Potential symmetry (for matrices) (1 Viewer)

[3.14159potato26]

A symmetric matrix is one that is equal to its transpose, i.e. A = A^T.
Consider the following 3-by-3 matrix:
{ 0 0 1 }
{ 1 0 0 }
{ 0 1 0 }
It can be seen that this is not a symmetric matrix. However, by interchanging rows (2) and (3), the matrix:
{ 0 0 1 }
{ 0 1 0 }
{ 1 0 0 }
is obtained. This matrix is symmetric.
The question is: is there a general test that can be applied to an n-by-n dimensional matrix to check whether it can be made into a symmetric matrix via elementary row/column operations (in other words, testing if it is potentially symmetrical), without testing all possible row/column operations?

 

[Iruka]

The Wikipedia entry on symmetric matrices might be useful.

http://en.wikipedia.org/wiki/Symmetric_matrix

In particular note, part way down the page:

"Every real non-singular matrix can be uniquely factored as the product of an orthogonal matrix and a symmetric positive definite matrix, which is called a polar decomposition. Singular matrices can be also factored, but not uniquely."

 

[3.14159potato26]

The Wikipedia entry on symmetric matrices might be useful.

http://en.wikipedia.org/wiki/Symmetric_matrix

In particular note, part way down the page:

"Every real non-singular matrix can be uniquely factored as the product of an orthogonal matrix and a symmetric positive definite matrix, which is called a polar decomposition. Singular matrices can be also factored, but not uniquely."

xD....i'm not up to that level yet, have mainly been concentrating on algorithms (i.e. direct/iterative methods for solving systems of linear equations), but thanks anyways, i'll try to look through it when i'm free.

 

[3.14159potato26]

Um...yeah, but consider the following:
Let A be the (non-symmetric) matrix:
{0 4}
{1 0}
The characteristic polynomial is (t - 2)(t + 2), therefore t = -2,2.
Eigenvectors are given by the equation:
{t -4}{a} = 0
{-1 t}{b}
Solving it generally gives the eigenvectors as x = (-2b,b) and (2b,b), which are not perpendicular (use multiplication of gradients to diff points != -1). Therefore, A is not symmetric.
But by applying a row interchange to A:
{1 0}
{0 4}
gives a symmetric matrix.
The characteristic polynomial is (t - 1)(t - 4), therefore t = 1,4.
Solving the equation for eigenvectors gives x = (k,0) and (0,k), which are perpendicular. Therefore, it is symmetric.
So, the original question was if you could tell if transforming one non-symmetric matrix using elementary row/column operations (row interchange in this case) in any way could give you a symmetric matrix. Since the eigenvectors/eigenvalues are changed if you use elementary row/column operations, you can't tell if one matrix can be made into a symmetric one by calculating eigenvectors/eigenvalues.