How does Jordan form connect to the kernel of a matrix?

[JamesGoh]

For some of my homework exercises, the lecturer has specified the values for a kernel of a square matrix minus the scalar product of the eigenvector and the identity matrix.

Mathematically, I am given

kernel(A - λI) = some integer value where A is a nxn square matrix

lambda is the eigenvalue obtained from the characteristic polynomial

I is the identity matrix

My question is, what is the logical and mathematical connection between kernel(A - λI) and finding the equivalent jordan block ?

 

[Pi01]

The kernel of an operator (or matrix) is a vectorspace. The operator transforms the elements of the kernel to the null vector. For the Jordan form, the following dimensions should be known

dim kernel (A-λI)^k,

where k=1...r and r is the algebraic multiplicity of λ.

For k=1,

dim kernel (A-λI)

gives the number of blocks of λ in the J form.

If and only if,

dim kernel (A-λI)=r

for all λ, then the matrix can be diagonalized, since each block is 1-dimensional.