[Math] Showing that a lower triangular matrix is positive definite

Question

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?

Answer 1

Hint:

The determinant of all the principal minors is positive (in fact, it is $\;1\;$ in all cases).