# pos def matrix with diagonal=1 -> to lower dimensional pos def matrix

• Nov 24th 2010, 12:40 AM
AlexHuber
pos def matrix with diagonal=1 -> to lower dimensional pos def matrix
Hello!

I have a nxn symmetric pos def matrix and the diagonal elements are all equal 1.

Is there a possibility to construct a n-1xn-1 matrix pos def matrix which contains all the entries from the nxn matrix except the diagonal 1 element?

I thought about constructing the n-1xn-1 matrix by deleting the diagonals and the the first upper diagonal of the nxn matrix, but the resulting matrices are not necessarily pos def...

Has anybody an idea? Or is it feasible at all?

Thank you!
Alex
• Nov 24th 2010, 10:08 AM
Drexel28
Quote:

Originally Posted by AlexHuber
Hello!

I have a nxn symmetric pos def matrix and the diagonal elements are all equal 1.

Is there a possibility to construct a n-1xn-1 matrix pos def matrix which contains all the entries from the nxn matrix except the diagonal 1 element?

I thought about constructing the n-1xn-1 matrix by deleting the diagonals and the the first upper diagonal of the nxn matrix, but the resulting matrices are not necessarily pos def...

Has anybody an idea? Or is it feasible at all?

Thank you!
Alex

Can you explain further what you're looking for? You want to create a $\displaystyle (n-1)\times(n-1)$ 'submatrix' which excludes one of the diagonal entries?
• Nov 24th 2010, 12:27 PM
Opalg
The first place to test this idea would be to look at the smallest possible value of n, namely n=2.

The $\displaystyle 2\times2$ matrix $\displaystyle \begin{bmatrix}1&x\\x&1\end{bmatrix}$ is positive definite whenever |x|<1. The only $\displaystyle 1\times1$ matrix containing all the non-diagonal elements is then $\displaystyle [x]$. Since x can be anything from –1 to +1, there is no way to ensure that x is positive.

So I am not optimistic that a construction of this sort is possible.
• Nov 24th 2010, 05:26 PM
AlexHuber
Hi,
thank you! Basically I have a matrix process $\displaystyle $$A(l)$$$ l=1...L of corr matrices and I assume that it follows a lower dimensional factor model: $\displaystyle $$A(l)=B^{'}a(l)B+a_{0},$$$where $\displaystyle $$a_{l}$$$ is lower dimensional matrix process and $\displaystyle a_{0}$ and B are constants. But it is not good also to fit the diagonal of $\displaystyle $$A(l)$$$ since it is constant one. That's why I am looking for a spd matrix$\displaystyle $$\tilde{A(l)}$$$ which does not contain the diagonal elements 1 anymore. In order to guarantee the pos def of a(l) I require pos def of $\displaystyle $$\tilde{A}(l)$$$.

Hmm, but I am afraid this is not possible... Maybe I have to alter my approach..