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We say $A$ is a positive definite matrix if and only if $x^T A x > 0$ for all nonzero vectors $x$. Then why does every positive definite matrix have strictly positive eigenvalues?

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  • $\begingroup$ Write down the definition! what does it mean for a matrix to be strictly positive definite? (assuming having positive eigenvalues is not the definition though!) $\endgroup$ Commented Aug 21, 2015 at 3:09
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    $\begingroup$ What exactly is your definition of a positive definite matrix? $\endgroup$
    – davidlowryduda
    Commented Aug 21, 2015 at 3:10
  • $\begingroup$ @mixedmath A is a positive definite matrix, if and only if X'AX is greater than 0 for all the non zero entry of X..... $\endgroup$
    – EHMJ
    Commented Aug 21, 2015 at 3:12

3 Answers 3

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Suppose our matrix $A$ has eigenvalue $\lambda$.

If $\lambda = 0$, then there is some eigenvector $x$ so that $Ax = 0$. But then $x^T A x = 0$, and so $A$ is not positive definite.

If $\lambda < 0$, then there is some eigenvector $x$ so that $Ax = \lambda x$. But then $x^T A x = \lambda \lvert x \rvert^2$, which is negative since $\lvert x \rvert^2 > 0$ and $\lambda < 0$. Thus $A$ is not positive definite.

And so if $A$ is positive definite, it only has positive eigenvalues.

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    $\begingroup$ Great explanation, can't be more descriptive and clear than this. $\endgroup$
    – EHMJ
    Commented Aug 21, 2015 at 3:29
  • $\begingroup$ Why must an eigenvalue a real number? $\endgroup$
    – user1551
    Commented Aug 21, 2015 at 10:21
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    $\begingroup$ It doesn't need to be, but complex eigenvalues fit into the second case too. $\endgroup$
    – davidlowryduda
    Commented Aug 21, 2015 at 14:37
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    $\begingroup$ The OP's definition of positive definiteness concerns only about $x^{\color{red}{T}}Ax$ for real vector $x$. For complex eigenvector we don't have $x^Tx=|x|^2$. While showing that $x^\ast Ax>0$ for all complex vectors $x$ is just a one-liner, in the context of the OP, I think this is the least obvious part. $\endgroup$
    – user1551
    Commented Aug 22, 2015 at 10:12
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Hint: If $\lambda$ is an eigenvalue of $A$, let $x$ be the associated eigenvector, and consider $x'Ax$.

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As you said, we say $A \in \mathbb{R}^{n\times n}$ is a positive definite matrix if and only if $x^T A x > 0$ for all nonzero vectors $x \in \mathbb{R}^n$.

Consider the eigenvector $v \neq 0$ and its associated eigenvalue $\lambda$. It means $Av = \lambda v$.

Since $A$ is positive definite, it means $v^T A v > 0$, wich also means $\lambda v^Tv > 0$. We have that:

$$v^Tv = \sum\limits_{i=1}^{n} v_i^2$$

which is always strictly positive. So $\lambda$ must be strictly positive.

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