If $A$ is a $k \times k$ submatrix of $n \times n$ unitary matrix and $2k>n$. why that some singular value of $A$ is equal to $1$ If $A$ is a $k \times k$ submatrix of $n \times n$ unitary matrix and $2k>n$. why does some singular value of $A$ is equal to $1$? 
 A: Let $Q\in\mathbb{C}^{n\times k}$, $n\geq k$, such that $Q^*Q=I$ and consider the partitioning
$$
Q=\left(\begin{array}{c}Q_1\\Q_2\end{array}\right)\begin{array}{l}\}\;k\\\}\;n-k\end{array}.
$$
Since $Q$ has orthonormal columns, for any $x\in\mathbb{C}^k$ such that $\|x\|_2=1$,
$$
1=\|x\|_2^2=\|Qx\|_2^2=\|Q_1x\|_2^2+\|Q_2x\|_2^2.
$$
Hence
$$\tag{1}
\sigma_{\max}(Q_1)=\max_{\|y\|_2=1}\|Q_1y\|_2=\left(1-\min_{\|y\|_2=1}\|Q_2y\|_2^2\right)^{1/2}.
$$
So if $Q_2$ has a nontrivial nullspace, the minimum on the right-hand side is zero and hence the maximal singular value of $Q_1$ is equal to 1. A sufficient condition for this is that $Q_2$ has more columns than rows, that is, $k>n-k$, or, $2k>n$. By a more careful use of the variational characterization of singular values instead of (1), one can show that the multiplicity of this singular value is given by the dimension of the nullspace of $Q_2$.
A: Let $Q\in\mathbb{C}^{n\times k}$, $n\geq k$, such that $Q^*Q=I$ and consider the partitioning
$$
Q=\begin{bmatrix}
Q_1  \\
Q_2      
\end{bmatrix}\begin{matrix}
\}&k\\
\}&n-k\\
\end{matrix}
$$
Since $Q$ has orthonormal columns,  we see that
$$
I_k=Q^*Q=Q_1^*Q_1+Q_2^*Q_2.
$$
Let the SVD of $Q_1$ be
$$
Q_1=V\begin{bmatrix}
\begin{smallmatrix}
\sigma_1                     &                          &            &            \\
                               &\sigma_2                &            &            \\
                               &                          &\ddots      &            \\
                               &                          &            &\sigma_k  \\
\end{smallmatrix}
\end{bmatrix}W^*,
$$
so we have
$$
Q_2^*Q_2= I_k-Q_1^*Q_1=W\begin{bmatrix}
\begin{smallmatrix}
1-\sigma_1^2                     &                          &            &            \\
                               &1-\sigma_2^2                &            &            \\
                               &                          &\ddots      &            \\
                               &                          &            &1-\sigma_k^2  \\
\end{smallmatrix}
\end{bmatrix}W^*.
$$
Notice that $\mathrm{rank}(Q_2)=\mathrm{rank}(Q_2^*Q_2)$. Thus,
$$\mathrm{rank}(Q_2)=\mathrm{rank}(\mathrm{diag}[1-\sigma_1^2 ,1-\sigma_2^2 ,\ldots, 1-\sigma_k^2])=k-m$$
i.e.
$$ m=k-\mathrm{rank}(Q_2)$$
where $m$ is  the number of singular values $\sigma_i=1$. If $k>n-k$, then
this equality says that
$$m=k-\mathrm{rank}(Q_2)\geq k-(n-k)=2k-n$$
