If one subtracts one positive definite matrix from another, will the result still be positive definite, or not? How can one prove this？
Asked
Active
Viewed 4,660 times
2
-
7The one-by-one matrices $(1)$ and $(2)$ are positive-definite. What about the matrix $(1)-(2)=(-1)$? – whuber Mar 29 '18 at 23:28
-
Much more elegant via Erdös's criteria. – Gregg H Mar 30 '18 at 01:03
-
Related: https://stats.stackexchange.com/questions/81285/appropriate-measure-to-find-smallest-covariance-matrix/385902#385902 which also answers the question. – kjetil b halvorsen Dec 09 '19 at 00:56
2 Answers
6
Let $G$ and $H$ be positive definite, and let $v$ be any vector. Because the matrices are positive self definite, $\exists$ $a$ and $b$ such that $v^T G v = a >0$ and $v^T H v = b>0$. Without loss of generality, assume $a \gt b$. Then $H-G$ is not positive definite: $$v^T(H-G)v^T = v^THv - v^TGv = b - a ≤ 0$$
Gregg H
- 3,571
- 6
- 25
-
I don}t think this is correct as written, because there do exist posdef $G, H$ such that either $G-H$ or $H-G$ is posdef. See https://stats.stackexchange.com/questions/81285/appropriate-measure-to-find-smallest-covariance-matrix/385902#385902 – kjetil b halvorsen Dec 09 '19 at 00:56
3
In general $H-G$ is not positive definite, but $H-G$ will be positive definite if we further assume the smallest singular of $H$ is larger than the largest singular of $G$. To see this we let $H= U_1S_1V_1^T $ and $G= U_2S_2V_2^T$ be the svds of $H$ and $ G$ respectively, and let x be arbitrary. Then
$|x^T H x| = |x^T(U_1S_1V_1^T)x| \geq |x^T min(\sigma_i(S_1)x|$
and
$|x^T G x| = |x^T(U_1S_2V_1^T)x| \leq |x^T max(\sigma_i(S_2))x| $
Robert Schmid
- 31
- 3