# Understanding the implication in linear algebra regarding vectors

Let $$V$$ be a subspace of $$\mathbb{R}^n$$ with the usual dot product, and let $$\mathbf{z}, \mathbf{w} \in V$$ be fixed vectors. If for every $$\mathbf{v} \in V$$ it holds that $$\mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v}$$, then $$\mathbf{z} = \mathbf{w}$$.

I'm trying to understand why inserting the $$\mathbf{v} = 0$$ is not the correct approach. This is considered an implication right, so if I take $$\mathbf{z}, \mathbf{w}$$ as different vectors, then it holds that $$\mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v}$$ is 0 = 0 and the consequent $$\mathbf{z} = \mathbf{w}$$ is false when $$\mathbf{v} = 0$$?

Why is this the wrong approach?

• The key words: "for every!" The zero vector is just one choice. Commented Jun 21 at 13:30
• Seems like a confusion in basic logic. The claim is that if for every $v$ the equality holds, then $z = w$. It does, however, not say that if for some $v$ the equality holds, then $z=w$. That would be a different statement ("for all" versus "there exists"). Commented Jun 21 at 13:31
• This is extremely unclear. What does "inserting the v=0" mean? "if I take z,w as different vectors" What do you mean, "take"? "Why is this the wrong approach?" What is "this approach"? "This approach" to what? PS "if I take z,w as different vectors, then it holds that z⋅v=w⋅v is 0 = 0 and the consequent z=w is false"--"if I take z,w as different vectors then" "z=w is false" always, without using anything in this post. Commented Jun 21 at 23:21

The proposed statement reads: If for every $$\mathbf{v} \in V$$ it holds that $$\mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v}$$, then $$\mathbf{z} = \mathbf{w}.$$

This can be rewritten as follows: Let $$\mathbf{z}, \mathbf{w}\in V.$$ If it holds that $$\mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v}$$ regardless of chosen vector $$\mathbf{v}\in V,$$ then $$\mathbf{z}=\mathbf{w}.$$

Therefore, in order to reach a contradiction, one must be able to come up with two vectors $$\mathbf{z},\mathbf{w}\in V$$ which are not equal (i.e. $$\mathbf{z}\neq\mathbf{w})$$ where $$\mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v}$$ for all $$\mathbf{v}\in V.$$

You suggest there exists a contradiction to this claim because there exist vectors $$\mathbf{z}\neq\mathbf{w}$$ which $$\mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v}$$ for some vector $$\mathbf{v}=\mathbf{0}.$$ But since some vector isn’t all vectors in $$V,$$ you’re conclusion is incorrect.

Does this help?

It is the wrong approach, because inserting $$\mathbf{v} = \mathbf{0}$$ is the not the given condition on which the conclusion is based. The condition is that for all $$\mathbf{v}$$, we have $$\mathbf{v \cdot w } = \mathbf{v \cdot z}$$, not just one vector $$\mathbf{v} = \mathbf{0}$$.

Let $$\mathbf{v}=(x_1,x_2,\cdots,x_n) \in V$$ be any vector such that $$\mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v}$$. Writing, $$\mathbf{z}=(z_1,z_2,\cdots,z_n) \quad \text{and} \quad \mathbf{z}=(w_1,w_2,\cdots,w_n)$$ we obtain $$z_1x_1+z_2x_2+\cdots+z_nx_n=w_1x_1+w_2x_2+\cdots+w_nx_n$$ which implies $$(z_1-w_1)x_1+(z_2-w_2)x_2+\cdots+(z_n-w_n)x_n=0.\tag{1}$$ Since $$\mathbf{v}$$ is arbitrary (and, hence, also $$x_1,x_2,\cdots,x_n \in \mathbb{R}$$, see lemma below) we conclude that $$z_1=w_1, z_2=w_2, \cdots, z_n=w_n$$ that is $$\mathbf{z} = \mathbf{w}$$, as desired.

Lemma. Let $$x_1,x_2,,\cdots,x_n \in \mathbb{R}$$, for some $$n \in \mathbb{N}$$. If $$a_1x_1+a_2x_2+\cdots+a_nx_n=0$$ for all $$a_1,a_2,\cdots, a_n \in \mathbb{R}$$, then $$x_1=x_2=\cdots=x_n=0$$.

