# 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 '18 at 23:28
• Much more elegant via Erdös's criteria. Mar 30 '18 at 1:03
• Related: stats.stackexchange.com/questions/81285/… which also answers the question. Dec 9 '19 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 '19 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
$$|x^T H x| = |x^T(U_1S_1V_1^T)x| \geq |x^T min(\sigma_i(S_1)x|$$
$$|x^T G x| = |x^T(U_1S_2V_1^T)x| \leq |x^T max(\sigma_i(S_2))x|$$