## Pythagorean theorem

Hi Sir,

I’ve got a question on Math2201, Further Linear Algebra.

On page 52 of your online notes, there was a proof of Pythagoras’ Theorem. Is this Theorem only on a Real Vector Space? Because if it is not, when you get to the part where you have to equate Re() = 0, you cannot conclude that = 0. For example, if you are in a complex inner product space, say. Take v = (2-i,2) and w = (1,-1). These 2 vectors will satisfy Pythagoras’ Theorem but their inner product will not be zero.

Shi Hui Woon.

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Reply:

This is a good correction. I’ve changed it now, so that it reads: If and are orthogonal, then

That is, the orthogonality is a sufficient condition in general. As you point out, it is also a necessary condition when we are working in a real inner-product space. In the complex case, the necessary and sufficient condition to have the equality above is , as the proof makes clear.

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