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?
$\begingroup$
$\endgroup$
3
-
$\begingroup$ 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!) $\endgroup$– Ehsan M. KermaniAug 21, 2015 at 3:09
-
1$\begingroup$ What exactly is your definition of a positive definite matrix? $\endgroup$– davidlowryduda ♦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..... $\endgroup$– EHMJAug 21, 2015 at 3:12
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
4
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.
answered Aug 21, 2015 at 3:16
-
1$\begingroup$ Great explanation, can't be more descriptive and clear than this. $\endgroup$– EHMJAug 21, 2015 at 3:29
-
-
$\begingroup$ It doesn't need to be, but complex eigenvalues fit into the second case too. $\endgroup$– davidlowryduda ♦Aug 21, 2015 at 14:37
-
$\begingroup$ 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. $\endgroup$– user1551Aug 22, 2015 at 10:12
$\begingroup$
$\endgroup$
Hint: If $\lambda$ is an eigenvalue of $A$, let $x$ be the associated eigenvector, and consider $x'Ax$.
answered Aug 21, 2015 at 3:18
$\begingroup$
$\endgroup$
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.
