### 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.

# linear-algebra

# inner-product

# norm

# vector-space

# orthogonality

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