Diagonal Scaling of a 2x2 Positive Definite Matrix

Tags:
1. Nov 16, 2015

Drazick

Given a Positive Definite Matrix $A \in {\mathbb{R}}^{2 \times 2}$ given by:

$$A = \begin{bmatrix} {A}_{11} & {A}_{12} \\ {A}_{12} & {A}_{22} \end{bmatrix}$$

And a Matrix $B$ Given by:

$$B = \begin{bmatrix} \frac{1}{\sqrt{{A}_{11}}} & 0 \\ 0 & \frac{1}{\sqrt{{A}_{22}}} \end{bmatrix}$$

Now, defining the Diagonal Scaling of $A$ given by $C = B A B$.
One could see the main diagonal elements of $C$ are all $1$.
Actually $C$ is given by:

$$C = B A B = \begin{bmatrix} 1 & \frac{{A}_{12}}{\sqrt{ {A}_{11} {A}_{22} }} \\ \frac{{A}_{12}}{\sqrt{ {A}_{11} {A}_{22} }} & 1 \end{bmatrix}$$

This scaling, intuitively, makes the Matrix closer to the identity Matrix (Smaller off diagonal values, 1 on the main diagonal) and hence improve the Condition Number.
Yet I couldn't prove it.

How could one prove $\kappa \left( C \right) = \kappa \left( B A B \right) \leq \kappa \left( A \right)$?

2. Nov 16, 2015

Staff: Mentor

Can you explain to me what a 'Condition Number' and what $\kappa \left( C \right)$ is?

3. Nov 16, 2015

Drazick

The Condition Number is the division of the largest Eigen Value by the Smallest (Under L2 Norm):

https://en.wikipedia.org/wiki/Condition_number

The symbol $\kappa \left( A \right)$ stands for the condition number of $A$.

Thank You.

4. Nov 17, 2015

Staff: Mentor

The eigenvalues of (2,2)-matrices can be explicitly calculated and therefore the condition numbers. The resulting expression depends on the entries of elements of $A$. After two dozens of steps or so I think I have proven it with a brute-force attack.

Last edited: Nov 17, 2015
5. Nov 17, 2015

Drazick

Yea, this is what I tried.
I was after something more elegant.

6. Nov 17, 2015

Staff: Mentor

I don't know whether it hepls or I correctly transformed it. However, defining

$T = trace (A) , D = det (A)$ and $det(B) = µ$ or $µ^2 = det (C) \cdot det (A^{-1})$

I've been able to reduce the inequation to

$\bar{λ}^2 ≤ μ^2 λ^2$ with the larger eigenvalues $\bar{λ}$ of $C$ and $λ$ of $A$

by using that fact the determinant of a (2,2)-matrix is the product of its eigenvalues and therefore eliminating one of them. Still a little work to do but keeping $T, D, μ$ as long as possible and taking $trace (C) = 2$ it should be somehow shorter than brute-force.

7. Nov 17, 2015

Drazick

Hi,
Could you show your steps?

Thank You.

8. Nov 17, 2015

Staff: Mentor

If I understood the conditional number correct then with eigenvalues $λ_1 ≤ λ_2$ we get:

$κ(A) = λ_2 λ^{-1}_1$.

The characteristic polynomial of $A$ is $det (A - x \cdot 1) = (x - λ_1)(x - λ_2) = x^2 - x \cdot T + D$ and $D = λ_1 λ_2$. Similar is true for $K(C)$.

Therefore $D \cdot κ(A) = λ^{2}_2$ and $\bar{D} \cdot κ(C) = \bar{λ}^{2}_2$ if the barred constants are those of $C$.

Now I've calculated $\bar{D} = det (B) det(A) det (B) = det^2 (B) \cdot D = μ^2 D$. Dropping now the indices of $λ$ and dividing by $D$ the inequality reads

$\bar{λ}^2 ≤ μ^2 λ^2$. Note that also $λ_1 + λ_2 = trace (A)$ is true. Same with $B$ or $C$. I didn't use this but maybe it shortens even more by using it to prove the rest.

9. Nov 17, 2015

Drazick

You're saying it is equivalent to show your eigen values inequality to the statement above?

10. Nov 17, 2015

Staff: Mentor

Yes, as long as the maximal eigenvalues aren't 0. Just substitute $λ_1$ by $D \cdot λ^{-1}_2$. Same with $C = BAB$ and finally define $μ = det(B)$, i.e. $\bar{D} = \bar{λ}_1 \bar{λ}_2 = det(C) = det^2 (B) \cdot det (A) = μ^2 D$. Note: I did not check your matrix multiplications. And don't try to take square roots. The squares above help a lot.