Can this $3 \times 3$ tridiagonal Toeplitz matrix be rank-$1$?

Question

I am trying to determine whether the following tridiagonal $3 \times 3$ matrix can have a rank of $1$.

$$\begin{bmatrix}a&b&0\\b&a&b\\0&b&a\end{bmatrix}$$

For $a = b = 0$, the rank is clearly $0$. For $a = 0$ and $b \neq 0$, the rank is $2$, and for $a \neq 0$ and $b=0$, the rank is $3$. However, I am not able to determine for what values of $a$ and $b$ the rank is $1$. Does that mean it is not possible for the rank to equal $1$?

Answer 1

If $a=0$, then your matrix is$$\begin{bmatrix}0&b&0\\b&0&b\\0&b&0\end{bmatrix},$$whose rank is $2$ if $b\ne0$ and $0$ if $b=0$.

I will assume now that $a\ne0$. If the rank of the matrix is $1$, then it is smaller than $3$, and therefore the determinant of the matrix is $0$. But that determinant is $a^3-2ab^2$, and, since $a\ne0$,$$a^3-2ab^2=0\iff a=\pm\sqrt2b.$$So, the matrix is$$\begin{bmatrix}\pm\sqrt2b&b&0\\b&\pm\sqrt2b&b\\0&b&\pm\sqrt2b\end{bmatrix}.$$But$$\begin{vmatrix}\pm\sqrt2b&b\\b&\pm\sqrt2b\end{vmatrix}=b^2.$$So, if $b\ne0$, the rank is $2$ (since you have a $2\times2$ submatrix whose determinant is not $0$), and if $b=0$, $a=0$, which I am assuming that it does not happen.

Therefore, indeed, the problem has no solution.

Answer 2

The eigenvalues$^\color{magenta}{\star}$ are

$$ \begin{bmatrix} \lambda_1 \\ \lambda_2 \\ \lambda_3 \end{bmatrix} = a \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + \sqrt{b^2} \begin{bmatrix} \sqrt{2} \\ 0 \\ -\sqrt{2} \end{bmatrix} $$

Since this half-plane does not intersect any of the axes (except at the origin), the given tridiagonal Toeplitz matrix cannot be rank-$1$.

---

_{$\color{magenta}{\star}$ Silvia Noschese, Lionello Pasquini, and Lothar Reichel, Tridiagonal Toeplitz matrices — properties and novel applications, 2006.}