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

  • 7
    $\begingroup$ The one-by-one matrices $(1)$ and $(2)$ are positive-definite. What about the matrix $(1)-(2)=(-1)$? $\endgroup$
    – whuber
    Mar 29 '18 at 23:28
  • $\begingroup$ Much more elegant via Erdös's criteria. $\endgroup$
    – Gregg H
    Mar 30 '18 at 1:03
  • $\begingroup$ Related: stats.stackexchange.com/questions/81285/… which also answers the question. $\endgroup$ 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: $$v^T(H-G)v^T = v^THv - v^TGv = b - a ≤ 0$$


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| $


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