -3,

the zero vector. The ordinary negative fails: $v \oplus (-v) = 3$ .

Solved: 0 / 4

Unique Zero and Inverses, 0v = 0, and the Subspace Test

IKey Formulas

IIWhich Axioms Does Each Proof Use?

IIIThe Subspace Test in R2\mathbf{R}^2R2

IVQuiz Questions

IIIThe Subspace Test in $\mathbf{R}^2$