Positive Definiteness of Band Matrix

Question

Let $A$ be the $n \times n$ matrix: $$ A = \begin{bmatrix} a & 1\\ 1 & a & 1\\ &1 & a &1\\ &&\ddots &\ddots & \ddots\\ &&&1 & a& 1\\ &&&&1&a \end{bmatrix} $$ Show that $A$ is positive definite for $a \ge 2$.

Answer 1

Hint: $2 |x_i x_{i+1}| \le x_i^2 + x_{i+1}^2$.

Answer 2

These illustrate, for $a=2,$ how to use Sylvester's Law of Inertia. In order to get $a > 2$ we are just adding a positive multiple of the identity.

$$\left( \begin{array}{rr} 1 & 0 \\ \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rr} 2 & 0 \\ 0 & \frac{ 3 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rr} 1 & \frac{ 1 }{ 2 } \\ 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rr} 2 & 1 \\ 1 & 2 \\ \end{array} \right) $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 \\ 0 & \frac{ 2 }{ 3 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & \frac{ 3 }{ 2 } & 0 \\ 0 & 0 & \frac{ 4 }{ 3 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & \frac{ 2 }{ 3 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \\ \end{array} \right) $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ 0 & \frac{ 2 }{ 3 } & 1 & 0 \\ 0 & 0 & \frac{ 3 }{ 4 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & \frac{ 3 }{ 2 } & 0 & 0 \\ 0 & 0 & \frac{ 4 }{ 3 } & 0 \\ 0 & 0 & 0 & \frac{ 5 }{ 4 } \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 1 & \frac{ 2 }{ 3 } & 0 \\ 0 & 0 & 1 & \frac{ 3 }{ 4 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 2 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 2 \\ \end{array} \right) $$ $$\left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 & 0 & 0 \\ 0 & \frac{ 2 }{ 3 } & 1 & 0 & 0 \\ 0 & 0 & \frac{ 3 }{ 4 } & 1 & 0 \\ 0 & 0 & 0 & \frac{ 4 }{ 5 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrrr} 2 & 0 & 0 & 0 & 0 \\ 0 & \frac{ 3 }{ 2 } & 0 & 0 & 0 \\ 0 & 0 & \frac{ 4 }{ 3 } & 0 & 0 \\ 0 & 0 & 0 & \frac{ 5 }{ 4 } & 0 \\ 0 & 0 & 0 & 0 & \frac{ 6 }{ 5 } \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & \frac{ 1 }{ 2 } & 0 & 0 & 0 \\ 0 & 1 & \frac{ 2 }{ 3 } & 0 & 0 \\ 0 & 0 & 1 & \frac{ 3 }{ 4 } & 0 \\ 0 & 0 & 0 & 1 & \frac{ 4 }{ 5 } \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrr} 2 & 1 & 0 & 0 & 0 \\ 1 & 2 & 1 & 0 & 0 \\ 0 & 1 & 2 & 1 & 0 \\ 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 1 & 2 \\ \end{array} \right) $$

and so on...

Answer 3

Option 1: Your matrix $A$ satisfies $$A = (n-1)^2 B + (a-2)I$$ where $B$ is the discrete Laplacian of dimension $n$. It is known that $B$ is symmetric positive definite. Hence, $A$ is symmetric positive definite.

Option 2: Gershgorin's circle theorem implies that $A$ is symmetric positive definite for all $a>2$. Gershgorin's theorem does not apply directly to the case of $a=2$ as it cannot eliminate $0$ as an eigenvalue.