Linear - from a charpoly, determine if T is one to one, and nullity T

[bennyska]

Homework Statement

if T is a linear operator on V and dim V = 4 and charpoly(T)=t²(t-1)(t+7), then is T one-to-one? What is the dimension of the nullspace of T?

Homework Equations

The Attempt at a Solution

So i know T has 3 eigenvalues, 0, 1, -7. Since 0 is an eigenvalue, i know there's at least one vector v in V, v =! 0, such that T(v) = 0v = 0. so i know T is not one to one.
now i need to figure out the dimension of the null space. but i don't know anything about T, other than the charpoly splits. if i knew dim eigenspace corresponding to 0 = 2, then i would be able to construct a basis, and show that null space of T has dim 2. so, I'm a little lost.

 

[Dick]

I think all you can know from the characteristic polynomial having a factor of t^2 is that the null space of T has either dimension 1 or 2. [[0,0],[0,0]] has characteristic polynomial t^2 and has null space dimension 2. [[0,1],[0,0]] has characteristic polynomial t^2 and has null space dimension 1. No way to tell the difference from the characteristic polynomial.

 

[bennyska]

cool. that's kind of what i was thinking.