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I'm unsure if this is adequate to prove the problem. This is all in $\mathbb{R}^2$ btw.

Proof

Let $X$ be any vector in $\mathbb{R}^2$. We want to prove that if $X \cdot A = O$, then $A=O$ for any vector $X$. Let $X = (x_1,x_2)$ and $A = (a_1,a_2)$. We have that $X \cdot A = x_1a_1+x_2a_2=O$. Because $X$ is any vector while $A$ is fixed and $x_1a_1=0$ and $x_2a_2=0$, then we have that $a_1=0$ and $a_2=0$. Since $a_1,a_2=0$, $A=(0,0)=O$.

End Proof

Edit: After some more thinking, I was wondering if I should add that $A$ is orthogonal to itself meaning $A \cdot A = O$, which says that $a_1a_1 + a_2a_2 = 0$. I don't know how I would incorporate that into the proof if I should add it.

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cypherfry
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  • Just because $x_1a_1 + x_2a_2 = 0$ it does not follow that $x_1a_1 = 0 = x_2a_2$. To guarantee that, you might want a clever choice of $\mathbf{x}$… –  Oct 28 '22 at 03:09

2 Answers2

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We cannot say that $x_1a_1+x_2a_2=O \implies x_1a_1=0$ and $x_2a_2=0$.

We want to show that the zero vector $O$ is orthogonal to every vector, and then prove that it is the only vector to satisfy that property.

Let $V \in \mathbb{R}^2$. It is clear that $$ V\cdot O =v_1\times 0+v_2\times 0=0+0=0 $$ for any values of $V$.

Now, assume there exists another vector $W \in \mathbb{R}^2$ such that $$V \cdot W=0 \implies v_1w_1+v_2w_2=0.$$ is true for all $V$.

Since $W=(w_1,w_2)$ is fixed, the only value that can satisfy the equation for all $V$ is $W=(0,0)=O$

This proves that the vector $O$ is orthogonal to all other vectors and that it is the only vector to do so. It is unique. $\Box$

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If we have $a_1 a_1 + a_2 a_2 =0$, then we can straightforwardly get $a_1 = a_2 = 0$ without involving the statement regarding $X$.

For your proof, I don't think it's adequate enough because I don't see why you can get $x_1 a_1=0$ and $x_2 a_2=0$. If I were you, I would define a function $f(x_1,x_2)=a_1 x_1 + a_2 x_2$ and argue that since $f(x_1,x_2)\equiv 0$ for any pair $(x_1,x_2)$, then we have $f\equiv0$ which means $a_1=a_2=0$.

Imnotarobot
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