Least Singular Value of bidiagonal matrix Consider the $n \times n$, bidiagonal matrix
$$
\left(\begin{array}{ccccc}
x & \\
1 & x \\
& 1& x \\
& & & \ddots \\
& & & 1 & x
\end{array} \right)
$$
It is claimed that the least singular value for this matrix is less than $O(2^{-n})$ when $x$ is fixed and $|x-1| > 1/2$.  Is there a simple way to see this?  I have thought about it as a diagonal shift of a nilpotent matrix, but this does not seem to help.
 A: Is there a typo?  The claim is obviously false.  Your matrix is relatively close to $x$ times the identity, so when $x$ is large
the singular values should be relatively close to $x$, not near $0$.
For an easy concrete example try $n=2$ with $x=2$.  The singular values are both greater than $1$.
EDIT: If $A$ is your matrix, $A^* A$ has diagonal elements $|x|^2+1$ (except for the last diagonal element $|x|^2$), $\overline{x}$ in the first sub-diagonal, $x$ in the first super-diagonal, and $0$ elsewhere.  By Gershgorin's theorem, all eigenvalues of $A^*A$ are within distance $2|x|$ of $|x|^2+1$ or within $|x|$ of $|x|^2$.  In particular, when $|x| > 3/2$ they are all greater than $1/4$, so all the singular values are greater than $1/2$.
A: This is not true. E.g. when $n=2$, the least singular value of the matrix is equal to $0.618$ when $x=-1$ and equal to $1.562$ when $x=2$.
The statement is true, however, when $|x|<\frac12$. The case $x=0$ is trivial because the matrix is singular. Suppose $0\ne|x|<\frac12$. Call your matrix $B$. Then
$$
B^{-1}=\pmatrix{
y\\
-y^2&y\\
y^3&-y^2&\ddots\\
\vdots&\ddots&\ddots&\ddots\\
(-1)^{n+1}y^n&\cdots&y^3&-y^2&y}
$$
where $y=\frac{1}{x}$. Therefore $\sigma_1(B^{-1})\ge\|B^{-1}e_1\|_2=\sqrt{y^2+y^4+y^6+\cdots+y^{2n}}>|y|^n>2^n$ and $\sigma_n(B)=\frac{1}{\sigma_1(B^{-1})}<\frac{1}{2^n}$.
A: In case I have not made mistake and your matrix has only two non-zero diagonals, since it is lower triangular, its eigenvalues and singular values would be $x$ and $|x|$ which are constant, not $O(2^{-n})$.
