Least Singular Value of bidiagonal matrix

Question

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.

Answer 1

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 $1/2$.

Answer 2

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 $\sigma_n(B)=\frac{1}{\sigma_1(B^{-1})}<\frac{1}{2^n}$.

Answer 3

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 $O(2^{-n})$.