Prove that a function in $\mathbb R^n$ is surjective

Question

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!

Answer 1

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?

Answer 2

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.

Answer 3

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.

Answer 4

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.