Given a Positive Definite Matrix ## A \in {\mathbb{R}}^{2 \times 2} ## given by:(adsbygoogle = window.adsbygoogle || []).push({});

$$ 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) ##?

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# Diagonal Scaling of a 2x2 Positive Definite Matrix

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