Why is the geometric multiplicity of an eigen value equal to number of jordan blocks corresponding to it? Geometric multiplicity of an eigen value is $$ \dim \mathrm{null} (A -\lambda I)\tag 1.$$
Suppose $A$ is in jordan normal form and has two Jordan forms with eigen value $\lambda$, one of size   $2 \times 2$ and other of size $3\times 3$. Then, why is $\dim \mathrm{null} (A -\lambda I)$ necessarily equal to  $2 $ i.e. why is geometric multiplicity of $\lambda =2$?
From the concept of generalised eigen  vectors , I know the following :
$(T-\lambda I)^3$ will produce a Zero Matrix in place of both these sub-blocks.
 A: Because (the subspace corresponding to) each Jordan block for$~\lambda$ has an eigenspace for$~\lambda$ of dimension$~1$. In the traditional (upper triangular) form of Jordan blocks, this eigenspace is spanned by the first basis vector in the block. The decomposition of the generalised eigenspace $V_\lambda$ for$~\lambda$ into Jordan blocks is a direct sum, so $v\in V_\lambda$ is in the eigenspace for$~\lambda$ if and only if its projections on all the Jordan blocks are in their respective eigenspaces for$~\lambda$, and this makes the eigenspace for$~\lambda$ the direct sum of those $1$-dimensional eigenspaces of the blocks. The eigenvectors are those that may only have nonzero coordinates on each first row of a Jordan block.
A: You are trying to solve $(A-\lambda I)v=0$.  The matrix $A-\lambda I$ has three 1s in it - one in the $2\times 2$ block and two in the $3\times 3$ block.  So it has three pivots, each in a different column.  There are only two free variables: one from column of the $2\times 2$ block without a 1 in it; and one from the $3\times 3$ block.  So the set of solutions is two dimensional.
If there are any other eigenvalues, they will all have non-zero numbers on the diagonal of $A-\lambda I$, so they won't contribute any more free variables.
