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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?

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    $\begingroup$ Positive definite and non-symmetric are a dangerous mix. $\endgroup$ – Rodrigo de Azevedo Mar 28 '18 at 12:48
  • $\begingroup$ What do you mean by "positive definite"? $\endgroup$ – user1551 Mar 28 '18 at 14:51
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Matrix A has a positive definite because it's lower triangular matrix and the definite of upper(lower) triangular matrix is equal to the product of its diagonal elements.

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  • $\begingroup$ what is this theorem called? $\endgroup$ – james black Mar 28 '18 at 15:22
  • $\begingroup$ Well, it is not a theorem as well, but it’s a property of diagonal matrix $\endgroup$ – Yauhen Mardan Mar 28 '18 at 15:43
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Hint:

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

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In this case it's easier to use Sylvester's criterion.

All the principal minors are easy to evaluate and they are all $1 > 0 \implies$ the matrix is positive definite.

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