MHB Is the Matrix Positive Definite?

[evinda]

Hello! (Wave)

We have that $q(x) \geq q_0>0, x \in [0,1], h>0$.

Suppose that we have this $(N+1) \times (N+1) matrix$:

$\begin{bmatrix}
\frac{1}{h^2}+\frac{1}{h}+\frac{q(x_0)}{2} & -\frac{1}{h^2} & 0 & \cdots & 0 \\
-\frac{1}{h^2} & \frac{2}{h^2}+q(x_1) & -\frac{1}{h^2} & \cdots &0 \\
0 & & & & \\
& & & & \\
0 & & 0 & -\frac{1}{h^2} &\frac{2}{h^2}+q(x_N)
\end{bmatrix}$

I want to show that it is invertible.

I thought to write it as $Q+T$, where $Q=diag (\frac{1}{2}q(x_0),q(x_1), \dots, q(x_N))$ and

$$T=\begin{bmatrix}
\frac{1}{h^2}+\frac{1}{h} & -\frac{1}{h^2} & 0 & & 0\\
-\frac{1}{h^2} & \frac{2}{h^2} & -\frac{1}{h^2}& &0 \\
0 & -\frac{1}{h^2} & \frac{2}{h^2} & & \\
& & & & -\frac{1}{h^2}\\
& & & -\frac{1}{h^2} & \frac{2}{h^2}
\end{bmatrix}$$But how can we show that $T$ is positive defined?

I have found that $x^T T x=\frac{x_0^2}{2}+\frac{1}{h^2} \sum_{i=0}^{N-1} (x_i-x_{i+1})^2+ \frac{x_N^2}{h^2}$. Is it right? (Thinking)

 

[Opalg]

Hello! (Wave)

We have that $q(x) \geq q_0>0, x \in [0,1], h>0$.

Suppose that we have this $(N+1) \times (N+1)$ matrix:

$\begin{bmatrix}
\frac{1}{h^2}+\frac{1}{h}+\frac{q(x_0)}{2} & -\frac{1}{h^2} & 0 & \cdots & 0 \\
-\frac{1}{h^2} & \frac{2}{h^2}+q(x_1) & -\frac{1}{h^2} & \cdots &0 \\
0 & & & & \\
& & & & \\
0 & & 0 & -\frac{1}{h^2} &\frac{2}{h^2}+q(x_N)
\end{bmatrix}$

I want to show that it is invertible.

I thought to write it as $Q+T$, where $Q=diag (\frac{1}{2}q(x_0),q(x_1), \dots, q(x_N))$ and

$$T=\begin{bmatrix}
\frac{1}{h^2}+\frac{1}{h} & -\frac{1}{h^2} & 0 & & 0\\
-\frac{1}{h^2} & \frac{2}{h^2} & -\frac{1}{h^2}& &0 \\
0 & -\frac{1}{h^2} & \frac{2}{h^2} & & \\
& & & & -\frac{1}{h^2}\\
& & & -\frac{1}{h^2} & \frac{2}{h^2}
\end{bmatrix}$$But how can we show that $T$ is positive defined?

I have found that $x^T T x=\frac{x_0^2}{2}+\frac{1}{h^2} \sum_{i=0}^{N-1} (x_i-x_{i+1})^2+ \frac{x_N^2}{h^2}$. Is it right? (Thinking)

For convenience, I'll write $a = 1/h$. Let $S$ be the $(N+1) \times (N+1)$ matrix with $a$'s all down the main diagonal, $-a$'s all down the first super-diagonal, and zeros everywhere else: $$S = \begin{bmatrix}
a & -a & 0 & & 0\\ 0 & a & -a & &0 \\ 0 & 0 & a & & \\ & & & \ddots & -a\\ & & & 0 & a \end{bmatrix}.$$ You can check that $$S^{\textsf{T}}S = \begin{bmatrix}a^2 & -a^2 & 0 & & & 0\\ -a^2 & 2a^2 & -a^2 & & &0 \\ 0 & -a^2 & 2a^2 & & & \\ &&& \ddots && \\ & & & & 2a^2& -a^2\\ & & & & -a^2 & 2a^2 \end{bmatrix}.$$ That is a positive definite matrix. It is almost the same as your matrix $T$, which you can obtain by adding $a$ to the element in the top left corner of $S^{\textsf{T}}S$. That will preserve the positive-definiteness (alternatively you could add $a$ to the first element of the diagonal matrix $Q$). This shows that $T$ is positive definite.

 

[evinda]

Is the way I did it wrong? (Thinking)

 

[Opalg]

Is the way I did it wrong? (Thinking)

Your method is fine. I had not noticed that you already found a solution.

 

[evinda]

Your method is fine. I had not noticed that you already found a solution.

Great! And $Q$ is positive defined since $x^T Q x=\frac{q(x_0)}{2} x_0^2+q(x_1) x_1^2+ \dots+ q(x_N) x_N^2>0$ for $x \neq \overline{0}$ where $x \in R^{N+1}$, right? (Thinking)

 

[Opalg]

Great! And $Q$ is positive defined since $x^T Q x=\frac{q(x_0)}{2} x_0^2+q(x_1) x_1^2+ \dots+ q(x_N) x_N^2>0$ for $x \neq \overline{0}$ where $x \in R^{N+1}$, right? (Thinking)

Yes! (Star)

Linguistic note: the name for such a matrix is "positive definite", not "positive defined".

 

[evinda]

Yes! (Star)

Great! Thank you! (Smirk)

Linguistic note: the name for such a matrix is "positive definite", not "positive defined".

A ok... (Nod)