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)
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.
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).