Diagonal Scaling of a 2x2 Positive Definite Matrix

[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 $\kappa \left( C \right) = \kappa \left( B A B \right) \leq \kappa \left( A \right)$?

 

[fresh_42]

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

 

[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.

 

[fresh_42]

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.

 

[Drazick]

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

 

[fresh_42]

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.

 

[Drazick]

Hi,
Could you show your steps?

Thank You.

 

[fresh_42]

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.

 

[Drazick]

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

 

[fresh_42]

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.