Aug 19

Main theorem: Jordan normal form

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.

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