Help understanding why the dot product of two unit ≤ 1

Question

I am a 16 year old High school student and I have just started using Gilbert Strang's Linear algebra book. I have reached the second problem set in the first chapter which is about dot products and angles between vectors and I am stuck on this particular question rather I don't understand. I have tried thinking about this for some time and I can't wrap my head around this proof.

24. One line proof of the $|u \cdot U| \le 1$ for unit vectors $u = (u_1, u_2)$ and $U = (U_1, U_2)$

$|u\cdot U| ≤ |u_1|\cdot |U_1| + |u_2|\cdot |U_2| ≤ \frac{u_1^2 + U_1^2}{2} + \frac{u_2^2 + U_2^2}{2} = 1$

I understand $|u \cdot U| \le |u_1| \cdot |U_1| + |u_2| \cdot |U_2|$ but I dont understand the $\frac{u_1^2 + U_1^2}{2} + \frac{u_2^2 + U_2^2}{2} = 1$ part of the inequality.

Answer 1

This follows from $2ab ≤ a^2 + b^2$, which follows from $(b-a)^2≥0$. Specifically, you use $a = |u_i|$ and $b=|U_i|$ to see that $$ |u_i||U_i| ≤ \frac{1}{2} ( |u_i|^2 + |U_i|^2 )$$

The combination is 1 because \begin{align} |u_1||U_1| + |u_2||U_2| & ≤ \frac{1}{2} ( \color{red}{|u_1|^2} + \color{blue}{|U_1|^2} ) + \frac{1}{2} ( \color{red}{|u_2|^2} + \color{blue}{|U_2|^2} ) \\&= \frac{1}{2} ( \color{red}{|u_1|^2 + |u_2|^2} ) + \frac{1}{2} ( \color{blue}{|U_1|^2 + |U_2|^2} ) \\&= \frac12 \|u\|^2 + \frac12 \|U\|^2 \\&= \frac12 + \frac 12 \\&= 1,\end{align} where $\|u\|=1$ and $\|U\|=1$ are the vector norms of $u$ and $U$.

Answer 2

For a dot product, $$\left\lvert {\bf a} \cdot {\bf b}\right\rvert = \left\lvert {\bf a} \right\rvert \left\lvert {\bf b}\right\rvert \cos \theta$$

Since $\cos \theta$ is always between $0$ and $1$ for interior angles, and since we are working with unit vectors, we know that we know that

$$\left\lvert {\bf a} \cdot {\bf b}\right\rvert = \cos\theta \le 1$$

For $\hat{\bf u}$ and $\hat{\bf U}$ to be unit vectors (the hat helps us remember that), the following must be true: $$u_1^2+u_2^2=1=\left\lvert \hat{\bf u}\right\rvert \qquad U_1^2+U_2^2=1=\left\lvert\hat{\bf U}\right\rvert$$

Since $1=1$, we can say that $$u_1^2+u_2^2=U_1^2+U_2^2 =1$$ Dividing by $2$ gives $$\frac{u_1^2+u_2^2}{2}=\frac{U_1^2+U_2^2}{2} =\frac12$$ so of course $$\frac{u_1^2+u_2^2}{2}+\frac{U_1^2+U_2^2}{2} =1$$

Putting all of this together gives us your (in)equalities.

Answer 3

In this context, it looks like they are using the AM-GM inequality.

$(|u_1| - |U_1|)^2 \ge 0\\ u_1^2 + U_1^2 - 2|u_1||U_1| \ge 0\\ \frac {u_1^2 + U_1^2}2 \ge |u_1||U_1|$

Answer 4

This comes from the inequalities geometric mean $\le $ arithmetic mean $\le$ quadratic mean: $$\sqrt{uU}\le \dfrac{u+U}2\le \sqrt{\frac{u^2+U^2\strut}2}.$$