Show that the tridiagonal matrix is positive definite

[mathmari]

Hey!

We have the tridiagonal matrix $A=\begin{pmatrix}2 & 1 & \ldots & 0 \\ 1 & 2 & 1 & \ldots \\ \ldots & \ldots & \ldots & \ldots \\ 0 & \ldots & 1 & 2\end{pmatrix}$. I want to show that it is positive definite.

For that it is given the following hint:
1) $\langle x, Ax\rangle \geq 0$
2) $\langle x, Ax\rangle =0 \Rightarrow x=0$ I have done the following:

The $i$-th component of the vector $Ax$ is \begin{equation*}(Ax)_i=x_{i-1}+2x_i+x_{i+1} , \ i=1, 2, \ldots , n \ \text{ with } x_0=x_{n+1}=0\end{equation*}
Then we have the following: \begin{equation*}\langle x, Ax\rangle=\sum_{i=1}^nx_i(Ax)_i =\sum_{i=1}^nx_i\left (x_{i-1}+2x_i+x_{i+1}\right )=\sum_{i=1}^nx_i x_{i-1}+2\sum_{i=1}^n x_i^2+\sum_{i=1}^n x_ix_{i+1}\end{equation*}

Is everything correct so far? (Wondering)

Now we have to show that it is positive if the vector $x$ is not the zero vector and equal to $0$ if the vector is the zero vector, right? But how can we do that? (Wondering)

 

[I like Serena]

Hey!

We have the tridiagonal matrix $A=\begin{pmatrix}2 & 1 & \ldots & 0 \\ 1 & 2 & 1 & \ldots \\ \ldots & \ldots & \ldots & \ldots \\ 0 & \ldots & 1 & 2\end{pmatrix}$. I want to show that it is positive definite.

For that it is given the following hint:
1) $\langle x, Ax\rangle \geq 0$
2) $\langle x, Ax\rangle =0 \Rightarrow x=0$ I have done the following:

The $i$-th component of the vector $Ax$ is \begin{equation*}(Ax)_i=x_{i-1}+2x_i+x_{i+1} , \ i=1, 2, \ldots , n \ \text{ with } x_0=x_{n+1}=0\end{equation*}
Then we have the following: \begin{equation*}\langle x, Ax\rangle=\sum_{i=1}^nx_i(Ax)_i =\sum_{i=1}^nx_i\left (x_{i-1}+2x_i+x_{i+1}\right )=\sum_{i=1}^nx_i x_{i-1}+2\sum_{i=1}^n x_i^2+\sum_{i=1}^n x_ix_{i+1}\end{equation*}

Is everything correct so far? (Wondering)

Now we have to show that it is positive if the vector $x$ is not the zero vector and equal to $0$ if the vector is the zero vector, right? But how can we do that? (Wondering)

Hey mathmari!

How about trying to write it as a sum of squares?
That is, something like $(x_1+x_2)^2 + (x_2 + x_3)^2 + ...$.
Squares are always $\ge 0$ aren't they?
With equality only if all squares are $0$? (Wondering)

 

[mathmari]

How about trying to write it as a sum of squares?
That is, something like $(x_1+x_2)^2 + (x_2 + x_3)^2 + ...$.
Squares are always $\ge 0$ aren't they?
With equality only if all squares are $0$? (Wondering)

