Let A be a vector perpendicular to every vector X. Show that A = 0. Let A be a vector perpendicular to every vector $X$. Show that $A = 0$.
Suppose $A = X$.
$$A \cdot X = 0 \implies  a_1^2 + \cdots + a_n^2 = 0.$$
Since the square of any real number is nonnegative and the only sum of nonnegatve reals that equals $0$ is the sum of $0$s, $a_i^2 = 0$. Then, $a_i = 0$. So, $A = 0$.
How do we prove it for X $ \neq $ A? Or is there a better proof for this statement that doesn't assume the knowledge of matrices and anything even more advanced? Thanks. 
 A: Hint: Firstly, take in the place of $x$ the basis vectors $e^j$ with $$e^j_i=\begin{cases}1, & i=j \\0, & i \neq j\end{cases}=(0,0,0,\ldots,0\underbrace{1}_{j-\text{th position}},0,
\ldots,0)$$ Now A is perpendicular to every $e^j$. What does this imply? 
Write, subsequently every vector $x=(x_1,\ldots, x_n)$ as a linear combination of the $e^j$, that is $$x=\sum_{j=1}^{n}x_je^j$$ to conclude.
A: Your proof is complete.  There is no need to prove anything for $X\ne A$.  It's not like you were supposed to prove something for all $X$ and only proved it for $X=A$.  Quite the opposite: the hypothesis told you something was true for all $X$, and you only used it for one particular $X$.  That's not incomplete, it's thrifty.
A: Let $A=(a_1,a_2,....,a_n)$
and $X = (x_1,x_2,...,x_n)$
Then $A\cdot X=a_1x_1+a_2x_2+.....a_nx_n = 0$
Now, suppose there is $Y=(2x_1,x_2,x_3,....x_n)=(x_1,x_2,x_3,....x_n)+(x_1,0,0,...0)$
Then $A\cdot Y=(a_1x_1+a_2x_2+.....a_nx_n)+(a_1x_1,0,0....,0)=A\cdot X+(a_1x_1,0,...,0)$
i.e. $A\cdot Y=0=0+a_1x_1$
As $x_1$ is arbitrary, we have
$a_1=0$
Similarly show this for other $a_i$'s
