Prove that a function in $\mathbb R^n$ is surjective I have a function $f$: $\mathbb R^n$$\,\to\,$$\mathbb R^n$ defined by $f(\hat{x}) = \hat{x} - 2(\hat{x} \cdot\hat{v})\hat{v}$, with $\mid \hat{v}\mid$ $= 1$. I'm looking to prove that this function is surjective.
I'm a bit rusty on vectors, however, so I'm struggling to solve for $\hat{x}$, to show that for any $y \in \mathbb R^n$ there exists some $\hat{x} \in \mathbb R^n$ such that $f(\hat{x}) = \hat{y}$.
Is there another way to go about proving surjectivity? Or is this it—in which case, what's the right vector manipulation?
Thanks!
 A: So, I don't know that you're expected to recognize this, but if $v\cdot v =1$, then this is the formula for reflecting the vector $x$ in the plane that is perpendicular to $v$. Which means that if you repeat the reflection you get the same vector back again, i.e.
$$ f(f(x)) = x $$
for all vectors $x$. You can verify this by grinding it out. And once you know $f(f(x)) = x$, do you see how you can easily prove surjectivity?
A: Hint: $f$ is linear,  $f(x)=0$ implies that $x=av$ and $av-2(av.v)v=0$ and $a=2a\|v\|^2$ implies $a=2a$ since $\|v\|=1$ we deduce that $a=0$ a linear injective function defined on a finite dimensional vector space is surjective.
A: This explicit derivation may be helpful.  Everywhere it is assume that the vectors are unit length.
$$
\hat{x}-2\left(\hat{x}\cdot\hat{v}\right)\hat{v}=\hat{y} \qquad (1)
$$
$$
\hat{x}=\hat{y}+2\left(\hat{x}\cdot\hat{v}\right)\hat{v}  \qquad (2)
$$
Now, using (1) and taking the dot product with $\hat{v}$:
$$
\hat{x}\cdot \hat{v}-2\left(\hat{x}\cdot\hat{v}\right)\hat{v} \cdot \hat{v}=\hat{y} \cdot \hat{v}
$$
$$
-\left(\hat{x}\cdot\hat{v}\right)=\hat{y} \cdot \hat{v}
$$
Making the substitution in (2):
$$
\hat{x}=\hat{y}-2\left(\hat{y}\cdot\hat{v}\right)\hat{v}
$$
The below figure shows the motivating geometry:

I hope this helps.
A: If you didn't know that this function is a reflection, you could use the fact that it's linear and then show it's null space is just the zero vector.
