Singular value decomposition of a 4 times 5 matrix

Consider the matrix

   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).

A singular value decomposition of this matrix is given by A = U tilde{ {S}} V^T, with

     U = left(begin{array}{cccc}  0 & 0 & 1 & 0 0 & 1 & 0 & 0 0 & 0 & 0 & -1 1 & 0 & 0 & 0 end{array}right) , ;; tilde{ {S}} = left(begin{array}{ccccc}  4 & 0 & 0 & 0 & 0 0 & 3 & 0 & 0 & 0 0 & 0 & sqrt{5} & 0 & 0  0 & 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) .

Notice above that tilde{ {S}} has non-zero values only in its diagonal, and can be written as

 tilde{ {S}} = mbox{bf diag}( {S}, 0, 0), ;;  {S} := mbox{bf diag}(sigma_1,sigma_2,sigma_3),

with sigma_1 = 4, sigma_2 = 3, sigma_3 = sqrt{5}. The rank of A (which is the number of non-zero elements on the diagonal matrix tilde{ {S}}) is thus r = 3 < min(m,n). We can check that V^TV = VV^T = I_5, and UU^T = U^TU= I_4.

See also: