What causes a complex symmetric matrix to change from invertible to non-invertible?

[Tybalt]

I'm trying to get an intuitive grasp of why an almost imperceptible change in the off-diagonal elements in a complex symmetric matrix causes it to change from being invertible to not being invertible. The diagonal elements are 1, and the sum of abs values of the off-diagonal elements in each row is 1.34 in the invertible case and 1.3778 in the non-invertible case.

 

[Office_Shredder]

It's a generally true fact that if you take a non invertible matrix and just randomly change the entries by a small amount, it's probably now invertible.

Without more context about how you constructed this specific matrix I'm not sure there's more that can be said

 

[Haborix]

I think it would be fair to say that whatever change you made, you moved one of the eigenvalues to zero (or near enough to upset a computer numerically calculating the inverse).

 

[martinbn]

Why do you expect the sum of the off diagonal elements to be important? The invertibility of the matrix is determined by the determinant of the matrix, which involves products of those elements. You can change the elements while keeping their sum. Take for example a 2 by 2 matrix with 1's along the main diagonal. The determinant is $1-bc$ if you keep the sum of the off diagonal elements fixed, say equal to 2 i.e. $b+c=2$ you can still change the determinant. The determinant will be $1-2b+b^2$. Any choice of $b$ will give a value for the other element by $c=2-b$, which doesn't change their sum, but changes the determinant.

 

[Tybalt]

It's a generally true fact that if you take a non invertible matrix and just randomly change the entries by a small amount, it's probably now invertible.

Without more context about how you constructed this specific matrix I'm not sure there's more that can be said

This is how I constructed the matrix (M): I start with a complex nxn rank-n matrix (A) and remove a row to make (n-1)xn matrix (B). So B is an operator that maps n data symbols to (n-1) coded symbols, so obviously it has a zero eigenvalue. The inter-symbol interference matrix is M = (B^-1)B. The matrix A that I'm using makes M symmetric and tri-diagonal.

An interesting aside: real(M) is invertible, whereas imag(M) is not.

Matrices K = (A^-1)A and M = (B^-1)B are exactly the same except for the slightest difference in the off-diagonal values. I'm trying to get some insight as to how this difference in values causes the matrix to go from invertible to non-invertible.

I'm hoping that an understanding of this will help me understand the mechanics behind Mazo's limit.

 

[Tybalt]

That's the necessary condition of a proof that a "strictly" diagonally dominant matrix is invertible. The matrices I'm looking at are symmetric tri-diagonal.