## 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

Consecutively define

(2)

...

(3)

**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

such that

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

(1)

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.

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