Question

Find the Singular Value Decomposition of A, as done in Example 7.68 in the Kuttler textbook.

A = $\begin{bmatrix} 1 & 2&-1 \\ -2& 1& 2 \end{bmatrix}$

Answer 1

Find SVD ? A = [1 2 -1 -2 1 2] = U sigma v^T step1: first we compute the singular values of A i.e squareroot of eigen values of AA^T AA^T = [1 2 -1 -2 1 2] [1 -2 2 1 -1 2] = [6 -2 -2 9] det (AA^T-lambda I) = (6- lambda)(9-lambda)-4 = 0 implies 54 + lambda^2 -6 lambda - 9 lambda -4 = 0 implies lambda^2-15 lambda + 50 = 0 implies lambda = 15 plusminus squareroot 225-200/2 = 15 plusminus 5/2 = 10, 5 so 10 and 5 are two singular values of AA^T sigma_1 = squareroot 10 sigma_2 = squareroot 5 (say) step: 1 finding right vectors (the column of v) by finding an orthogonal set of eigen vectors of A^T A (we can also proceed by finding the columns of u instead) \r\n the eigen values of A^T A 5, 10, 0 (because non-zero eigen values of AA^T

This is genral method for finding SVD.

Kuttler textbook is not available so i don't know what is that method.

if anything other then this let me know in comments.

Thank You !?