## Matrix of order n

### Task number: 4439

Depending on $$n$$, decide whether the following matrix is positive definite $$\boldsymbol A_n= \begin{pmatrix} 2 & -1 & 0 & \cdots & 0 \\ -1 & 2 & -1 & \ddots & \vdots \\ 0 & \ddots & \ddots & \ddots & 0 \\ \vdots & \ddots & -1 & 2 & -1 \\ 0 & \cdots & 0 & -1 & 2\\ \end{pmatrix}$$
• #### Hint

Use the Sylvester criterion.
• #### Solution

It holds that $$\det\boldsymbol A_1=|2|=2$$, $$\det\boldsymbol A_2= \begin{vmatrix} 2 & -1 \\ -1 & 2 \end{vmatrix} =3$$, $$\det\boldsymbol A_3= \begin{vmatrix} 2 & -1 & 0\\ -1 & 2 & 1 \\ 0 & -1 & 2 \end{vmatrix} =4$$.

By Laplace expansion according to the first row we get: $\det\boldsymbol A_n= \begin{vmatrix} 2 & -1 & 0 & \cdots & \cdots & 0 \\ -1 & 2 & -1 & 0 & \cdots & 0 \\ 0 & -1 & 2 & -1 & \ddots & \vdots \\ \vdots & 0 & \ddots & \ddots & \ddots & 0 \\ \vdots & \vdots & \ddots & -1 & 2 & -1 \\ 0 & 0 & \cdots & 0 & -1 & 2\\ \end{vmatrix} = 2 \begin{vmatrix} 2 & -1 & 0 & \cdots & 0 \\ -1 & 2 & -1 & \ddots & \vdots \\ 0 & \ddots & \ddots & \ddots & 0 \\ \vdots & \ddots & -1 & 2 & -1 \\ 0 & \cdots & 0 & -1 & 2\\ \end{vmatrix} - (-1) \begin{vmatrix} -1 & -1 & 0 & \cdots & 0 \\ 0 & 2 & -1 & \ddots & \vdots \\ 0 & \ddots & \ddots & \ddots & 0 \\ \vdots & \ddots & -1 & 2 & -1 \\ 0 & \cdots & 0 & -1 & 2\\ \end{vmatrix}$

We expand the last determinant is along the first column and we get a recurrent relation:

$$\det\boldsymbol A_n= 2 \det \boldsymbol A_{n-1} - \det \boldsymbol A_{n-2}$$.

From there, we derive the expected rule $$\det\boldsymbol A_n=n+1$$ by inductionand by substituting into the recurrent relation:

$$\det\boldsymbol A_n= 2 \det \boldsymbol A_{n-1} - \det \boldsymbol A_{n-2} = 2 (n-1+1)-(n-2+1)=n+1$$.

Now, by a straightforward application of Sylvester criterion, we get that all determinants of the matrices $$\boldsymbol A_1, \ldots, \boldsymbol A_n$$ are positive, so every matrix $$\boldsymbol A_n$$ is positive definite.

• #### Answer

The matrix $$\boldsymbol A_n$$ is always positive definite.