In 4x4 matrix when does row swapping not affect eigenvaluess?

[NotEuler]

The title pretty much says it... I know that in general eigenvalues are not necessarily preserved when matrix rows or columns are swapped. But in many cases it seems they are, at least with 4x4 matrices.

So is there some specific rule that says when eigenvalues are preserved if I swap two rows in such a matrix (or two columns)?

 

[jambaugh]

You can express a row swaps (and general row operations) by left multiplication by a matrix (the identity matrix with the same row operation performed on it). Likewise column operations can be expressed similarly by right multiplication. I suggest you consider your question in this context. "When does MA have the same eigen-values as A?"

Your next step might then be to consider the characteristic equation corresponding to the eigen-values recalling that the matrix itself satisfies this polynomial equation.

 

[NotEuler]

Thanks jambaugh,
I didn't get very far with this yet but will keep trying.

Actually I now realised my problem is perhaps slightly different: it consists of swapping the two middle columns of the matrix and then swapping the two middle rows of the resulting matrix. Maybe this process of two sequential swaps is why the eigenvalues seem to remain unchanged.

At least I can see that these swaps preserve both the trace and the determinant of the matrix. But I am not sure if that necessarily implies that eigenvalues are preserved.

 

[NotEuler]

You can express a row swaps (and general row operations) by left multiplication by a matrix (the identity matrix with the same row operation performed on it). Likewise column operations can be expressed similarly by right multiplication. I suggest you consider your question in this context. "When does MA have the same eigen-values as A?"

Your next step might then be to consider the characteristic equation corresponding to the eigen-values recalling that the matrix itself satisfies this polynomial equation.

Still haven't got around to your method, but I think I found another way. I could show that det(A-xI) is identical to det(B-xI) where B is the manipulated matrix. Then, the eigenvalues must also be identical. I suppose this ultimately amounts to the same thing as you wrote, but done in a different way.

 

[WWGD]

I'd say, since the Determinant can be seen as the signed volume of a k-dimensional parallelepiped with sides given in rows, the sign would change when swapping sides changes the orientation of the resulting figure.

 

[Vanadium 50]

I would say that a NxN matrix can be expressed as N linear polynomials. Since the order cannot possibly matter in one expression, it cannot matter in the other.

 

[jambaugh]

Thanks jambaugh,
I didn't get very far with this yet but will keep trying.

Actually I now realised my problem is perhaps slightly different: it consists of swapping the two middle columns of the matrix and then swapping the two middle rows of the resulting matrix. Maybe this process of two sequential swaps is why the eigenvalues seem to remain unchanged.

At least I can see that these swaps preserve both the trace and the determinant of the matrix. But I am not sure if that necessarily implies that eigenvalues are preserved.

Since this is a combination of a row operation and a column operation you can express it as the product of a matrix on the left and another on the right. As it happens since these matrices are row/column swaps of identity matrices they are unipotent (they square to the identity) or said another way they are their own inverses.

What is more, if it is the same index pair for row swaps and column swaps they are the same matrix. Your transformation is of the form: A maps to UAU^-1. Where U is the matrix formed by swapping the j and k^th row of the identity matrix or equivalently swapping the j and k^thcolumn.

It is thus, a similarity transformation and will not affect eigen-values. The Eigen-vectors will get mapped by:
v maps to Uv.