Main theorem: Jordan normal form
By Exercise 1, it is sufficient to show that in each root subspace the matrix takes the Jordan form.
Step 1. Take a basis
Exercise 1. The vectors in (2) are linearly independent relative to the vectors in (3) are linearly independent.
Proof. Consider (2), for example. Suppose that Then
The conclusion that contradicts assumption (1).
Exercise 2. The system of vectors listed in (1)-(3) is linearly independent, so that its span is of dimension
Proof. Suppose Then by inclusion relations
which implies for by relative independence stated in (1). This process can be continued by Exercise 1 to show that all coefficients are zeros.
Next we show that in each of we can find a basis relative to the lower indexed subspace According to (1), in we already have such a basis. If the vectors in (2) constitute such a basis in , we consider
Step 2. If not, by Exercise 3 we can find vectors
represent a basis relative to
Then we can define
By Exercise 2, the 's defined here are linearly independent. But we can show more:
Exercise 3. All 's from Step 1 combined with the 's from Step 2 are linearly independent.
The proof is analogous to that of Exercise 2.
Denote the span of vectors introduced in Step 2. because they have different bases. Therefore we can consider a direct sum Repeating Step 2 as many times as necessary, after the last step we obtain a subspace, say, such that The restrictions of onto the subspaces on the right is described by Jordan cells with the same and of possibly different dimensions. We have proved the following theorem:
Theorem (Jordan form) For a matrix in one can find a basis in which can be written as a block-diagonal matrix
Here are (square) Jordan cells, with possibly different lambdas on the main diagonal and of possibly different sizes, and all off-diagonal blocks are zero matrices of compatible dimensions.