2
Aug 19

## 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

(1) $x_{1,p},...,x_{k,p}\in N_{\lambda }^{(p)}$ independent relative to $N_{\lambda }^{(p-1)}.$

Consecutively define

(2) $x_{1,p-1}=(A-\lambda I)x_{1,p},\ ...,\ x_{k,p-1}=(A-\lambda I)x_{k,p}\in N_{\lambda }^{(p-1)},$

...

(3) $x_{1,1}=(A-\lambda I)x_{1,2},\ ...,\ x_{k,1}=(A-\lambda I)x_{k,2}\in N_{\lambda }^{(1)}.$

Exercise 1. The vectors in (2) are linearly independent relative to $N_{\lambda }^{(p-2)},...,$ the vectors in (3) are linearly independent.

Proof. Consider (2), for example. Suppose that $\sum_{j=1}^{k}a_{j}x_{j,p-1} \in N_{\lambda }^{(p-2)}.$ Then

$0=(A-\lambda I)^{p-2}\sum_{j=1}^{k}a_{j}x_{j,p-1}=(A-\lambda I)^{p-1}\sum_{j=1}^{k}a_{j}x_{j,p}.$

The conclusion that $\sum_{j=1}^{k}a_{j}x_{j,p}\in N_{\lambda}^{(p-1)}$ contradicts assumption (1).

Exercise 2. The system of $kp$ vectors listed in (1)-(3) is linearly independent, so that its span $L_{x}$ is of dimension $kp.$

Proof. Suppose $\sum_{j=1}^{k}a_{j,p}x_{j,p}+...+\sum_{j=1}^{k}a_{j,1}x_{j,1}=0.$ Then by inclusion relations

$\sum_{j=1}^{k}a_{j,p}x_{j,p}=-\sum_{j=1}^{k}a_{j,p-1}x_{j,p-1}-...- \sum_{j=1}^{k}a_{j,1}x_{j,1}\in N_{\lambda }^{(p-1)}$

which implies $a_{j,p}=0$ for $j=1,...,k,$ 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 $N_{\lambda }^{(p)},...,N_{\lambda}^{(1)}$ we can find a basis relative to the lower indexed subspace $N_{\lambda }^{(p-1)},...,N_{\lambda }^{(0)}=\{0\}.$ According to (1), in $N_{\lambda }^{(p)}$ we already have such a basis. If the vectors in (2) constitute such a basis in $N_{\lambda }^{(p-1)}$, we consider $N_{\lambda }^{(p-2)}.$

Step 2. If not, by Exercise 3 we can find vectors

$y_{1,p-1},...,y_{l,p-1}\in N_{\lambda }^{(p-1)}$

such that

$x_{1,p-1},...,x_{k,p-1},y_{1,p-1},...,y_{l,p-1}$ represent a basis relative to $N_{\lambda }^{(p-2)}.$

Then we can define

$y_{1,p-2}=(A-\lambda I)y_{1,p-1},\ ...,\ y_{l,p-2}=(A-\lambda I)y_{l,p-1}\in N_{\lambda }^{(p-2)},$

...

$y_{1,1}=(A-\lambda I)y_{1,2},\ ...,\ y_{l,1}=(A-\lambda I)y_{l,2}\in N_{\lambda }^{(1)}.$

By Exercise 2, the $y$'s defined here are linearly independent. But we can show more:

Exercise 3. All $x$'s from Step 1 combined with the $y$'s from Step 2 are linearly independent.

The proof is analogous to that of Exercise 2.

Denote $L_{y}$ the span of vectors introduced in Step 2. $L_{x}\cap L_{y}=\{0\}$ because they have different bases. Therefore we can consider a direct sum $L_{x}\dotplus L_{y}.$ Repeating Step 2 as many times as necessary, after the last step we obtain a subspace, say, $L_{z},$ such that $N_{\lambda }^{(p)}=L_{x}\dotplus L_{y}\dotplus ...\dotplus L_{z}.$ The restrictions of $A$ onto the subspaces on the right is described by Jordan cells with the same $\lambda$ and of possibly different dimensions. We have proved the following theorem:

Theorem (Jordan form) For a matrix $A$ in $C^{n}$ one can find a basis in which $A$ can be written as a block-diagonal matrix

(1) $A=\left(\begin{array}{ccc}A_{1} & ... & ... \\... & ... & ... \\... & ... & A_{m}\end{array}\right) .$

Here $A_{i}$ 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.