Prove that ##\langle x, y \rangle = 0 \iff ||x + cy|| \geq ||x||##.

[Hall]

Homework Statement

Prove that $\langle x, y \rangle = 0 \iff ||x =cy|| \geq ||x||$ for all real c.

Relevant Equations

A few of them.

(We are working in a real Euclidean space) So, we have to show two things: (1)the arrow goes from left to right, (2) the arrow comes from right to left.

(1) if we're given $\langle x, y \rangle = 0$
$$
|| x+ cy||^2 = \langle x,x \rangle + 2c\langle x,y\rangle +c^2 \langle y,y \rangle $$
$$
||x+cy||^2 = ||x||^2 + c^2||y||^2 +2c\langle x,y\rangle$$
$$||x+cy||^2 = ||x||^2 || + c^2||y||^2$$
As $c^2 ||y||^2 \geq 0$, we have $||x+cy||^2 \geq ||x||^2$.

(2) If we're given that $||x+cy||^2 \geq ||x||^2$, all that we can do is expand $||x+cy||^2$ and proceed
$$
||x||^2 +c^2 ||y||^2 + 2c \langle x, y \rangle \geq ||x||^2 $$
$$
c^2 ||y||^2 + 2c \langle x, y \rangle \geq 0$$

But how to conclude that $\langle x, y\rangle = 0$. This is one of the archetypical case when we have to move backwards in mathematics.

All I can do is to say, if $c^2 ||y||^2 + 2c \langle x, y \rangle \geq 0$ has to be true for all real c, the case may arise for some negative c such that $2c \langle x, y \rangle$ may exceed $c^2 ||y||^2$ and so to rule out that possibility we must make $\langle x, y \rangle = 0$. What do you say about that?

 

[PeroK]

$$
c^2 ||y||^2 + 2c \langle x, y \rangle \geq 0$$

That is a function of $c$, so you could use calculus.

 

[Hall]

That is a function of $c$, so you could use calculus.

Differentiating with respect to $c$,
$$
c ||y||^2 + \langle x,y \rangle \geq 0$$

 

[PeroK]

Differentiating with respect to $c$,
$$
c ||y||^2 + \langle x,y \rangle \geq 0$$

That's not like any calculus I've seen.
$$f(c) \ge 0 \ \Rightarrow \ f'(c) \ge 0 \ (\text{?})$$

 

[Hall]

That's not like any calculus I've seen.
$$f(c) \ge 0 \ \Rightarrow \ f'(c) \ge 0 \ (\text{?})$$

Oh! I simply differentiated both the sides, taking inequality, by mistake, as the equality.

 

[Hall]

That is a function of $c$, so you could use calculus.

Can you please amplify the hint a little more?

 

[pasmith]

$$
c^2 ||y||^2 + 2c \langle x, y \rangle \geq 0$$

But how to conclude that $\langle x, y\rangle = 0$.

Let $q(c) = \|y\|^2c^2 + 2c\langle x,y\rangle$. Now $q$ is a quadratic in $c$ with real coefficients. The coefficient of $c^2$ is positive, so the condition that $q(c) \geq 0$ for all $c$ is equivalent to the condition that $q$ has at most one distinct real root. It is obvious that $c = 0$ is a root. What is the other?

 

[PeroK]

Can you please amplify the hint a little more?

You know that $f(c) \ge 0$ for all $c$. Why not do an analysis of $f(c)$ regarding turning points, maxima and minima?

 

[Hall]

Let $q(c) = \|y\|^2c^2 + 2c\langle x,y\rangle$. Now $q$ is a quadratic in $c$ with real coefficients. The coefficient of $c^2$ is positive, so the condition that $q(c) \geq 0$ for all $c$ is equivalent to the condition that $q$ has at most one distinct real root. It is obvious that $c = 0$ is a root. What is the other?

$$
\frac{ 2 \langle x,y\rangle } {||y||^2}=0$$
$$
\langle x, y \rangle =0$$

 

[Hall]

You know that $f(c) \ge 0$ for all $c$. Why not do an analysis of $f(c)$ regarding turning points, maxima and minima?

There is some c such that f'(c)=0. But our function could be strictly increasing?

 

[PeroK]

There is some c such that f'(c)=0.

We have a quadratic in $c$ and can find the minimum by calculus (or by quadratic methods). But, we know that the minimum of $f(c)$ must be greater than or equal to zero. I'm hoping that tells us something about $\langle x, y \rangle$.

Never forget the basics!

 

[Hall]

We have a quadratic in $c$ and can find the minimum by calculus (or by quadratic methods). But, we know that the minimum of $f(c)$ must be greater than or equal to zero. I'm hoping that tells us something about $\langle x, y \rangle$.

Never forget the basics!

Yes, got that (post #9).

 

[PeroK]

$$
\frac{ 2 \langle x,y\rangle } {||y||^2}=0$$
$$
\langle x, y \rangle =0$$

I don't get this at all. How do you conclude that $\langle x, y \rangle = 0$?

 

[Hall]

I don't get this at all. How do you conclude that $\langle x, y \rangle = 0$?

@pasmith that said it has two two equal roots and one of them we know is zero, so, the other one must also be equal to zero.

 

[PeroK]

@pasmith that said it has two two equal roots and one of them we know is zero, so, the other one must also be equal to zero.

$$
\frac{ 2 \langle x,y\rangle } {||y||^2}=0$$
$$
\langle x, y \rangle =0$$

Okay, but I would have said that the other root is $-\frac{ 2 \langle x,y\rangle } {||y||^2}$. And, if we need that root to be zero, then $\langle x,y\rangle = 0$.

 

[Hall]

Okay, but I would have said that the other root is $-\frac{ 2 \langle x,y\rangle } {||y||^2}$. And, if we need that root to be zero, then $\langle x,y\rangle = 0$.

Quite the same logic of qudratics is used in this question:

It is given that $\langle x, x \rangle =0 \iff x=0$. Prove that either $\langle x,x \rangle \gt 0$ or $\langle x, x \rangle \lt 0$ for all $x \neq 0$.

 

[Orodruin]

There is a different argument that gives the result rather quickly:

We have $c^2 ||y||^2 + 2c\langle x,y\rangle \geq 0$ for all $c$. Assume that unless $\langle x,y\rangle \neq 0$. For small enough $c$, the first term in the inequality will be negligible compared to the second, which will therefore determine the sign of the LHS. However, as the sign of $c$ is arbitrary, we cannot have $c\langle x,y\rangle \geq 0$ for both positive and negative $c$, leading to a contradiction. Hence, the assumption that $\langle x,y\rangle \neq 0$ is false and therefore $\langle x,y\rangle = 0$.