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In the book of The Element of Real Analysis by Bartle, at page 265, it is given that

Le the real valued function $f$ have a continuous second partial derivatives on a neighbourhood of a critical point c in $\mathbb{R}^p$, and consider the second derivative $$P = (D^2f(c))(w)^2 = \sum_{i,j = 1}^p \frac{\partial^2 f}{\partial x_i \partial x_j} (c) w_i w_j \quad (21.18)$$ evaluated at $w = (w_1, ..., w_p).$

i-) If P > 0 for all $w\not =0$ in $\mathbb{R}^p$, f has a relative min. at $c$

ii-) If P < 0 for all $w\not =0$ in $\mathbb{R}^p$, f has a relative max. at $c$

iii-) If P take both positive and negative values for different $w \not = 0$ in $\mathbb{R}^p$, then $c$ is a saddle point of $f$.

and in the bottom of the page,

The preceding result indicates that the nature of the critical point $c$ is determined by the quadratic function given in $(21.18)$. In particular it is of importance to know whether this function can take on both positive and negative values or whether it is always of one sign. An important and well-known result of algebra can be used the determine this.for each $j = 1,2, ..., p$, let $\Delta_j$ be the determinant of the matrix $$\begin{bmatrix} f_{x_1 x_1} & ... & f_{x_1 x_j} \\ \vdots & \vdots & \vdots & \\ \\ f_{x_j x_1} & ... & f_{x_j x_j} \\ \end{bmatrix}$$

If the numbers $\Delta_1 ,\Delta_2, ..., \Delta_p$ are all positive, the second derivative $(21.18)$ takes only positive values, and similarly if the numbers $\Delta_1 ,\Delta_2, ..., \Delta_p$ are alternatively negative and positive, the expression (21.18) takes only negative values [...]

My question is that which "well-known result of algebra" is the author talking about ? and how can we prove this result in the book using that "well-known result" ?

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    $\begingroup$ Characterization 4. I don't really see the point of mentioning it. $\endgroup$ – reuns Dec 27 '17 at 6:25
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    $\begingroup$ en.wikipedia.org/wiki/Descartes%27_rule_of_signs $\endgroup$ – ziggurism Dec 27 '17 at 6:29
  • $\begingroup$ @reuns I can see that $\Delta_i$ for any $i =1,...,p$ is a Hermitian matrix, however, I'm still not able to see how does $\Delta_is$ determine the value of $P$. $\endgroup$ – onurcanbektas Dec 27 '17 at 6:33
  • $\begingroup$ @ziggurism I don't see the connection to the thing that we are talking. $\endgroup$ – onurcanbektas Dec 27 '17 at 6:34
  • $\begingroup$ The matrix is symmetric, and the question is if it is positive-semi-definite, ie. its characteristic polynomial has all its roots real and non-negative, which is seen by the rule of signs. $\endgroup$ – reuns Dec 27 '17 at 6:43
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Lets denote $A = (\alpha_{i,j}),$ where $\alpha_{i,j} = \frac{\partial^2 f}{\partial x_i \partial x_j}$. Then observe that the polynomial $$P = (D^2f(c))(w)^2 = \sum_{i,j = 1}^p \frac{\partial^2 f}{\partial x_i \partial x_j} (c) w_i w_j \quad (21.18)$$ can be expressed as

$$P = w^T A w, \quad \text{where} \quad w^T = (w_1, ..., w_p) \in \mathbb{R}^p .$$

Now, by the definition, the polynomial $P$ takes only positive values iff the matrix $A$ is positive definite.

Now by Sylvester's Criterion, the matrix $A$ is positive definite iff the determinant $\Delta_j$ for any $0 < j \leq p$ is positive, hence the first statement directly follows.

To show the second statement, we can apply the Sylvester's criterion to $-A$, hence we would get that

$$\Delta_i > 0\quad iff \quad \text{$i$ is even} \\ \Delta_i < 0\quad iff \quad \text{$i$ is odd}$$ iff the matrix $-A$ is negative definite.

Hence the desired result directly follows.

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