# Showing that a lower triangular matrix is positive definite

How can I show that the following square matrix is positive definite?

$$A = \begin{bmatrix} 1 & 0 & 0 & 0 & \dots & 0 & 0\\ -1 & 1 & 0 & 0 & \dots & 0 & 0\\ 0 & -1 & 1 & 0 & \dots & 0 & 0\\ 0 & 0 & -1 & 1 & \dots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \dots & -1 & 1\\ \end{bmatrix}$$

I can manually show the $1 \times 1$, $2 \times 2$, $3 \times 3$ squares have eigenvalue equal to $1$, but how do I generalize that?

• Positive definite and non-symmetric are a dangerous mix. – Rodrigo de Azevedo Mar 28 '18 at 12:48
• What do you mean by "positive definite"? – user1551 Mar 28 '18 at 14:51

The determinant of all the principal minors is positive (in fact, it is $\;1\;$ in all cases).
All the principal minors are easy to evaluate and they are all $1 > 0 \implies$ the matrix is positive definite.