# Circular definition confusion, linear algebra

1. Sep 12, 2011

### enfield

I'm no longer confused - nevermind.

So I need help sorting something out about one of the elementary theorems in linear algebra. y will respresent my eigenvalues, if that's okay.

wikipedia says det(Ax-yI)x = 0 follows from the matrix (Ax-yI) being non invertible, because the determinant of a non invertible matrix is always zero.

Now, I understand why the matrix has to be non invertible - if it's not then x=0 is a solution, because you can apply the inverse to both sides, but we don't want that - we want to know for what y's the equivalence could be true if x is non-zero.

What I can't figure out is how to prove the matrix being non invertible implies the determinant of it is zero, without assuming the above formula is valid.

If a matrix is non-invertible than for obvious reasons it must have at least 1 eigenvalue equal to 0. If we knew that the determinant of a matrix was the product of the eigenvalues we could say - okay - the determinant of a non-invertible matrix is zero, and be done.

But how do we show the determinant of a matrix is the product of it's eigenvalues? Well the only way I know involves using the fact that the determinant of an upper triangular matrix is the product of it's diagonal entries, and combining that with det(Ax-yI)x = 0.

For if A is an upper triangular matrix than that equation will only be valid when a diagonal entry is zero (because the det of it is the product of the diagonal entries), and that will only happen when an eigenvalue is equal to one of those entries.

But I can't use this to show why the determinant is the product of the eigenvalues, because that is what I am trying to show in the first place. So is there another way to show this, or to understand why the characteristic equation is valid?

Any help would be appreciated. Thanks.

Last edited: Sep 12, 2011