Calculating Eigenvalues: 0 Root Meanings

[wakko101]

This is just a general question:

If, when you are calculating the eigenvalues for a matrix, you get a root of 0 (eg. x^3 - x) --> x(x-1)(x+1), what does that mean for the eigenvectors?

thanks,
w.

 

[olgranpappy]

Nothing. It just means that one of the eigenvalues is zero, it doesn't mean anything special about eigenvectors... When diagonalized the matrix of the operator looks like
[tex]
\left(
\begin{array}{ccc}
-1 & 0 & 0 \\
0 & 0 & 0 \\
0& 0 & 1
\end{array}
\right)\;.
[/tex]
In this basis, the eigenvector with eigenvalue -1 is (1,0,0) and the eigenvector with eigenvalue 0 is (0,1,0) and the eigenvector with eigenvalue 1 is (0,0,1).

 

[CompuChip]

There is nothing wrong with an eigenvalue being zero, and it is not more special than an eigenvalue being -1, i or [itex]\pi[/itex].

Only an eigenvector cannot be zero. Which makes sense, because the zero vector trivially satisfies A 0 = [itex]\lambda[/itex] 0 for any number [itex]\lambda[/itex].

 

[D H]

A zero eigenvalue means the matrix in question is singular. The eigenvectors corresponding to the zero eigenvalues form the basis for the null space of the matrix.

 

[uiulic]

According to definition: Ax=cx,x is nonzero vector,then
we have Ax=0,
which means it has nonzero solutions,
also means A is signular,
also means the corresponding eigenvector is not uniquely determined.

 

[Hurkyl]

also means the corresponding eigenvector is not uniquely determined.

Eigenvectors are never¹ uniquely determined; at the very least, any nonzero scalar multiple of an eigenvector is an eigenvector.

1: Unless your scalar field is GF(2).

 

[quasar987]

And who, might I ask, is GF(2) ?

 

[uiulic]

Thank Hurkyl for pointing out my misunderstanding (I was thinking about sth else.)

 

[Hurkyl]

GF(2) is the finite field with two elements. It's isomorphic to the integers modulo 2.