Matrix of order n
Task number: 4439
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.
