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- anonymous

If a matrix U has vectors that are orthonormal, on an intuitive level why is the following true?
Transpose(U) times U = I (identity matrix)

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- anonymous

- schrodinger

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- anonymous

\(\huge\color{green}{Please~don't~hesitate~to~ask~questions!~:)}\)

- anonymous

\(\huge\color{green}{Please~don't~hesitate~to~ask~questions!~:)}\)

- anonymous

\(\huge\color{green}{Please~don't~hesitate~to~ask~questions!~:)}\)

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- anonymous

\(\Large\color{Red}{Please~don't~hesitate~to~ask~questions!~:)}\)

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(\Large\color{green}{Please~don't~hesitate~to~get~fluttered~in~the~retrice~:)}\)

- zzr0ck3r

https://en.wikipedia.org/wiki/Permutation_matrix

- Empty

An intuitive reason why a matrix full of vectors that are all orthonormal is that each element will be the dot products of each of the vectors together, and the only time they'll be 1 is when they're the same. So check this out:
|dw:1442544145805:dw| Hahaha my matrix got cut off here, but since the vectors are all orthonormal, you can see how this will be a diagonal matrix. If the vectors were just orthogonal and not normalized then you'd get other values.
This is arguably one of the most important concepts in linear algebra, for example the discrete Fourier transform relies on this property.

- anonymous

I got it Empty. Either the product of the vectors is 1 (when they're essentially the same vector), or they're 0 (because they're orthogonal to each other).

- Empty

Yup totally. :D

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