91Ó°ÊÓ

Let \(U_1 \Sigma_1 V_1^*\) and \(U_2 \Sigma_2 V_2^*\) be the singular value decompositions (SVD) of \(T_1\) and \(T_2\), respectively. Since \(T_1\) and \(T_2\) have the same singular values, we can say \(\Sigma_1 = \Sigma_2\). Now, let \(S_1 = U_1 U_2^*\) and \(S_2 = V_1 V_2^*\). Both \(S_1\) and \(S_2\) are isometries, since the product of unitary matrices (or their adjoints) is also a unitary matrix, and unitary matrices preserve the inner product and have the property \(U^* U = I\), where \(I\) is the identity matrix. Now, let's compute the product \(S_1 T_2 S_2\): \[(S_1 T_2 S_2) = (U_1 U_2^*) (U_2 \Sigma_2 V_2^*) (V_1 V_2^*)\] Since \(U_2^* U_2 = I\) and \(V_2 V_2^* = I\), we have: \[(S_1 T_2 S_2) = U_1 \Sigma_2 V_1^* = T_1\]. So, if the singular values of \(T_1\) and \(T_2\) are the same, there exist isometries \(S_1\) and \(S_2\) such that \(T_1 = S_1 T_2 S_2\).

Suppose we have isometries \(S_1\) and \(S_2\) such that \(T_1 = S_1 T_2 S_2\). Let's compute the singular values of \(T_1\) and \(T_2\). We know that singular values are the square roots of the eigenvalues of \(T^*T\). So, let's compute the adjoint of \(T_1 = S_1 T_2 S_2\): \[T_1^* = (S_1 T_2 S_2)^* = S_2^* T_2^* S_1^*\] Now, let's compute the product of \(T_1\) and \(T_1^*\): \[T_1^* T_1 = (S_2^* T_2^* S_1^*) (S_1 T_2 S_2)\] This can be simplified to: \[T_1^* T_1 = S_2^* (T_2^* T_2) S_2\] Now, let's compute the singular values of \(T_1\). Since singular values are the square roots of non-zero eigenvalues of \(T_1^* T_1\), we can observe that the eigenvalues of the matrix \(T_1^* T_1\) are the same as the eigenvalues of the matrix \(T_2^* T_2\), as they are related by the isometric transformation \(S_2^* S_2\). Thus, the singular values of \(T_1\) and \(T_2\) are the same. In conclusion, \(T_1\) and \(T_2\) have the same singular values if and only if there exist isometries \(S_1\) and \(S_2\) such that \(T_1 = S_1 T_2 S_2\).