Linear Algebra: Orthogonal Theorem presented in two different ways. Why and how is it possible?

Question

Book says:

According to the Pythagorean Theorem, two vectors are perpendicular if and only if:

Then, later is says: Let $u$ and $v$ be vectors in $\Bbb R^n$. Then u and v are orthogonal if and only if:

What I don't understand is how $\|v-u\|^2$ is the same as $\|u + v\|^2$. Why does the book present orthogonal in two different yet similar formulas?

Answer 1

Just look at the dot product version: We call two vectors $u$ and $v$ orthogonal if $u_1v_1+u_2v_2 = 0$.

This, however is true precisely if $(-u_1)v_1+(-u_2)v_2=0$, that is, if $-u$ and $v$ are orthogonal.

In other words, while in general $\|u-v\|^2\neq\|u+v\|^2$ it turns out that $\|u-v\|^2 = \|u\|^2+\|v\|^2$ precisely when $\|u+v\|^2=\|u-(-v)\|^2=\|u\|^2+\|-v\|^2=\|u\|^2+\|v\|^2$.

Actually, as a side remark, $\|u+v\|^2=\|u-v\|^2$ could serve as an alternative definition of orthogonality.

Answer 2

They are justifying the definition of orthogonality as $u\cdot v=0$.

With that definition, and the knowledge that $\|x\|^2=x\cdot x$, that computation can be rewritten:

$$ \|x-y\|^2=(x-y)\cdot(x-y)=x\cdot x-2(x\cdot y)+y\cdot y=\|x\|^2 -2(0)+\|y\|^2=\|x\|^2 +|y\|^2 $$

If you performed a similar computation with $\|x+y\|^2$:

$$ \|x+y\|^2=(x+y)\cdot(x+y)=x\cdot x+2(x\cdot y)+y\cdot y=\|x\|^2 +2(0)+\|y\|^2=\|x\|^2 +|y\|^2 $$

Draw the pictures of the vectors $x,y,x-y,x+y$ and slide them around to form the right trangles we are talking about.

Answer 3

In the image below, the first equation asks for the blue triangle to be right. The second equation asks for the pink triangle to be right. Since u and v are on the sides of both these triangles, this is all equivalent to $u \bot v$.