## Franklin's Notes

### Pythagorean Theorem

The Pythagorean Theorem can be expressed as an algebraic identity of vector norms which are induced by inner products. In an arbitrary inner product vector space $X$, if $x,y\in X$ are orthogonal, then The proof of this fact is straightforward and algebraic - we can just exploit the definition of the norm in terms of the inner product and use bilinearity as follows:

We may also consider the converse statement of the Pythagorean Theorem - that is, if $\lVert x+y\rVert^2 = \lVert x\rVert^2 + \lVert y\rVert^2$, is it necessarily true that $x,y$ are orthogonal in the inner product vector space $X$? The answer, it turns out, is more complicated than it seems. In $\mathbb{R}^n$, the answer is what we would expect:

Proposition 1. In $\mathbb{R}^n$, the converse of the Pythagorean Theorem holds.

Proof. It follows from algebraic manipulation of the inner product that so if $\lVert x+y\rVert^2 = \lVert x\rVert^2 + \lVert y\rVert^2$, it follows by subtracting this quantity from both sides that meaning that $\langle x,y\rangle=0$ and $x,y$ are orthogonal. $\blacksquare$

In $\mathbb{C}^n$, however, the converse does not hold. It is easy to find a counterexample:

Proposition 2. In $\mathbb{C}^n$, the converse of the Pythagorean Theorem does not hold.

Proof. Consider the vectors $x=(1,0,...,0)$ and $y=(i,0,...,0)$. We have that $\lVert x+y\rVert^2 = (1+i)(1-i)=2$ and $\lVert x\rVert^2 = \lVert y\rVert^2 = 1$, so we clearly have that $\lVert x+y\rVert^2 = \lVert x\rVert^2 + \lVert y\rVert^2$. However, $\langle x,y\rangle = i\ne 0$ so $x,y$ are not orthogonal. $\blacksquare$

This counterexample works, but it does not yield much real insight into the fundamental difference between $\mathbb{R}^n$ and $\mathbb{C}^n$ that causes this discrepancy. The real issue is that in $\mathbb{R}^n$, the inner product is symmetric, but in $\mathbb{C}^n$, it is only conjugate-symmetric - so, if we know that for two vectors $x,y$, in $\mathbb{R}^n$ we may conclude that both terms are equal and therefore equal to zero, but in $\mathbb{C}^n$, we only have that they are conjugates of each other, meaning that their sum could equal zero if they are nonzero but purely imaginary. This allows us to pin down when exactly the Pythagorean identity holds in $\mathbb{C}^n$:

Proposition 3. In $\mathbb{C}^n$, holds if and only if $\langle x,y\rangle$ is purely imaginary.