Prove there exists $y\neq 0$ but $x\cdot y=0$ I would like to know if I'm missing something with my solution - as an earlier version was wrong and I think I've managed to patch it up - to this problem from Rudin's Principles of Mathematical Analysis. Every solution I've read online always breaks it up into two cases: $k$ even and $k$ odd which makes me feel I might be wrong somewhere as I believe my solution is much simpler, if correct.

Problem 1.18 If $k \geq2$ and $x \in \mathbb{R}^k$, prove that there exists $y \in \mathbb{R}^k$ such that $y \neq 0$ but $x \cdot y =0$. 

Proof: Assume $x \neq 0$ as otherwise any non-zero $y$ would do. If $x = (x_1, \ldots, x_k)$ is non-zero then there is a pair $x_i \text{ and } x_j$ with $1 \leq i < j \leq k$ such that at least one is not zero. Letting $y = (0_1, \ldots, 0_{i-1}, -x_j, 0_{i+1},\ldots,0_{j-1}, x_i, 0_{j+1},\dots,0_k)$ we have $x\cdot y = 0$. 
Thank you. 
 A: Another inductive approach - Induction on $\,k\,$ :
$$\text{For}\;\;\,k=2:\,(0,0)\neq(a,b)\in\Bbb R^2\implies (a,b)\cdot(-b,a)=0\;\;\wedge\;\;(-b,a)\neq(0,0)$$
Assume for $\,k-1\ge1\,$ and show for $\,k\,$: let $\;(a_1,...,a_k)\neq (0,0,...,0)\,$ : 
If $\,\;\exists\,1\leq j\le k-1\;\;s.t.\;\;a_j\neq0\,$ we're done (why?), otherwise it must be that $\,a_k\neq 0\;\wedge\;\;a_j=0\;,\;j=1,2,...,k-1\,$ . Then simply choose $\,(1,0,...,0)\,$
Question: where did the inductive hypothesis kick in the above?
A: Another approach that uses neither coordinates nor $k$ (the dimension):
Assume $x\neq 0$, otherwise any $y$ is a solution.
Because the dimension is two or higher, there must be $z$ such that $z\not\in \mbox{span}(\{x\})$. Let
$$y=z-\frac{x\cdot z}{x\cdot x}x$$
Then $y$ is non-zero (otherwise $z$ would be a multiple of $x$) and
\begin{align}
x\cdot y &= x\cdot(z-\frac{x\cdot z}{x\cdot x}x) \\
&= x\cdot z-\frac{x\cdot z}{x\cdot x}(x\cdot x) \\
&= x\cdot z-x\cdot z \\
&= 0
\end{align}

This proof a simple geometric interpretation: Pick any vector not parallel to $x$ and subtract just enough of $x$ so that the result is perpendicular to it. (I'd like to add an image but I don't know a proper tool to make such image fast and nice.)
