 # [Homework 1] - Question 6 (Orthogonal Matrix Proof)

(Elliot Williams) #1

I’ve been wondering how I might go about solving this? I’m sure I’m missing something quite simple; it’s just that, after reading the Trefethen textbook for a while, I can’t seem to figure out how to do it.

Thanks!

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(Rachel Thomas) #2

To get started, let B = U S V be the SVD of B and try plugging that in.

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(Pavel Tyshevskyi) #3

I think, the first thing you should notice is that B, and hence A, are square matrices (from definition of Q being orthogonal). Second thing is, once you factorize B into U S V using SVD, both U and V are orthogonal too. Hence, using associativity we have A = (Q U) S (V.T Q.T). Now, multiplication of orthogonal matrices is still an orthogonal matrix, hence both (Q U) and (V.T Q .T) are orthogonal. Since S is left unchanged by this transformations, we can infer that it remains the same.

You can interpret it in the following way: imagine that Q defines a rotation of whole space around the origin. Now, when you only rotate the space, it doesn’t “inflate” or “contract”, meaning that scale of the basis vectors remains the same. That’s essentially what we have in this example - rotating one way, then doing transformation B, and rotating space back. Now, whatever change in scale was done by B, it wasn’t affected by Q nor by Q.T, hence A and B have the same singular values.

You can refer to this SO answer for more details.

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