Uniqueness of Singular Values Given a matrix A, one inductively constructs (and thus proving its very existence) the singular value decomposition as follows: take $ \sigma_{1}=||A||_{2} $, and consider a couple of vectors such that 
\begin{equation}
 A\textbf{v}_{1} = \sigma_{1} \textbf{u}_{1}
 \end{equation}
Now extend to two orhogonal bases and write down the first step:
\begin{equation}
  U_{1}^{T} A V_{1} = S = 
 \begin{bmatrix}
 \sigma_{1} & \textbf{0}^{T} \\ 
 \textbf{0} & A_{1}
 \end{bmatrix} 
 \end{equation} 
Now take $A_{1}$, and repeat inductively, costructing
\begin{equation}
U_{2}^{T} A_{1} V_{2} = S_{2} =
 \begin{bmatrix}
 \sigma_{2} & \textbf{0}^{T} \\ 
 \textbf{0} & A_{2}
 \end{bmatrix} 
\end{equation}
which gives rise to second singular value, by selecting $\sigma_{2}=||A_{1}||$.

My question is: suppose we choose another couple of extensions to
  orthogonal bases, call them $\hat{U_{1}},\hat{V_{1}}$. Doing so, we'll
  obtain another matrix $\hat{A_{1}}$: who guarantees this
  $\hat{A_{1}}$ has the same norm of ${A_{1}}$, thus giving the same
  second singular value $\sigma_{2}$?

Thanks.
 A: Note that if the columns of $U$ and $\hat{U}$ span the same subspace, then $\hat{U}=UM$ for some nonsingular $M$. If the columns of both $U$ and $\hat{U}$ are orthonormal, the matrix $M$ is orthogonal because $I=\hat{U}^T\hat{U}=M^TU^TUM=M^TM$.
So, if $U_1=[u_1,\tilde{U}_1]$ and $\hat{U}_1=[u_1,\tilde{\hat{U_1}}]$, then $\hat{U}_1=[u_1,\tilde{U}_1]M$ for some orthogonal matrix $$M=\begin{bmatrix}1&0\\0&\tilde{M}\end{bmatrix}.$$
Similarly, $\hat{V}_1=V_1N$ for an orthogonal matrix 
$$N=\begin{bmatrix}1 & 0\\0&\tilde{N}\end{bmatrix}.$$
Therefore,
$$
\hat{A}_1=\tilde{\hat{U}}_1^TA\tilde{\hat{V}}_1=M^T\tilde{U}_1^TA\tilde{V}_1N=\tilde{M}^TA_1\tilde{N}.
$$
The matrices $A_1$ and $\hat{A}_1$ have the same 2-norm since the 2-norm is orthogonally invariant, that is, $\|A_1\|=\|PA_1Q\|$ for any square orthogonal matrices $P$ and $Q$ because $\|PA_1Q\|^2=\rho((PA_1Q)^T(PA_1Q))=\rho(Q^TA_1^TA_1Q)=\rho(A_1^TA_1)=\|A_1\|^2$ and hence in the induction step the $\sigma_2$ is same no matter what orthonormal bases of the orthogonal complements of $u_1$ and $v_1$ you choose.
A: Note that
$$
\widehat U_1^T U_1 \pmatrix{\sigma_1&0^T\\0&A_1}V_1^T\widehat V_1
=
\pmatrix{\sigma_1&0^T\\0&\widehat  A_1}
$$
Break the unitary matrices into block matrices to see that $\widehat A_1 = U A_1 V^T$ for unitary matrices $U,V$.