Proof. Assume by contradiction that there exists at least one $$i \in \{1,2,\cdots, n\}$$ such that $$x_i \neq 0$$. Then, there exists a choice of coefficients $$a_1, a_2, \cdots, a_n \in \mathbb{R}$$ that makes $$a_1x_1 + a_2x_2 + \cdots + a_nx_n \neq 0$$. This would contradict the given condition that the equation holds for all $$a_1, a_2, \cdots, a_n \in \mathbb{R}$$. Hence, all $$x_1,x_2,\cdots, x_n$$ must be zero. $$\blacksquare$$

If for every $$\mathbf{v} \in V$$ it holds that $$\mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v}$$, then $$\mathbf{z} = \mathbf{w}$$.

In the syntax of mathematical logic, we can write this statement as follows:

$$\left( \forall \mathbf v\in V : \mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v}\right) \implies \left(\mathbf{z} = \mathbf{w}\right).$$

The left-hand side of the implication is the entire statement, "For every $$\mathbf{v} \in V$$ it holds that $$\mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v}.$$" If you wanted to disprove the implication, you would be looking for $$\mathbf z$$ and $$\mathbf w$$ in $$V$$ for which the left-hand side is true -- that is, $$\mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v}$$ for every $$\mathbf v$$ in $$V$$, not just for one $$\mathbf v$$ -- and for which the right-hand side is false, that is, $$\mathbf z\neq\mathbf w.$$

I'm trying to understand why inserting the $$\mathbf{v} = 0$$ is not the correct approach. This is considered an implication right, so if I take $$\mathbf{z}, \mathbf{w}$$ as different vectors, then it holds that $$\mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v}$$ is $$0 = 0$$ and the consequent $$\mathbf{z} = \mathbf{w}$$ is false when $$\mathbf{v} = 0$$?

It's very difficult to detect your train of thought in this paragraph.

• "This is considered an implication ... ." What do you think "this" is? Exactly what is the implication that we are looking at?
• "... so if I take $$\mathbf{z}, \mathbf{w}$$ as different vectors, then it holds that $$\mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v}$$ is $$0 = 0$$ ... ." The word "so" normally signifies that the thing that comes after the word "so" is true because of the thing that comes just before "so". But what you say after "so" is true just because $$\mathbf v = 0$$, not because of the "implication" (whatever you think the "implication" is).
• "... the consequent $$\mathbf{z} = \mathbf{w}$$ is false when $$\mathbf{v} = 0$$?" The statement $$\mathbf{z} = \mathbf{w}$$ (which is indeed the consequent in the implication in the theorem) is false if you take $$\mathbf{z}$$ and $$\mathbf{w}$$ as different vectors, as you did just after the "if" earlier in the same sentence. The way you have set this up, $$\mathbf{z} = \mathbf{w}$$ is false for any value of $$\mathbf v.$$ Yes, it's false "when $$\mathbf v = 0$$"; it's also false when $$\mathbf v \neq 0.$$ So why mention $$\mathbf v$$?

A reason why I questioned what you thought the implication was is that while you clearly identified a consequent, you never identified the antecedent. So I wonder if you thought you were being asked to prove this statement:

$$\forall \mathbf v\in V : \left( \mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v} \implies \mathbf{z} = \mathbf{w}\right).$$

This statement is clearly false, because there exists at least one $$\mathbf v$$ in $$V$$ such that the antecedent $$\mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v}$$ is true but the consequent $$\mathbf{z} = \mathbf{w}$$ is false.

Your second paragraph hints at a disproof of this statement. You found a way to set things up so that $$\mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v}$$ is true and $$\mathbf{z} = \mathbf{w}$$ is false.

But in the statement that you were supposed to prove, there is no implication whose antecedent is $$\mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v}$$. The correct antecedent is $$\forall \mathbf v\in V : \mathbf{z} \cdot \mathbf{v} = \mathbf{w} \cdot \mathbf{v}.$$