# Why eigenvalue specification reduces the no. of LI equations?

1. Jan 11, 2009

### neelakash

Hi everyone, I am stuck with the following for last couple of days.

Many books mention during the development in the idea of Eigenvalue problem: say, you have the equation
$$[\ A-\lambda\ I]\ X=\ 0$$ where A is an NxN matrix and X is an Nx1 vector.

The above consists of n equations.Say,all eigenvalues are non-degenerate.

If you specify one of the non-degenerate eigenvalues,the number of linearly independent equations will be (N-1).This is written in book.I am looking for the explanation.

The linear independence of the equations come from the vectors in the matrix $$[\ A-\lambda\ I]$$.Since,the matrix $$[\ A-\lambda\ I]$$ is singular,its rank can be at most (N-1).Means,the maximum number of the linearly independent vectors in the matrix A after specifying one of its eigenvalues is (N-1).At least one of the vectors can be expanded in terms of the (N-1) vectors.

I find it difficult to see how the specification of the eigenvalue results in this.It is clear that in specifying the eigenvalue,all the matrix elements become known to us.And we can readily calculate its deerminant=0.That way it is OK.But how do we know the rank is precisely (N-1) and not (N-2) or (N-3)...etc.?

-Please take part in the discussion so that the thing becomes clear.

Neel

Last edited: Jan 11, 2009
2. Jan 11, 2009

### lurflurf

I assume you have n eigenvalues. Your result only holds when they are all different. If x is an eigenvalue of (algebraic) multiplicity k (A-lambda*I)^k has rank N-k. It is also possible that (A-lambda*I)^l has rank N-k for some l=1,2,...,k-1. This helps to keep things interesting. We can see this because (A-x1*I)...(A-xN*I) has rank 0. When eigenvalues are distinct we conclude each (A-xk*I) has rank N-1.
example the matrix
{{x,1}
{0,x}}
The eigenvector x is multiplicity 2 (A-x*I) has rank 1 (A-x*I)^2 has rank 2

example the matrix
{{x,0}
{0,x}}
The eigenvector x is multiplicity 2 (A-x*I) has rank 2