# orthogonal projection: proving that length of the projected vector is less than or equal to the length of the original one

I was thinking about what does the Cauchy-Schwartz inequality tells us doing the following:

CS inequality: $|a\cdot b| \leq ||a||\ ||b||$ (a and b are vectors in R^n)

then I rewrote the dot product $|a\cdot b| = |cos\theta ||a||\ ||b|| |= \frac{||p(a)||}{||a||}||a|| \ ||b|| = ||p(a)|| \ ||b|| \\$

so now the CS ineq. becomes: $||p(a)|| \ ||b|| \leq ||a||\ ||b|| \\$

and

$||p(a)|| \leq ||a||$

Now this seems more obvious to me, that the length of the projected vector is less than or equal to the length of the original one. This is true only for orthogonal projection, right? How would one go about proving this?

I was thinking that because $p(a) = a - a^{\perp}$ I could just write $||a - a^{\perp}|| \leq ||a||$ and since there is something being subtracted from $a$ the quantity is smaller.

(if there's something very wrong with the reasoning above, I would be grateful for clarification)

It is important that the something being subtracted from $a$'' is orthogonal to the remaining vector. Write $$a = a-a^\perp + a^\perp$$ with $a-a^\perp$ and $a^\perp$ orthogonal to each other. Then using orthogonality (Pythagoras!) $$\|a\|^2 = \|a-a^\perp\|^2 +\| a^\perp\|^2\ge \|a-a^\perp\|^2=\|p(a)\|^2.$$