Does $(v - v_0) \perp v_0 \implies \|v_0\| \leq \|v\|$

Question

Sort of stuck on this fine point for a while now:

Suppose we have some vector space $V$ and some $v_0, v \in V$, such that $(v - v_0) \perp v_0$

Does $(v - v_0) \perp v_0 \implies \|v_0\| \leq \|v\|$ hold?

Answer 1

Finally figured this question out

Suppose $v-v_o \perp v_o$

Then $\|v\|^2 = \|(v - v_o) + v_o \|^2 = \|v-v_o\|^2 + \|v_o\|^2$

Note that in this case we have equality, this is a special case of the triangle inequality that all norms have to satisfy. This is only possible since $v-v_o \perp v_o$. Using equivalence between $\|\cdot\|^2$ and $<\cdot, \cdot>$ this can be shown. Actually this is called the pythagoras theorem but generalized to higher dimensional vectors.

Therefore, since all norms are positive (or 0), $\|v-v_o\|^2 \geq 0$, hence $\|v\|^2 \geq \|v_o\|^2$ and the conclusion follows by taking square root of both sides.

Answer 2

Yes it holds based on Pythagoras. The three vectors must create a right angle triangle in which $v$ is the hypotenuse.