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As Wikipedia tells us, a real $n \times n$ symmetric matrix $G = [g_{ij}]$ is positive definite if $v^TGv >0$ for all $0 \neq v \in \mathbb{R}^n$. By a well-known theorem of linear algebra it can be shown that $G$ is positive definite if and only if the eigenvalues of $G$ are positive. Therefore, this gives us two distinct ways to say what it means for a matrix to be positive definite.

In Amann and Escher's Analysis II, exercise 7.1.8 seems to provide yet another way recognize a positive definite matrix. In this exercise, $G$ is defined to be positive definite if there exists a positive number $\gamma$ such that $$ \sum\limits_{i,j = 1}^n g_{ij}v^iv^j \geq \gamma |v|^2 $$

I have not before seen this characterization of a positive definite matrix and I have not been successful at demonstrating that this characterization is equivalent to the other two characterizations listed above.

Can anyone provide a hint how one might proceed to demonstrate this apparent equivalence or suggest a reference that discusses it?

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up vote 3 down vote accepted

Let's number the definitions:

  1. $v^T G v > 0$ for all nonzero $v$.
  2. $G$ has positive eigenvalues.
  3. $v^T G v > \gamma v^T v$ for some $\gamma > 0$.

You know that 1 and 2 are equivalent. It's not hard to see that 3 implies 1. So it remains to show that either 1 or 2 implies 3. A short proof: 2 implies 3 because we can take $\gamma$ to be, say, half the smallest eigenvalue of $G$.

Another short proof: 1 implies 3 because 3 is equivalent to the condition that $v^T G v > \gamma$ for all $v$ on the unit sphere. But the unit sphere is compact, so if $v^T G v$ is positive on the unit sphere, it attains a positive minimum.

(I'd like to take the time to complain about definition 2. It is a misleading definition in that the statement it is describing makes sense for all matrices, but it is not equivalent to the first definition in this generality. The problem is that positive-definiteness is a property of a bilinear form $V \times V \to \mathbb{R}$, whereas eigenvalues are a property of an endomorphism $V \to V$, and in complete generality there's no natural way to turn one into the other. To do this you really need something like an inner product.)

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Perhaps, i didn't read it carefully. Anyhow remove the answer :) – user9413 May 25 '11 at 12:47
Thanks for explaining the equivalence. My oversight was in not recognizing that your number 3 is an equivalent way to express my number 3. – ItsNotObvious May 25 '11 at 14:27
@Qiaochu Yunan, can you please explain how half of smallest eigen value of G helps in proving 2) implies 3) – Sry Feb 28 '15 at 7:24

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