Let $V$ be a finite dimensional vector space and let $T: V \to V$ be a diagonalizable linear map with characteristic polynomial $$f ( x ) = x^2(x-3)^2(x-9)(x+2)$$
Find the dimension of the image of T.
What does the characteristic polynomial tell us about the dim(im(T))? Is it simply how many unrepeated eigenvalues there are (so in this case 4 because we have eigenvalues 0,3,9,-2)? Please let me know how I should think about it, thank you!!
Since the map is diagonalisable we know that the algebraic multiplicities of an eigenvalue is equal to the dimension of the eigenspace corresponding to that eigenvalue.
In particular for the eigenvalue $\lambda = 0$ the eigenspace is the space
$$ \ker (T - \lambda I) = \ker (T-0I) = \ker T. $$ It follows that $\dim \ker T$ must be equal to $2$ since this is also the algebraic multiplicity of the eigenvalue $\lambda = 0$.
Moreover $\dim V = 2 + 2 + 1 + 1 = 6$ so we can apply the rank nullity theorem to get
$$ \dim T(V) + \dim \ker T = \dim V \iff \dim T(V) +2 = 6 \iff \dim T(V) = 4$$
Alternatively if you denote by $V_{(\lambda)}$ the eigenspace for the eigenvalue $\lambda$ then you clearly have
$$ T(V_{(\lambda)}) = \begin{cases} V_{(\lambda)} & \text{ if } \lambda \neq 0 \\ 0 & \text{ if } \lambda = 0 \end{cases}. $$ Since $V$ is diagonalisable then $V = \bigoplus_{\lambda} V_{(\lambda)}$ so we have \begin{align*} T(V) &= T \left(\bigoplus_{\lambda \text{ eigenvalue}} V_{(\lambda)} \right) \\ &= \bigoplus_{\lambda \text{ eigenvalue}} T(V_{(\lambda)}) \\ &= \bigoplus_{0 \neq \lambda \text{ eigenvalue}} V_{(\lambda)}. \end{align*}
Therefore $$\dim T(V) = \sum_{ \lambda \neq 0} \dim V_{(\lambda)} = 2 + 1 + 1 = 4$$