Discover Linear Algebra: A first course in linear algebra

We will show that $(\mathbf{u}\mathbf{\cdot }\mathbf{v}{)}^2\le (\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert {)}^2\text{.}$ Once this is established, then for $\mathbf{u}\mathbf{\cdot }\mathbf{v}$ to have a smaller square than $\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \text{,}$ it must be smaller in magnitude. That is, $(\mathbf{u}\mathbf{\cdot }\mathbf{v}{)}^2\le (\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert {)}^2$ can only be true if $\left|\mathbf{u}\mathbf{\cdot }\mathbf{v}\right|\le \left|\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \right|$ is true. But since neither $\Vert \mathbf{u}\Vert$ nor $\Vert \mathbf{v}\Vert$ can be negative, we have $\left|\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \right|=\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \text{,}$ and so

will be established.

So, we will try to prove that $(\mathbf{u}\mathbf{\cdot }\mathbf{v}{)}^2\le (\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert {)}^2$ is always true for every pair of vectors $\mathbf{u}$ and $\mathbf{v}$ in ${\mathbb{R}}^n\text{.}$ We might as well assume that $\mathbf{v}$ is nonzero, since if it is zero then both $(\mathbf{u}\mathbf{\cdot }\mathbf{v}{)}^2$ and $(\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert {)}^2$ are $0\text{,}$ and the required inequality is true. In the case that $\mathbf{v}$ is nonzero, then also $\Vert \mathbf{v}\Vert \ne 0$ (Statement 1 of Proposition 12.5.1), and we can form the vector

without worry that we’ve accidentally divided by zero. We will find that ${\Vert \mathbf{w}\Vert }^2$ is related to the inequality we are trying to prove, so compute

with justifications

Now, ${\Vert \mathbf{w}\Vert }^2$ cannot be negative, so we have

where multiplying both sides of the second inequality by the nonnegative quantity ${\Vert \mathbf{v}\Vert }^2$ does not change the direction of the inequality.

Because $\mathbf{u}\mathbf{\cdot }\mathbf{v}$ could be negative, we will change our last inequality above to

In words, this inequality says that the square of one number is less than or equal to the square of another number. But when we square two numbers, the bigger number will always result in the bigger square (as long as neither number is negative). Since neither $\left|\mathbf{u}\mathbf{\cdot }\mathbf{v}\right|$ nor $\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert$ can be negative, the bigger number must be $\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert$ to result in a bigger square (or the two numbers could be equal). That is,

We have

with justifications

Now, keep in mind that $\mathbf{u}\mathbf{\cdot }\mathbf{v}$ is a number, and it may be positive, negative, or zero. But every number $x$ satisfies

since if $x$ is positive or zero then the two sides are equal, and if $x$ is negative then obviously the negative number $x$ must be less than the positive number $\left|x\right|\text{.}$ Applying this for $x=\mathbf{u}\mathbf{\cdot }\mathbf{v}\text{,}$ we have $\mathbf{u}\mathbf{\cdot }\mathbf{v}\le \left|\mathbf{u}\mathbf{\cdot }\mathbf{v}\right|\text{,}$ and so

with justifications

Following the chain of equalities and inequalities from beginning to end, we now have

In words, this says that the square of one number is less than or equal to the square of another number. But when we square two numbers, the bigger number will always result in the bigger square (as long as neither number is negative). Since neither $\Vert \mathbf{u}+\mathbf{v}\Vert$ nor $\Vert \mathbf{u}\Vert +\Vert \mathbf{v}\Vert$ can be negative, the bigger number must be $\Vert \mathbf{u}\Vert +\Vert \mathbf{v}\Vert$ to result in a bigger square (or the two numbers could be equal). That is,