Does every vector of arbitrary dimension have an orthogonal vector?

Question

Suppose I have at my disposal, a non zero vector $v$ of dimension $N>1$ with complex coefficients. $N$ can be any number possible, so long as it makes sense mathematically.

Now I claim, that for any possible $v$ of any possible $N$, I can always find another non zero vector $u$ of dimension $N$ such that they are orthogonal.

$$u•v=0$$

Is my claim true and if not, to what extent is it false ie, only true if dimension is a positive integer less than infinity and/or values of coeficients must be real?

And can I also perhaps have a proof of it if possible?

Answer 1

For $N>1$ this is possible. Suppose $v$ is only non-zero in one coordinate, then let $m$ be such that $v_{m}=0$ and pick $u=(u_{n})$ where $u_{n}=0$ if $n\neq m$ and $n=1$ if $n=m$. If $v$ is non-zero in at least two coordinates let $k,m$ be such that $v_{k}\neq 0$ and $v_{m}\neq 0$. Let $u=(u_{n})$ where $u_{n}=0$ if $n\neq m,k$, $u_{n}=1$ if $n=k$ and $u_{n}=-\frac{v_{k}}{v_{m}}$ if $n=m$.

Answer 2

When $N>1$ is a natural number the answer is yes. First, for simplicity, I'm assuming that by $u\circ v$ you mean the standard inner product: $$u\circ v = \sum_{i=1}^Nu_i\overline{v_i}.$$ Then we can prove this claim by induction. The base case is when $N=2$: Let $(u_1,u_2) \in \mathbb{C}^2$. If $u_2=0$ take $v=(0,1)$; otherwise take $v=(1,\frac{\overline{u_1}}{\overline{u_2}})$. Now, suppose the claim holds for $N$, and let $u = (u_1,\ldots,u_{N+1}) \in \mathbb{C}^{N+1}$. If $(u_1,\ldots,u_N) \neq 0$ then by the assumption there exists some $0 \neq (v_1,\ldots,v_N) \in \mathbb{C}^N$ orthogonal to it in $\mathbb{C}^N$; we may then take $v = (v_1,\ldots,v_N,0)$. Otherwise, we may take $v = (1,0,\ldots,0) \in \mathbb{C}^{N+1}$.

Answer 3

Consider an inner product space $X$ of dimension greater than or equal to $2$ (including, possibly, infinite dimensional), and $v \in X$.

Suppose first that $v = 0$. Then any non-zero $u \in X$ will suffice, and one must exist because otherwise $\dim X = 0$ by definition.

Otherwise, suppose $v \neq 0$. Suppose there wasn't some $w \in X$ such that $\{v, w\}$ was linearly independent. Then there exists some scalar $\alpha$ such that $w = \alpha v$. In other words, $w \in \operatorname{span} \{v\}$, which means $\{v\}$ spans $X$. Since $v \neq 0$, it is linearly independent, and hence is a basis for $X$. That is, $\dim X = 1$, which is a contradiction.

So, for this linearly independent $w$, construct $$u = w - \frac{\langle w, v \rangle}{\langle v, v \rangle}v.$$ Then $$\langle u, v\rangle = \langle w, v \rangle - \frac{\langle w, v \rangle}{\langle v, v \rangle} \langle v, v \rangle = 0.$$

All we need to do is check that $u \neq 0$. If $u = 0$, then this would imply $\{ v, w \}$ is linearly dependent. This cannot be the case, by construction, so $u \neq 0$ and $u \perp v$.