How do I prove $|\left \langle x,y \right \rangle|=\left \| x \right \|\cdot \left \| y \right \|\Leftrightarrow y=cx,c\in F$

Question

Proving $\Leftarrow$ is easy enough, it's just a matter of plugging it right in. For $\Rightarrow$, I tried changing the right side to $\left (\left \langle x,x \right \rangle \cdot\left \langle y,y \right \rangle \right )^{\frac{1}{2}}$, and then having $|\left \langle x,y \right \rangle|=\left \langle x,\left \langle y,y \right \rangle x \right \rangle^{\frac{1}{2}}$, but that looks off to me, and I don't know where to proceed even if it's right.

Answer 1

Define $z:=x-\frac{\left\langle y,\,x \right\rangle}{\left\langle x,\,x \right\rangle}y$. Rearrange $\left\langle z,\,z \right\rangle \geq 0$, with equality iff $z = 0$, to $\left|\left\langle x,\,y \right\rangle\right|^2 \leq \left\langle x,\,x \right\rangle \left\langle y,\,y \right\rangle$. This inherits the above equality condition, which implies $x,\,y$ are parallel (if we regard the zero vector as parallel to all vectors). Strictly speaking, we can violate $y = cx$ with $x=0\neq y$, in which case we have $x = cy$ instead with $c = 0$.

Answer 2

Note: the correct statement you should prove is $$|\left \langle x,y \right \rangle|=\left \| x \right \|\cdot \left \| y \right \|\Leftrightarrow x,y\text{ are not linearly independent}$$ (there is a distinction of cases to be made for $x=0$, for instance)

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A classic approach: Consider the polynomial (in $\lambda$) $$ P(\lambda) = \lVert x+\lambda y\rVert^2 = \langle x+\lambda y, x+\lambda y\rangle = \lVert x\rVert^2 + 2\lambda \langle x,y\rangle + \lambda^2 \lVert y\rVert^2. $$ It is a non-negative quadratic polynomial, so its determinant $\Delta = 4\langle x,y\rangle^2 - 4\lVert x\rVert^2\lVert y\rVert^2$ keeps a constant (non-positive) sign. This gives you Cauchy-Schwarz inequality.

Now, if you have equality above, the polynomial has determinant equal to zero (not strictly negative), so it must have a root... consider this root: it is a value $\lambda^\ast$ such that $0=P(\lambda^\ast) = \lVert x+\lambda^\ast y\rVert^2$.

Answer 3

Let $S=\mathrm{span}\{x\}$ and $P$ the orthogonal projection onto $S$. Then $Py=cx$ for some $c\in F$.

Thus

$\begin{multline}\|Py\|^2=|\langle Py,Py\rangle|=|\langle Py,y\rangle|\\=|\langle cx,y\rangle|=|c\langle x,y\rangle|=|c|\|x\|\|y\|=\|cx\|\|y\|=\|Py\|\|y\|...(*)\end{multline}$

If $Py=0$, done. if $Py\neq 0$ , then by $(*)$ we have $\|Py\|=\|y\|$. Now, by Pythagoras

$$\|y-Py\|^2+\|Py\|^2=\|y\|^2=\|Py\|^2$$

Thus, $\|y-Py\|=0$, or $Py=y$. But $Py=cx$ and done.