How does one prove that a bilinear form is positive definite?

Question

$\mathrm{V}$ is a $\mathbb{Q}$ - vector space with base $\mathrm{B} = (v_1,....v_n)$. A symmetric bilinear form is given with:

$$\mathrm{F}(v_i,v_j) = \{ \text{1 for i = j}, \frac{1}{2} \text{ for i = j-1 or i=j+1}, \text{0 else}\}$$

How can I prove, that $\mathrm{F}$ is positive definite?

Answer 1

Hint: Try considering matrices of the form $$\pmatrix{1 & \frac 12 & 0 & \cdots & 0 & 0 \\ \frac 12 & 1 & \frac 12 & \cdots & 0 & 0 \\ 0 & \frac 12 & 1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 1 & \frac 12 \\ 0 & 0 & 0 & \cdots & \frac 12 & 1}$$

One option is then to proceed by induction.

Answer 2

The following $n \times n$ tridiagonal matrix

$$\begin{bmatrix} 1 & \frac 12 & 0 & \cdots & 0 & 0 \\ \frac 12 & 1 & \frac 12 & \cdots & 0 & 0 \\ 0 & \frac 12 & 1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 1 & \frac 12 \\ 0 & 0 & 0 & \cdots & \frac 12 & 1\end{bmatrix}$$

is symmetric and Toeplitz. Hence, its eigenvalues are real and given by [0]

$$\lambda_k = 1 + \cos \left(\frac{k \pi}{n+1}\right) > 0, \qquad k \in \{1,2,\dots,n\}$$

As all the eigenvalues are real and positive, the matrix is positive definite.

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[0] Silvia Noschese, Lionello Pasquini, and Lothar Reichel, Tridiagonal Toeplitz Matrices: Properties and Novel Applications, 2006.