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If A is a real symmetric matrix congruent to $I_n$, show that all eigenvalues of A are positive.

This exercise induces me to use the Sylvester's Law of Inertia: https://en.wikipedia.org/wiki/Sylvester's_law_of_inertia. The fact that they are congruent implies that they have the same inertia. And since the inertia of $I_n$ has only positive eigenvalues, this induces that A have only positive eigenvalues. But i think that i could solve it in a different way. Here it is:

Consider: $x^tAx$. Since A is congruent to the identity matrix, there exists an invertible matrix P such that: $x^tAx=x^t(PIP^t)x = x^t(PP^t)x = (P^tx)^tP^tx= \langle P^tx, P^tx \rangle.$ Since $P^tx \neq0$ and$ \langle P^tx, P^tx \rangle \gt0 $, this shows that $x^tAx \gt0$. Hence, A is a positive-definite matrix and then all its eigenvalues are positive.

Is it correct?

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  • $\begingroup$ In the sentence of "Consider...", what is $x$? $\endgroup$ – Jack Dec 12 '16 at 0:32
  • $\begingroup$ $x$ is an element of $R^n$. $\endgroup$ – math.h Dec 12 '16 at 0:33
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    $\begingroup$ just take $x$ to be an eigenvector, get positive eigenvalue $\endgroup$ – m-agag2016 Dec 12 '16 at 0:37
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You are on the right track (using inner product) except that eigenvalues of $A$ do not appear in your argument at all. You might want $x$ to be an eigenvector of $A$.

You proof is correct. (Except that I would write "Consider $x^tAx$ with $x\neq 0$".) Essentially you are using the following fact:

If $A$ is positive definite, then all its eigenvalues are positive.


To show the fact above, one just needs to calculate $\langle v,Av\rangle$ with $Av=\lambda v$.

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  • $\begingroup$ All i did was to use a already known fact: if A is symmetric and it is positive definite, all its eigenvalues are positive. $\endgroup$ – math.h Dec 12 '16 at 0:37
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    $\begingroup$ No. This is no true. Consider the zero matrix. You also need $A$ to be of full rank. $\endgroup$ – Jack Dec 12 '16 at 0:38
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    $\begingroup$ I showed that A was positive definite: $\langle Ax,x \rangle \gt 0$, right? This excludes the case that A is the zero matrix. Now, i can conclude that A have only positive eigenvalues. (A theorem of the book i'm using). Now i think it's ok $\endgroup$ – math.h Dec 12 '16 at 0:44
  • $\begingroup$ You are right. I misread you statement. If A is symmetric and it is positive definite, then all its eigenvalues of $A$ are positive. True. $\endgroup$ – Jack Dec 12 '16 at 0:46

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