Sufficient conditions for difference of matrices A-B to be positive definite

Question

Define $A>B\iff A-B\succ 0$, i.e. $A-B$ is positive definite. A well-known sufficient condition for $A>B$ is that the largest eigenvalue of $B$ is smaller than the smallest eigenvalue of $A$.

Are there any other useful sufficient conditions for $A>B$? Are there any equivalent properties (i.e. $A>B$ if and only if $A,B$ satisfy some property $P$)?

Answer 1

What follows addresses the case $B>0$.

So, let $A,B\in \mathbb{C}^{n\times n}$ such that $A^*=A, B^*=B$ and $B>0$. In this case, it is known that the equation $\det(A-\lambda B)=0$ has exactly $n$ complex solutions if multiplicity is taken in account, say $\lambda_1,...,\lambda_n$, and actually these solution are also real. Furthermore, there is a $B$-orthogonal base of $\mathbb{C}^n$ made of generalized eigenvector of $A$ and $B$, say $v_1,...,v_n\in\mathbb{C}^n$. In other words:

• ${v_1,...,v_n}$ is a base of $\mathbb{C}^n$;
• $\forall k,h\in\{1,...,n\}, (k\neq h)\implies(v_h^*Bv_k=0)$;
• $\forall k\in\{1,...,n\}, Av_k=\lambda_kBv_k$.

Now, suppose $A-B>0$. If $k \in\{1,...,n\}$, then $0<v_k^*(A-B)v_k= (\lambda_k-1)v_k^*Bv_k$ and so, dividing by $v_k^*Bv_k$ (it is possible to divide because $B>0$ and $v_k\neq0$), we obtain $\lambda_k>1$. So, if $\lambda\in\mathbb{R}$ is the smallest solution of $\det(A-\lambda B)=0$, then $\lambda>1$.

On the other hand, suppose that the smallest solution of $\det(A-\lambda B)=0$ is greater then $1$. Then, if $v\in\mathbb{C}^n\backslash \{0\}$ and $v=\sum_{k=1}^n\alpha_kv_k$, we get: $$v^*(A-B)v=\sum_{h=1}^n\sum_{k=1}^n\bar\alpha_h\alpha_kv_h^*(A-B)v_k=\sum_{h\neq k}\bar\alpha_h\alpha_kv_h^*(A-B)v_k+\sum_{k=1}^n|\alpha_k|^2v_k^*(A-B)v_k=\\=\sum_{h\neq k}\bar\alpha_h\alpha_k(\lambda_k-1)v_h^*Bv_k+\sum_{k=1}^n|\alpha_k|^2(\lambda_k-1)v_k^*Bv_k=\sum_{k=1}^n|\alpha_k|^2(\lambda_k-1)v_k^*Bv_k>0,$$ and so $A-B>0$.

In conclusion: $A-B>0$ if and only if the smallest solution $\lambda$ of the polynomial equation (of degree n) $$\det(A-\lambda B)=0$$ is greater the 1.

A final remark: notice that, if $A,B\in\mathbb{R}^{n\times n}$, the same result can be obtained through the following (intuitively simple but a little bit heavy to make rigorous) geometric argument. Define: $$S_A:=\{v\in\mathbb{R}^n\ |\ v^tAv=1 \}, S_B:=\{v\in\mathbb{R}^n\ |\ v^tBv=1 \}, H:=\{v\in\mathbb{R}^n\ |\ v^tBv<1 \}.$$ By $B>0$, the hypersurface $S_B$ is an hyperellipsoid and $H$ is the bounded connected component of $\mathbb{R}^n\backslash S_B$. The fact that the smallest solution of $\det(A-\lambda B)=0$ is greater then 1 is equivalent to the following two things:

• $S_A$ is an hyperellipsoid (because this condition implies that $A$ is positive definite too);
• We have to stretch $S_A$ by a factor of $\lambda>1$ in order to get $S_A$ touch "for the first time" $S_B$ (because $\lambda S_A$ and $S_B$ have a point of tangency if and only if $\det(A-\lambda B)=0$. So, if for some $\mu\le1$ it happens that $\mu S_A$ and $S_B$ have a point in common, for some $\lambda<\mu$, $\lambda S_A$ and $S_B$ should have a point of tangency);

However, these last two conditions are obviously equivalent to the fact that $S_A$ is an hyperellipsoid contained in $H$, which in fact is equivalent (after translating this geometric condition into the language of linear algebra) to $A-B>0$.