Question

25. (Singular Value Decomposition) Suppose \[A = UD{V^T}\], where U and Vare \[n \times n\] matrices with the property that \[{U^T}U = I\] and \[{V^T}V = I\], and where D is a diagonal matrix with positive numbers \[{\sigma _1}, \ldots ,{\sigma _n}\] on the diagonal. Show that A is invertible, and find a formula for \[{A^{ - {\bf{1}}}}\].

Short answer

Here,Ais the product of invertible matrices. Thus, Ais invertible, and the formula for \[{A^{ - 1}}\] is \[{A^{ - 1}} = V{D^{ - 1}}{U^T}\].

Step-by-step solution

Use the fact of an invertible matrix

Given, \[A = UD{V^T}\] with \[{U^T}U = I\] and \[{V^T}V = I\].

This implies that U and \[{V^T}\] are invertible matrices with \[{U^{ - 1}} = {U^T}\], and \[{\left( {{V^T}} \right)^{ - 1}} = V\].

Note that the determinant of a diagonal matrix is the product of its diagonal entries, i.e., \[\det \,D = {\sigma _1} \ldots {\sigma _n} \ne 0\]. Hence, D is also invertible.

So, here,Ais the product of invertible matrices. Thus, Ais invertible.

Find the inverse matrix

\[\begin{array}{c}{A^{ - 1}} = {\left( {UD{V^T}} \right)^{ - 1}}\\ = {\left( {{v^T}} \right)^{ - 1}}{\left( {UD} \right)^{ - 1}}\\{A^{ - 1}} = {\left( {{v^T}} \right)^{ - 1}}{D^{ - 1}}{U^{ - 1}}\end{array}\]

Find the formula for the inverse of A

Here, \[{U^{ - 1}} = {U^T}\], and \[{\left( {{V^T}} \right)^{ - 1}} = V\]. Therefore, \[{A^{ - 1}} = V{D^{ - 1}}{U^T}\].