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

• Hall
In summary, we have shown that in a real Euclidean space, if we are given the condition that ##||x+cy||^2 \geq ||x||^2## for all real ##c##, then we can conclude that ##\langle x, y \rangle = 0##. This is proven by analyzing the quadratic function ##q(c) = \|y\|^2c^2 + 2c\langle x,y\rangle## and determining that in order for ##q(c)## to be non-negative for all ##c##, the only possible value for the other root (besides ##c=0##) is when ##\langle x, y \rangle = 0##. Thisf

#### 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?

$$c^2 ||y||^2 + 2c \langle x, y \rangle \geq 0$$
That is a function of ##c##, so you could use calculus.

That is a function of ##c##, so you could use calculus.
Differentiating with respect to ##c##,
$$c ||y||^2 + \langle x,y \rangle \geq 0$$

Differentiating with respect to ##c##,
$$c ||y||^2 + \langle x,y \rangle \geq 0$$
That's not like any calculus I've seen.
$$f(c) \ge 0 \ \Rightarrow \ f'(c) \ge 0 \ (\text{?})$$

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.

PeroK
That is a function of ##c##, so you could use calculus.
Can you please amplify the hint a little more?

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

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?

Infrared
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?

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

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?

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!

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).

$$\frac{ 2 \langle x,y\rangle } {||y||^2}=0$$
$$\langle x, y \rangle =0$$
I don't get this at all. How do you conclude that ##\langle x, y \rangle = 0##?

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.

@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##.

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##.

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##.

Last edited:
PeroK