# Two independent eigenvectors

Hi, im a little confused how do you find out when a matrix has two independent eigenvectors or when it has one or when it has more than two, or is it possible it can have no eigenvectors.

It depends what field you are working over. You can have a real matrix that has no real eigeinvectors. However, every complex (and therefore real) matrix has complex eigenvectors. This is because the complex numbers are algebraically closed (every polynomial in C has a solution in C) so that the characteristic equation necessarily has a root. Distinct eigenvalues will have linearly independent eigenvectors.

A theorem is that an eigenvalue's geometric multiplicity (the dimension of the eigenspace) is less than or equal to its algebraic multiplicity (its multiplicity in the characteristic equation). For example, if the characteristic equation is (x-1)^2(x+2) then since 1 has algebraic multiplicity 2, you know that there are at most 2 linearly independent eigenvectors with eigenvalue 1. Since -2 has multiplicity 1, there is exactly one eigenvector (up to scale).

If there is a 2x2 matrix for example [0,0] in the first row and [0,1] in the 2nd row? how many independent eigenvectors does it have?

saadsarfraz, can you write down the characteristic equation of the matrix? That is a good place to start.

the characteristic equation is r(r-1)=0 which gives r=0 and r=1, for r=0 i get x2=0 and for r=1 i get x1=0,i think there might be two independent eigenvectors, but i would be grateful if someone could tell me what those eigenvectors be in this case.

HallsofIvy
Homework Helper
That's an easy case: if two eigenvector correspond to distinct eigenvalues, then they are independent.

Suppose $Au= \lambda_1 u$ and $Av= \lambda_2 v$ where $\lambda_1\ne\lambda_2$, u and v non-zero. That is, that u and v are eigenvectors of A corresponding to distinct eigenvalues. Let $a_1u+ a_2v= 0$. Applying A to both sides of the equation, $a_1A(u)+ a_2A(v)= 0$ or $a_1\lambda_1 u+ a_2\lambda_2 v= 0$.

First, if $\lambda_1= 0$, then we have $a_2\lambda_2 v= 0. Further,[itex]\lambd_2$ is non- zero because the eigenvalues are distinct so it follows that $a_2= 0$. If $\lambda_1\ne 0$, we can divide by it and get
$$a_1u+ \frac{\lambda_2}{\lambda_1}a_2 v= 0[/itex]. Since we also have that $a_1u+ a_2v= 0$, it follows that [tex]\frac{\lambda_2}{\lambda_1}a_2 v= a_2 v$$
again giving $\lambda_2= 0$.

If two eigenvectors correspond to the same eigenvalue, they are not necessarily distinct.

If it has distinct eigenvalues then it has linearly independent eigenvectors. However if the eigevalues are not distinct then you cannot guarantee linearly independent eigenvectors.