Question

For any nxn matrix A, use the SVD to show that there is an nxn orthogonal matrix Q such that AAT = QT(ATA)Q.

Answer 1

let A be the n $\times$ n matrix of n-dimensional points

By the singular value decomposition:

A - $\cup$ $\times$$\cap$$\times$ V ^ T

here

$\cup$ ( n$\times$n ) is orthogonal $\cup$ ^T $\cup$ = I

$\cap$ ( n$\times$n) has r positive singular value is descending order units diagonal

V( n$\times$n ) is orthogonal V^T V = I

columns of value the orthogonal eigenvector of A^T A

= (
UX
$\times$ V^T) ^T (
UX
$\times$ V^T)

=  
UX
^T $\times$   $\cap$$\times$  V^T

=
UX
^2 $\times$ V^T

columns of value the orthogonal eigenvector of A A^T

=(
UX
$\times$ V^T) (
UX
$\times$ V^T) ^T

=
UX
$\times$  $\cap$^T$\times$  V^T

=
UX
^2 $\times$ V^T

A^T A=(QR)^T (QR)

=R^TR [$\because$ Q^TQ = I ]

we have , A A^T = Q(RR^T)Q^T  

so, if it  is an eigen vector for AA^T,then Q^TQ is an eigenvector for RRT.

Hence ,the matrix V becomes Q^T $\cup$ .