# is the difference of two positive definite matrices also positive definite?

If one subtracts one positive definite matrix from another, will the result still be positive definite, or not? How can one prove this？

• The one-by-one matrices $(1)$ and $(2)$ are positive-definite. What about the matrix $(1)-(2)=(-1)$?
– whuber
Mar 29, 2018 at 23:28
• Much more elegant via Erdös's criteria. Mar 30, 2018 at 1:03
• Related: stats.stackexchange.com/questions/81285/… which also answers the question. Dec 9, 2019 at 0:56

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: $$v^T(H-G)v^T = v^THv - v^TGv = b - a ≤ 0$$
• 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 stats.stackexchange.com/questions/81285/… Dec 9, 2019 at 0:56
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
$$|x^T H x| = |x^T(U_1S_1V_1^T)x| \geq |x^T min(\sigma_i(S_1)x|$$
$$|x^T G x| = |x^T(U_1S_2V_1^T)x| \leq |x^T max(\sigma_i(S_2))x|$$