# Why does positive definite matrix have strictly positive eigenvalue?

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?

• 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!) Aug 21, 2015 at 3:09
• What exactly is your definition of a positive definite matrix? Aug 21, 2015 at 3:10
• @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..... Aug 21, 2015 at 3:12

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.

• Great explanation, can't be more descriptive and clear than this. Aug 21, 2015 at 3:29
• Why must an eigenvalue a real number? Aug 21, 2015 at 10:21
• It doesn't need to be, but complex eigenvalues fit into the second case too. Aug 21, 2015 at 14:37
• 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. Aug 22, 2015 at 10:12

Hint: If $\lambda$ is an eigenvalue of $A$, let $x$ be the associated eigenvector, and consider $x'Ax$.

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.