Singular Value Decomposition

Algebra Level 4

Let A = ( 5 − 1 2 − 1 5 2 ) . A=\left(\begin{array}{ccc} 5&-1&2\\ -1&5&2\end{array}\right). Then, by singular value decomposition , there is a diagonal matrix Σ \Sigma and orthogonal matrices U U and V V such that A = U Σ V A=U\Sigma V . The sum of the entries on the main diagonal of Σ \Sigma can be written in the form a + b c a+b\sqrt{c} , where a , b , c a,b,c are positive integers and c c square-free. Find a + b + c a+b+c .


The answer is 14.

Samir Khan by Samir Khan , 23, USA

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1 solution

Samir Khan
Jun 29, 2016

The entries of Σ \Sigma will be the singular values, which are the square roots of the eigenvalues of A A T AA^T . Then, A A T = [ 5 − 1 2 − 1 5 2 ] [ 5 − 1 − 1 5 2 2 ] = [ 30 − 6 − 6 30 ] , AA^T=\left[\begin{array}{ccc} 5&-1 &2 \\ -1&5&2\end{array}\right]\left[\begin{array}{cc} 5&-1\\ -1&5\\ 2&2 \end{array}\right]=\left[\begin{array}{cc} 30&-6\\ -6&30 \end{array}\right], and the eigenvalues of this matrix are given by ( 30 − λ 2 ) − 36 = 0 ⟹ λ 1 = 36 , λ 2 = 24. (30-\lambda^2)-36=0\implies \lambda_1= 36, \lambda_2=24. Thus, the sum of the entries is 6 + 2 6 6+2\sqrt{6} , and the answer is 14.

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