TITLE: Rank preservation of Hankel matrix by adding constrained sample
QUESTION [5 upvotes]: Let some $x_i \in \mathbb{R}$ for every $i$ such that the Hankel matrix
$$H_0=\begin{bmatrix} x_0 & x_1 & x_2 & x_3\\
x_1 & x_2 & x_3 & x_4\\
x_2 & x_3 & x_4 & x_5
\end{bmatrix} $$
is full rank, with the rank equal to 3. Show that there exists some $y\in [-1,1]$ such that the following Hankel matrix is full rank too
$$H_1=\begin{bmatrix}  x_1 & x_2 & x_3 & x_4\\
 x_2 & x_3 & x_4 & x_5\\
 x_3 & x_4 & x_5 & y
\end{bmatrix} $$
The matrix $H_1$ is obtained by deleting the first column of $H_0$ and adding a new column as $[x_4 \ x_5 \ y]^\top$.
I tried to solve this by showing that there exists some $a_0,a_1,a_2,a_3\in \mathbb{R}$ with $a_0\neq 0$ and some $y_0\in [-1,1]$ such that
\begin{equation}
\begin{bmatrix}
x_4\\ x_5\\ y_0
\end{bmatrix}= \begin{bmatrix} x_0 & x_1 & x_2 & x_3\\
x_1 & x_2 & x_3 & x_4\\
x_2 & x_3 & x_4 & x_5
\end{bmatrix} \begin{bmatrix}
a_0\\ a_1 \\ a_2 \\ a_3
\end{bmatrix}
\end{equation}
But still couldn't come up with a formal proof for it. Do you have any suggestion?

REPLY [4 votes]: Suppose the contrary that $\operatorname{rank}(H_1)\le2$ for all $y\in[-1,1]$. Let $S$ be the square submatrix consisting of the first three columns of $H_1$. Then $S$ must be singular. Since it is also comprised of the last three columns of $H_0$, we have $3=\operatorname{rank}(H_0)\le\operatorname{rank}(S)+1\le\operatorname{rank}(H_1)+1\le3$. Therefore $\operatorname{rank}(S)=\operatorname{rank}(H_1)=2$.
It follows that $\operatorname{span}\{(x_4,x_5,y)^T: y\in[-1,1]\}$ must lie inside the column space of $S$. In particular, both $(x_4,x_5,0)^T$ and $(0,0,1)^T$ lie inside the column space of $S$. Therefore the augmented matrix
$$
A=\begin{bmatrix}
x_1 & x_2 & x_3 & x_4 & 0\\
x_2 & x_3 & x_4 & x_5 & 0\\
x_3 & x_4 & x_5 & 0 & 1
\end{bmatrix}
$$
has rank $2$. By elementary column operations, we can reduce $A$ to
$$
B=\begin{bmatrix}
x_1 & x_2 & x_3 & x_4 & 0\\
x_2 & x_3 & x_4 & x_5 & 0\\
0 & 0 & 0 & 0 & 1
\end{bmatrix}.
$$
Hence $\operatorname{rank}(B)=2$ too and the first two rows of $B$ must be linearly dependent. It follows that the last two rows of $H_0$ are also linearly dependent. Now we arrive at a contradiction because $H_0$ has full row rank. Thus we conclude that $H_1$ must have full row rank for some $y\in[-1,1]$.