Low-rank approximation of a 4 times 5 matrix via its SVD

Returning to this example, involving a matrix with row size m = 4 and column size n = 5:
 A =   left(begin{array}{ccccc}  1 & 0 & 0 & 0 & 2 0 & 0 & 3 & 0 & 0  0 & 0 & 0 & 0 & 0 0 & 4 & 0 & 0 & 0 end{array}right).

As seen here, the SVD is given by A = U tilde{ {S}} V^T, with
 U = left(begin{array}{cccc}0 & 0 & 1 & 00 & 1 & 0 & 00 & 0 & 0 &-11 & 0 & 0 & 0 end{array}right) , ;; tilde{ {S}} = left(begin{array}{ccccc}  4 & 0 & 0 & 0 & 00 & 3 & 0 & 0 & 00 & 0 &sqrt{5} & 0 & 00 & 0 & 0 & 0 & 0 end{array}right) , ;; V^T = left(begin{array}{ccccc}0 & 1 & 0 & 0 & 0 0 & 0 & 1 & 0 & 0  sqrt{0.2} & 0 & 0 & 0 & sqrt{0.8}0 & 0 & 0 & 1 & 0-sqrt{0.8} & 0 & 0 & 0 &sqrt{0.2} end{array}right) .

The matrix is rank r = 3. A rank-two approximation is given by zeroing out the smallest singular value, which produces
 begin{array}{rcl} hat{A}_2 &=& left(begin{array}{cccc}0 & 0 & 1 & 00 & 1 & 0 & 00 & 0 & 0 &-11 & 0 & 0 & 0 end{array}right)left(begin{array}{ccccc}  4 & 0 & 0 & 0 & 00 & 3 & 0 & 0 & 00 & 0 & 0 & 0 & 00 & 0 & 0 & 0 & 0 end{array}right)left(begin{array}{ccccc}0 & 1 & 0 & 0 & 0 0 & 0 & 1 & 0 & 0  sqrt{0.2} & 0 & 0 & 0 & sqrt{0.8}0 & 0 & 0 & 1 & 0-sqrt{0.8} & 0 & 0 & 0 &sqrt{0.2} end{array}right)  &=&  left(begin{array}{cc}0 & 0 0 & 1 0 & 0 1 & 0  end{array}right) left(begin{array}{cc}  4 & 0 0 & 3 end{array}right)left(begin{array}{ccccc}0 & 1 & 0 & 0 & 0 0 & 0 & 1 & 0 & 0  end{array}right)  &=& left(begin{array}{ccccc}  0 & 0 & 0 & 0 & 0 0 & 0 & 3 & 0 & 0  0 & 0 & 0 & 0 & 0 0 & 4 & 0 & 0 & 0 end{array}right). end{array}

We check that the Frobenius norm of the error |A-hat{A}_2|_F is the sum of singular values we have zeroed out, which here reduces to sigma_3 = sqrt{5}:
 E := A-hat{A}_2 = left(begin{array}{ccccc}  1 & 0 & 0 & 0 & 2 0 & 0 & 0 & 0 & 0  0 & 0 & 0 & 0 & 0 0 & 0 & 0 & 0 & 0 end{array}right), ;; |E|_F^2 = 1^2+2^2 = 5.