The Pythagorean Theorem for Inner Product Spaces Examples 1

# The Pythagorean Theorem for Inner Product Spaces Examples 1

Recall from The Pythagorean Theorem for Inner Product Spaces page that if $V$ is an inner product space and if $u, v \in V$ are such that $u$ and $v$ are orthogonal to each other, that is, $<u, v> = 0$, then:

(1)
\begin{align} \quad \| u + v \|^2 = \| u \|^2 + \| v \|^2 \end{align}

We will now look at some examples regarding the Pythagorean theorem for inner product spaces.

## Example 1

Let $V$ be an inner product space. Let $u, v \in V$. Prove that $u$ and $v$ are orthogonal if and only if $\| u \| ≤ \| u + c v \|$ for every $c \in \mathbb{F}$.

$\Rightarrow$ Suppose that $u$ and $v$ are orthogonal to each other. Then $<u, v> = 0$. Therefore by the Pythagorean theorem we have that:

(2)
\begin{align} \quad \| u + cv \|^2 = \| u \|^2 + \| cv \|^2 \end{align}

Clearly we have that $\| u \|^2 ≤ \| u \|^2 + \| cv \|^2 = \| u + cv \|^2$, and in squaring both sides we get that $\| u \| ≤ \| u + cv \|$.

$\Leftarrow$ Suppose that $\| u \| ≤ \| u + cv \|$ for every $c \in \mathbb{F}$. If we square both sides of this inequality then we have that:

(3)
\begin{align} \quad \| u \| ≤ \| u + cv \| \\ \quad \| u \|^2 ≤ \| u + cv \|^2 \\ \quad \| u \|^2 ≤ <u + cv, u + cv> \\ \quad \| u \|^2 ≤ <u, u> + <u, cv> + <cv, u> + <cv, cv> \\ \quad \| u \|^2 ≤ \| u \|^2 + \bar{c} <u, v> + c<v, u> + c\bar{c} <v, v> \\ \quad \| u \|^2 ≤ \| u \|^2 + \bar{c} <u, v> + c\overline{<u, v>} + \mid c \mid^2 \| v \|^2 \\ \quad \| u \|^2 ≤ \| u \|^2 + 2 \Re (\bar{c} <u, v>) + \mid c \mid^2 \| v \|^2 \\ \quad 0 ≤ 2 \Re (\bar{c} <u, v>) + \mid c \mid^2 \| v \|^2 \\ \quad -2 \Re (\bar{c} <u, v>) ≤ \mid c \mid^2 \| v \|^2 \end{align}

Now let $c = -b<u, v>$ where $b > 0$. Then we have that:

(4)
\begin{align} \quad -2 \Re (\overline{-b<u, v>} <u, v>) ≤ \mid -b<u, v> \mid^2 \| v \|^2 \\ \quad 2b \overline{<u, v>} <u, v> ≤ b^2 \mid <u, v> \mid^2 \| v \|^2 \\ \quad 2b \mid <u, v> \mid^2 ≤ b^2 \mid <u, v> \mid^2 \| v \|^2 \\ \quad 2 \mid <u, v> \mid^2 ≤ b \mid <u, v> \mid^2 \| v \|^2 \end{align}

If $v = 0$ then then we automatically have that $<u, v> = 0$. If $v \neq 0$, then let $b = \frac{1}{\| v \|^2}$ to get that:

(5)
\begin{align} \quad 2 \mid <u, v > \mid^2 ≤ \mid <u, v> \mid^2 \end{align}

Therefore $\mid <u, v> \mid^2 = 0$ so $<u, v> = 0$.