For the SVD $$\textbf{A=U}\boldsymbol\Sigma \textbf{V}^{*}$$
Where $\textbf{U}$ and $\textbf{V}$ are unitary
By partitioning the matrix $\textbf{A}$, we have the following:
$$\textbf{A} = \left[\begin{array}{c|c}
\textbf{U}_{1} &\textbf{U}_{2}
\end{array}
\right]
\left[
\begin{array}{c|c}
\tilde{\boldsymbol\Sigma} & \textbf{0} \\
\hline
\textbf{0} & \textbf{0}
\end{array}
\right]
\begin{bmatrix}
\textbf{V}_{1}^{\ast} \\
\textbf{V}_{2}^{\ast}
\end{bmatrix}$$
\begin{equation*}
\textbf{A} =
\textbf{U}_{1} \tilde{\boldsymbol\Sigma} \textbf{V}_{1}^{\ast}.
\end{equation*}
Where $\tilde{\boldsymbol\Sigma}=diag(\sigma_1,..,\sigma_k)$
Prove that $$\textbf{A}^\dagger=\textbf{V}\boldsymbol\Sigma^\dagger \textbf{U}^*$$
is the Moore-Penrose pseudoinverse of $\textbf{A}$
Where $$\boldsymbol\Sigma^{\dagger} = diag(\sigma_{1}^{-1},\sigma_{2}^{-1},..,\sigma_{k}^{-1},0,..,0)$$
After some simple calculations we have that $\textbf{AA}^\dagger\textbf{=U}_{1}\textbf{U}_{1}^{*}$ which shows that it is conjugate symmetric.
However, when I want to prove that $\textbf{AA}^{\dagger}\textbf{A=A}$, I will have $\textbf{AA}^{\dagger}\textbf{A=U}_1\textbf{U}_1^{*}\textbf{A}$ but is $\textbf{U}_1\textbf{U}_1^{*}=\begin{bmatrix}
\textbf{I}_{k} &\textbf{0} \\
\textbf{0} & \textbf{0}
\end{bmatrix}$? And why?
Best Answer
I don't think you can easily say things about $U_1U_1^*$. But you can calculate $$ A^* A=V_1\tilde\Sigma^* U_1^*U_1\tilde\Sigma V_1^*. $$ Since $U$ is a unitary, its columns are orthonormal. If we denote the columnss by $u_1,\ldots,u_k$, then the entries of $U_1^*U_1$ are $u_k^*u_j=\delta_{k,j}$, so $U_1^*U_1=I_kV_1^*V_1$. Then $$ A^*A=V_1V_1^*, $$ and $$ AA^*A=AV_1V_1^*=U_1\tilde\Sigma V_1^*V_1V_1^*=U_1\tilde\Sigma V_1^*=A. $$