Re: Proving that [inv(A)](k,k) >= 1/[A(k,k)]

In article <20060207.121141@xxxxxxxx>, Rob Johnson <rob@xxxxxxxxxxxxxx>
wrote:

In article <24632048.1139318862526.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
vukad <vukad@xxxxxxxxx> wrote:
Does anyone have an idea how to prove that the following inequality
holds for the coefficients of any positive definite matrix A:

[inv(A)](k,k) >= 1/[A(k,k)]

This is the same as asking if

-1
(u,Au) (u,A u) >= 1 [1]

for any unit vector u. For the question you ask, let u be the k^th
coordinate vector.

Suppose that { v_k } are eigenvectors of A with eigenvalues { w_k }.
Since A is positive definite, w_k > 0. Suppose that

---
u = > a v [2]
--- k k

Then

---
(u,Au) = > a a (v ,v ) w [3]
--- j k j k k

-1 ---
(u,A u) = > a a (v ,v ) 1/w [4]
--- j k j k k

Since u is a unit vector, we have

---
> a a (v ,v )
--- j k j k

= (u,u)

= 1 [5]

and since 1/x is convex for x > 0, we can apply Jensen's inequality
to [3] and use [4] to get

-1
1/(u,Au) <= (u,A u) [6]

Inequality [6] is a restatement of [1] as long as u can be written as
a linear combination of eigenvectors of A. All vectors in R^n can be
written as linear combinations of eigenvectors if and only if A is
diagonalizable. I tried to remove the need for diagonalizability, but
then I realized that the matrix

+- -+
| 1 1 |
A = | -1 1 |
+- -+

whose inverse is

+- -+
-1 | 1/2 -1/2 |
A = | 1/2 1/2 |
+- -+

is positive definite, but not diagonalizable, and fails to satisfy the
conclusion above. So the condition of diagonalizability can not be
removed completely.

I suspect he means SYMMETRIC positive-definite.

--Ron Bruck
.

Relevant Pages

  • Re: Proving that [inv(A)](k,k) >= 1/[A(k,k)]
    ... Since u is a unit vector, ... written as linear combinations of eigenvectors if and only if A is ... So the condition of diagonalizability can not be ...
    (sci.math)
  • Re: Proving that [inv(A)](k,k) >= 1/[A(k,k)]
    ... Since u is a unit vector, ... written as linear combinations of eigenvectors if and only if A is ... So the condition of diagonalizability can not be ... I suspect he means SYMMETRIC positive-definite. ...
    (sci.math)