What's the relationship between geometric multiplicity of eigenvalues and dim ker(T)? Let $T: \mathbb{R}^7 \rightarrow \mathbb{R}^7$ be a diagonalizable linear operator with characteristic polynomial give by $p(t) = t(t-1)^2(t+2)^3(t -3)$.
Calculate $\dim(\ker(T - Id), \dim(Im(T + 2Id)), \dim(Im(T))$
I'm thinking whats the relationship between knowing that the algebraic multiplicity of each eigenvalue equals the geometric multiplicity and the dimension of the kernel and the image.

Since $T$ is diagonalizable, it has an eigenbasis (in other words, you can find a linearly independent collection of eigenvectors which span $\mathbb{R}^7$).  Let $p(t)$ be the characteristic polynomial of $T$ (or the matrix associated to $T$).  Then, 


*

*$\ker(T-\lambda I)$ is a linear space of dimension equal to the multiplicity of $\lambda$ as a root of $p(t)$.  The idea is that if $A$ is the matrix corresponding to $T$, then diagonalizability means that there is an invertible matrix such that $P^{-1}AP=D$ is diagonal.  Then, $A-tI$ can be diagonalized as $PDP^{-1}-tPIP^{-1}=P(D-tI)P^{-1}$.  If $\lambda$ is an eigenvalue then $D-\lambda I$ will have as many zeros on the diagonal as the multiplicity of the root of $p(t)$ at $\lambda$.  Moreover, if the $i$-th position is zero, then $Pe_i$ will be in the kernel (and be an eigenvector).  All $Pe_j$'s where the $j$-th position is nonzero are not eigenvectors.

*The eigenbasis vectors which correspond to $\lambda$ are a basis for $\ker(T-\lambda I)$.  The construction above describes how to get these.

*The dimension of the images can be found through the rank-nullity theorem.
The first two facts fail when $T$ is not diagonalizable.
