Is $\| u + v \|^2 - \| u - v \|^2= 2uv - (-2uv)$?

Question

I'm trying to prove an equation. My professor took up the answer and I copied it down. One of the steps was this (let u and v be vectors and let || u + v || mean the norm / magnitude of u + v):

line $1: \| u + v \|^2 - || u - v ||^2$

line $2: = 2uv - (-2uv)$

line $3: = 4uv$

But it doesn't look correct, even if $uv$ in line $2$ and $3$ is the dot product of $u$ and $v$, right?

Because

$$\| u + v \|^2 = \sum_{i=1}^n(u_i + v_i)^2$$

which is not the same thing as $2uv$ or $2\langle u , v\rangle$. Did I copy it down wrong? What could it have meant? (I'm going to go ask him during office hours but I'm trying to figure it out before I go).

Answer 1

$$\|u+v\|^2=(u+v)\cdot(u+v)=u\cdot u+v\cdot u+u\cdot v+v\cdot v=\|u\|^2+2u\cdot v+\|v\|^2$$Similarly, $$\|u-v\|^2=(u-v)\cdot(u-v)=\|u\|^2-2u\cdot v+\|v\|^2$$ So it is correct.

Here, we used the symmetry of the dot product - that $u\cdot v=v\cdot u$. We also used the distributive rule for vector fields, when expanding the quadratic-type expression.

Answer 2

\begin{align} &\| u+v\|^2 -\|u-v\|^2 \\ &= \|u\|^2 + 2u^Tv + \|v\|^2-(\|u\|^2 - 2u^Tv + \|v\|^2)\\ &=2u^Tv-(-2u^Tv) \end{align}

Your teacher cancels out $\|u\|^2+\|v\|^2$ as it appears in both terms.

Answer 3

If you have a Hilbertspace, then you have $\|x\|^2=xx$ with the dot product.

Hence, $$ \|u+v\|^2=(u+v)(u+v)=uu+uv+vu+vv = \|u\|^2+2uv+\|v\|^2 $$ and $$ \|u-v\|^2=(u-v)(u-v)=uu-uv-vu+vv = \|u\|^2-2uv+\|v\|^2. $$ Together you get \begin{align} \|u+v\|^2-\|u-v\|^2 &= (\|u\|^2+2uv+\|v\|^2)-(\|u\|^2-2uv+\|v\|^2)\\ &=2uv-(-2uv)=4uv. \end{align}