Is it valid to write, $\|x-y\|^2=\|x\|^2+\|y\|^2-2\|x\|\|y\|$?

Question

I dont know if it has been asked before in this forum, but I couldnt find any such thread hence I am asking it here! I am simplifying an expression involving matrices and vectors, in my simplification, I have utilized, $\|x-y\|^2=\|x\|^2+\|y\|^2-2\|x\|\|y\|$. Is it correct way to proceed? If not, then how else can I write $\|x-y\|^2$?

Answer 1

In general, the answer is no.

Assuming the given norm doesn't result from an inner product, we can approach it as next.

The RHS can be rewritten as \begin{align*} \|x\|^{2} + \|y\|^{2} - 2\|x\|\|y\| = (\|x\|-\|y\|)^{2} \end{align*} Consequently, the proposed result holds when \begin{align*} \|x-y\|^{2} = (\|x\| - \|y\|)^{2} \Longleftrightarrow \|x - y\| = |\|x\| - \|y\|| \end{align*} Moreover, according to the triangle inequality, we have that \begin{align*} \|x\| = \|x -y + y\| \leq \|x-y\| + \|y\| \Longrightarrow \|x-y\| \geq \|x\|-\|y\| \end{align*} Similarly, one obtains that $\|x-y\| \geq \|y\|-\|x\|$, that is to say, $\|x-y\| \geq |\|x\|-\|y\||$.

Therefore the proposed relation holds when $\|x-y\|$ attains its minimum value $|\|x\| - \|y\||$.

On the other hand, if it does come from an inner product, we have that \begin{align*} \|x-y\|^{2} = \langle x-y,x-y\rangle & = \langle x,x\rangle - \langle x,y\rangle - \langle y,x\rangle + \langle y,y\rangle\\\\ & = \|x\|^{2} - \langle x,y\rangle - \overline{\langle x,y\rangle} + \|y\|^{2}\\\\ & = \|x\|^{2} - 2\text{Re}\langle x,y\rangle + \|y\|^{2} \end{align*}

Hopefully it helps.

Answer 2

What is true is $$\|x-y\|^2 = \|x\|^2 + \|y\|^2 - 2 \langle x,y \rangle.$$ In general $\langle x, y \rangle \le \|x \| \|y\|$ (Cauchy-Schwarz inequality), but equality only holds if $x$ and $y$ are [positive] scalar multiples of each other.

Answer 3

No. $\vert \vert x-y \vert \vert ^2 = (x-y) \cdot (x-y) = x \cdot x -2x \cdot y + y \cdot y = \vert \vert x \vert \vert ^2 -2x \cdot y + \vert \vert y \vert \vert ^2 $