How to check this statements about the inner product?

Question

I need help with this problem:

Determine which of the following statements are true or false, justify your answer. If $x$ and $y$ are vectors on $\mathbb{R}^n$, then whe have:

1. $x\cdot y=0$ if and only if $\| x+y\|=\| x-y\|$.
2. $x\cdot y=0$ if and only if $\| x+y\|^2=\| x\|^2+\| y\|^2$.
3. $x\cdot y=0$ if and only if $\| x+cy\|\geq\| x\|$ for all $c\in\mathbb{R}$.
4. $(x+y)\cdot(x-y)=0$ if and only if $\| x\|=\| y\|$

I tried the first one first by assuming that $\| x+y\|=\| x-y\|$, then I rewrote the dot product as $$x^\mathrm{T} y=0=\begin{bmatrix} x_1&\cdots&x_n \end{bmatrix}\begin{bmatrix} y_1\\ \vdots \\y_n \end{bmatrix}=x_1y_1+x_2y_2+\cdots+x_ny_n=0,$$ after that I don't know what to do. The same happens when I triedo to do the other way, can you please help me?

Answer 1

Observe that $\|x+y\|^2 = \|x\|^2+\|y\|^2+2\ x \cdot y.$

So all the options are correct.

For the third one observe that

$$\|x+cy\|^2 = \|x\|^2 + |c|^2 \|y\|^2 + 2c\ x \cdot y.$$

Since $x \cdot y = 0$ we have

$$\|x+cy\|^2 = \|x\|^2 + |c| \|y\|^2 \geq \|x\|^2$$ since $|c|^2\|x\|^2 \geq 0.$

So by taking positive square root in both sides of the above inequality we have

$$\|x+cy\| \geq \|x\|.$$

For the fourth one observe that

$$(x+y) \cdot (x-y) = \|x\|^2 - \|y\|^2.$$

So if $(x+y) \cdot (x-y) = 0$ then we have

$$\|x\|^2 = \|y\|^2.$$

Taking the positive square root in both the sides we have

$$\|x\|=\|y\|.$$

Note $:$ Try to do these things on your own without seeing the spoilers. Otherwise you cannot able to learn mathematics.

Answer 2

To expand on my comment under Dbchatto67's answer, here's the picture I'm picturing:

The light blue line is the line generated by $x + cy$ as $c$ ranges over $\Bbb{R}$. The inequality $\|x + cy\| \ge \|x\|$ says that $x + 0y$ is the point along the line closest to the origin $0$. As you can see in this picture, this is not the case (nor is $x$ and $y$ perpendicular), as the point marked $x + cy$ is closer to $0$ than $x$. As I recommended in my comment, we should find this closest point, using projections.

We should start by projecting the line segment between $0$ and $x$ onto the light blue line. One end of this projection should be $x$, and the other is the point we want (marked $x + cy$). We take the vector from $x$ to $0$ (which is $-x$) and project it onto the vector $y$, which defines the direction of the line. That is, we compute: $$\operatorname{proj}_y(-x) = \frac{(-x) \cdot y}{\|y\|^2} y = -\frac{x \cdot y}{\|y\|^2} y.$$ Adding this vector to $x$ will produce the point we want, marked $x + cy$: $$x - \frac{x \cdot y}{\|y\|^2} y,$$ which is of the desired form.

Basically, what I'm saying is that the most troublesome value of $c$ ought to be $c = -\frac{x \cdot y}{\|y\|^2}$. Let's try applying the inequality $\|x + cy\| \ge \|x\|$ with this particular value of $c$. I'm hoping the results are interesting!

We have \begin{align*} \|x\|^2 &\le \left\|x - \frac{x \cdot y}{\|y\|^2} y\right\|^2 \\ &= \|x\|^2 + \left\|\frac{x \cdot y}{\|y\|^2} y\right\|^2 - 2 x \cdot \left( \frac{x \cdot y}{\|y\|^2} y \right) \\ &= \|x\|^2 + \frac{(x \cdot y)^2}{\|y\|^4}\|y\|^2 - 2 \frac{x \cdot y}{\|y\|^2} x \cdot y \\ &= \|x\|^2 + \frac{(x \cdot y)^2}{\|y\|^2} - 2 \frac{(x \cdot y)^2}{\|y\|^2} \\ &= \|x\|^2 - \frac{(x \cdot y)^2}{\|y\|^2}, \end{align*} and hence, cancelling $\|x\|^2$ from both sides, $$\frac{(x \cdot y)^2}{\|y\|^2} \le 0 \implies (x \cdot y)^2 \le 0$$ But, $(x \cdot y)^2$ is a square number, and must be at least $0$. Thus, $x \cdot y = 0$.