Given that Q is an orthogonal nxn matrix and B is an mxn matrix, how can we find the sum of squares of all elements of QB in terms of the sum of squares of all elements of B?
I know that the sum of the diagonal elements of (QB)*(QB)^T would yield the answer, but I am at a loss with regard to how represent this value in terms of the sum of square of all elements of B.
The sum of squares of a matrix is known as the square of the Frobenius norm. The Frobenius norm is known to be invariant under multiplication by an orthogonal matrix. Hence the answer is that the sum of the squares of $ Q B $ equals the sum of the squares of $ B $.
More formally: $ \| Q B \|_F^2 = \| B \|_F^2 $.
How do you prove this?
First, you prove that for any vector $ x $ and orthogonal matrix $ Q $ it is the case that $$ \| Q x \|_2^2 = \| x \|_2^2 .$$ (Or, equivalently, $ \| Q x \|_2 = \| x \|_2 $.)
Then your recognize that $$ \| B \|_F^2 = \sum_i \| b_i \|_2^2 $$ where $ b_i $ equals the $ i $th column.
Finally, you recognize that the $ i $th column of $ Q B $ equals $ Q b_i $.
Put all the pieces together, and you get your answer.