Ahh so we have the following, don't we?
\begin{align*}\langle x, Ax\rangle&=\sum_{i=1}^nx_i(Ax)_i \\ & =\sum_{i=1}^nx_i\left (x_{i-1}+2x_i+x_{i+1}\right )\\ & =\sum_{i=1}^nx_i x_{i-1}+2\sum_{i=1}^n x_i^2+\sum_{i=1}^n x_ix_{i+1} \\ & =x_1 x_{0}+\sum_{i=2}^nx_i x_{i-1}+2\sum_{i=1}^n x_i^2+\sum_{i=1}^{n-1} x_ix_{i+1} +x_nx_{n+1} \\ & =\sum_{i=2}^nx_i x_{i-1}+2\sum_{i=1}^n x_i^2+\sum_{i=1}^{n-1} x_ix_{i+1} \\ & = \sum_{i=1}^{n-1}x_i x_{i+1}+2\sum_{i=1}^n x_i^2+\sum_{i=1}^{n-1} x_ix_{i+1} \\ & = 2\sum_{i=1}^{n-1}x_i x_{i+1}+2\sum_{i=1}^n x_i^2 \\ & = \left (\sum_{i=1}^{n-1} x_i^2+x_n^2\right )+2\sum_{i=1}^{n-1}x_i x_{i+1}+\left (x_1^2+\sum_{i=2}^n x_i^2\right ) \\ & = \sum_{i=1}^{n-1} x_i^2+2\sum_{i=1}^{n-1}x_i x_{i+1}+\sum_{i=2}^n x_i^2+x_1^2+x_n^2 \\ & = \sum_{i=1}^{n-1} x_i^2+2\sum_{i=1}^{n-1}x_i x_{i+1}+\sum_{i=1}^{n-1} x_{i+1}^2+x_1^2+x_n^2 \\ & = \sum_{i=1}^{n-1} \left (x_i^2+2x_i x_{i+1}+ x_{i+1}^2\right )+x_1^2+x_n^2 \\ & = \sum_{i=1}^{n-1} \left (x_i+ x_{i+1}\right )^2+x_1^2+x_n^2 \end{align*}
So we get $$\langle x, Ax\rangle \geq 0 \ \text{ and } \ \langle x, Ax\rangle=0 \iff x_i=0 \ \forall i \iff x=0$$

(Wondering)

 

[I like Serena]

Ahh so we have the following, don't we?
\begin{align*}\langle x, Ax\rangle&=\sum_{i=1}^nx_i(Ax)_i \\ & =\sum_{i=1}^nx_i\left (x_{i-1}+2x_i+x_{i+1}\right )\\ & =\sum_{i=1}^nx_i x_{i-1}+2\sum_{i=1}^n x_i^2+\sum_{i=1}^n x_ix_{i+1} \\ & =x_1 x_{0}+\sum_{i=2}^nx_i x_{i-1}+2\sum_{i=1}^n x_i^2+\sum_{i=1}^{n-1} x_ix_{i+1} +x_nx_{n+1} \\ & =\sum_{i=2}^nx_i x_{i-1}+2\sum_{i=1}^n x_i^2+\sum_{i=1}^{n-1} x_ix_{i+1} \\ & = \sum_{i=1}^{n-1}x_i x_{i+1}+2\sum_{i=1}^n x_i^2+\sum_{i=1}^{n-1} x_ix_{i+1} \\ & = 2\sum_{i=1}^{n-1}x_i x_{i+1}+2\sum_{i=1}^n x_i^2 \\ & = \left (\sum_{i=1}^{n-1} x_i^2+x_n^2\right )+2\sum_{i=1}^{n-1}x_i x_{i+1}+\left (x_1^2+\sum_{i=2}^n x_i^2\right ) \\ & = \sum_{i=1}^{n-1} x_i^2+2\sum_{i=1}^{n-1}x_i x_{i+1}+\sum_{i=2}^n x_i^2+x_1^2+x_n^2 \\ & = \sum_{i=1}^{n-1} x_i^2+2\sum_{i=1}^{n-1}x_i x_{i+1}+\sum_{i=1}^{n-1} x_{i+1}^2+x_1^2+x_n^2 \\ & = \sum_{i=1}^{n-1} \left (x_i^2+2x_i x_{i+1}+ x_{i+1}^2\right )+x_1^2+x_n^2 \\ & = \sum_{i=1}^{n-1} \left (x_i+ x_{i+1}\right )^2+x_1^2+x_n^2 \end{align*}
So we get $$\langle x, Ax\rangle \geq 0 \ \text{ and } \ \langle x, Ax\rangle=0 \iff x_i=0 \ \forall i \iff x=0$$

(Wondering)

(Nod)

 

[mathmari]

(Nod)

Thank you! (Yes)