14
$\begingroup$

I've found this while googling some properties of positive semidefinite matrices. (Unfortunately, I cannot remember where I've discovered it.) If this is true, it'll greatly save my time in my work. Is it true? How can you prove it?


Let's say I have two real symmetric and positive definite matrices $\mathbf A$ and $\mathbf B$ of the same size. Also, assume that $\mathbf A - \mathbf B$ is positive semidefinite too. Then, $\mathbf B^{-1} - \mathbf A^{-1}$ is also positive semidefinite.

$\endgroup$
27
$\begingroup$

Yes. Here is one way to prove it:

Since $A-B\geq0$, we have (conjugating with $B^{-1/2}$) that $B^{-1/2}AB^{-1/2}-I\geq0$. This tells you that all the eigenvalues of the positive definite matrix $B^{-1/2}AB^{-1/2}$ are $\geq1$.

Now note that $B^{-1/2}AB^{-1/2}=(B^{-1/2}A^{1/2})(A^{1/2}B^{-1/2})$. Since commuting the product of two matrices does not change the eigenvalues, we get that $A^{1/2}B^{-1/2}B^{-1/2}A^{1/2}=A^{1/2}B^{-1}A^{1/2}$ has also all eigenvalues $\geq1$. So $$ A^{1/2}B^{-1}A^{1/2}-I\geq0. $$ Now conjugating with $A^{-1/2}$ we get $B^{-1}-A^{-1}\geq0$.

$\endgroup$
8
$\begingroup$

I like Martin Argerami's answer better, but here is a slightly different approach which does not use that the nonzero eigenvalues of $CD$ are the same as those of $DC$.

Note: this approach applies verbatim to the more general case of a $C^*$-algebra. Note also that in view of the notion of operator monotone function, we could interpret this result as saying that $x\longmapsto x^{-1}$ is operator nonincreasing.

The rather lengthy following paragraph shows that we can assume $B=I_n$ without loss of generality. Then it is obvious.

Since $B$ is positive definite, we have its square root $B^{1/2}$ which is also positive definite with positive definite inverse $B^{-1/2}$. Now since $A-B$ is positive semidefinite, so is $B^{-1/2}(A-B)B^{-1/2}=B^{-1/2}AB^{-1/2}-I_n=A_0-I_n$. And to prove $B^{-1}-A^{-1}$ positive semidefinite is equivalent to proving $B^{1/2}(B^{-1}-A^{-1})B^{1/2}=I_n-A_0^{-1}$ is positive semidefinite. Since $A_0=B^{-1/2}AB^{-1/2}$ is still positive definite, this shows that we can assume $B=I_n$ withtout loss of generality.

Now using the usual notation $C\geq D$ for symmetric matrices to mean that $C-D$ is positive semidefinite, or equivalently that the spectrum of $C-D$ is nonnegative, we get $$ A-I_n\geq 0\Rightarrow A^{-1/2}(A-I_n)A^{-1/2}=I_n-A^{-1}\geq 0. $$ So the result you mentioned holds, together, of course, with its converse.

$\endgroup$
2
$\begingroup$

Let me offer a constructive proof.

LEMMA 1 For invertible and symmetric matrix $A$, and vectors $x$ and $y$, we have

$$ x^{'}A^{-1}x = \underset{y}{max\, \, } 2{x}'y - {y}'Ay $$

LEMMA 2 For invertable matrices $A$ and $B$, we have

$$ (A + C) ^{-1} = A^{-1} - A^{-1} (A ^{-1} + C ^{-1}) ^{-1} A ^{-1} $$

THEOREM If $A > 0$, $B > 0$, ($A$ $B$ symmetric) and $A > B$, then $A > B$.

Proof By Lemma 1, we have

$$x ^{'} ( B ^{-1} - A ^{-1} ) x = \underset{y}{max\, \, } 2{x}'y - {y}' ( B ^{-1} - A ^{-1} ) ^{-1} y$$

Let $ y = A ^{-1} x $, we have $$ x ^{'} ( B ^{-1} - A ^{-1} ) x \ge 2x ^{'} A ^{-1} x - x ^{'} A ^{-1} ( B ^{-1} - A ^{-1} ) ^ {-1} A ^{-1} x $$

Then by Lemma 2, we have

$$ x ^{'} ( B ^{-1} - A ^{-1} ) x \ge x ^{'} A ^{-1} x + x ^{'} ( A - B ) ^{-1} x $$

By assumption, $ A > 0 $, so $ A ^{-1} > 0 $. Also $ A > B $, i.e. $ A - B > 0 $, so $ (A - B ) ^ {-1} > 0$. Thus the LHS of above equation is greater than 0, so $ B ^{-1} - A ^{-1} > 0 $, i.e. $ B ^{-1} > A ^{-1} $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.