# Pythagorean theorem in unitary vector spaces

In euclidean (i.e. real) vector spaces the pythagorean theorem holds, i.e. $$||x+y||^2 = ||x||^2 + ||y||^2 \Leftrightarrow x \perp y.$$ But for unitary (i.e. complex) vector spaces it fails because $$\langle x , y \rangle + \langle y, x \rangle = \langle x , y \rangle + \overline{\langle x, y \rangle} = 2\operatorname{Re}(x,y).$$ And I have a counterexample, let $X = \mathbb{C}$ be the 1-dimensional $\mathbb{C}$-vector space. Consider $x = 1$ and $y = i$. Then $2\operatorname{Re}(x,y) = 0$ but it is not the case that $x \perp y$. But here is my question, sure it is the case that $i$ is perpedicular to $1$, so why it is said that its not? Or is there another notion of perpendicularity in a unitary vector space, one that is not as intuitive? Another example is $1+i$ and $1-i$, when I draw them they are of course perpendicular, but \begin{align*} \langle 1+i,1-i \rangle & = \langle 1,1 \rangle + \langle i,1 \rangle + \langle 1,-i \rangle + \langle i,-i \rangle \\ & = 1 + (-i)\langle 1, 1 \rangle + (-i) \langle 1, 1 \rangle + i^2 \langle 1, 1 \rangle \\ & = 1 - i -i - 1 = -2i \ne 0 \end{align*} (using that $\langle 1,1 \rangle$, guess that holds) so they are not perpendicular?

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"Unitary space" does not usually mean complex vector space. It's an archaic way to refer to what mathematicians today prefer to call a "complex inner product space" or "complex pre-Hilbert space" (possibly with a finite-dimensionality requirement). –  kahen Nov 22 '12 at 17:29

Actually, if you let $\mathbb C$ be a $1$-dimensional space, then you are saying that you are considering as a complex space. Then $$\langle 1,i\rangle = \overline{i}\,\langle 1,1\rangle = \overline{i}=-i.$$ so $1$ and $i$ are not orthogonal to each other. As you showed, the real part of the inner product is zero, but that doesn't imply that the whole inner product is zero.

You say "when I draw them they are perpendicular". But you are assuming that your "picture-perpendicular" and "inner-product" perpendicular are the same thing, which they aren't. If you want to see $1$ and $i$ as perpendicular, you need to consider $\mathbb C$ as a two-dimensional real space, where the inner product will be $\langle x,y\rangle = \text{Re} x\overline{y}$.

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The short answer is that, over $\Bbb C$, $1$ and $i$ are linearly dependent, so geometrically it means they want to be parallel (over $\Bbb C$). Same for $1+i$ and $1-i$. As they are parallel, getting $$\big|\langle 1+i,1-i\rangle\big| = \big|(1+i)(1-i)\big|$$ is not a surprise, if you think $1$ dimensional vectors over $\Bbb R$.

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