0
$\begingroup$

Let $\dim(V)=n$ over the field $\mathbb{C}$. The Jordon canonical form of a linear transformation $T\colon V\rightarrow V$ can be obtained in the following way.

1) Let $c_T(x)=(x-\lambda_1)^{m_1}(x-\lambda_2)^{m_2}\cdots (x-\lambda_r)^{m_r}$ be the characteristic polynomial of $T$. Then

2) $V=\ker(T-\lambda_1I)^{n_1} \oplus \ker(T-\lambda_2I)^{n_2}\oplus \cdots\oplus\ker(T-\lambda_rI)^{n_r}$.

3) It is sufficient to find a canonical basis for each component $\ker(T-\lambda_iI)^{n_i}$. For it, consider the chain $$0 \subset \ker(T-\lambda_iI) \subset \ker(T-\lambda_iI)^{2} \subset \cdots \subset \ker(T-\lambda_iI)^{n_i}=\ker(T-\lambda_iI)^{n_i+1}$$

Denote this chain by $$0 \subset K_{\lambda_i}^1 \subset K_{\lambda_i}^2 \cdots \subset K_{\lambda_i}^{n_i}.$$

4) Let $e_{n_i,1},e_{n_i,2},\ldots, e_{n_i,q_i}$ be a basis of $K_{\lambda_i}^{n_i}$ modulo $K_{\lambda_i}^{n_i-1}$. Apply $T-\lambda_iI$ on these vectors, to obtain $n_i$ linearly independent vectors in $K_{\lambda_i}^{n_i-1}$: $$(T-\lambda_iI)e_{n_i,1},(T-\lambda_iI)e_{n_i,2},\ldots, (T-\lambda_iI)e_{n_i,q_i}.$$

Extend this set to a basis of $K_{\lambda_i}^{n_i-1}$ modulo $K_{\lambda_i}^{n_i-2}$, say $$(*)\qquad (T-\lambda_iI)e_{n_i,1},(T-\lambda_iI)e_{n_i,2},\ldots, (T-\lambda_iI)e_{n_i,q_i}, e_{n_i,q_i+1}, \ldots, e_{n_i,q_i+t_i}$$

5) Repeat step 4 for the set $(*)$. The unions of basis of $K_{\lambda_i}^{n_i}/K_{\lambda_i}^{n_i-1}, K_{\lambda_i}^{n_i-1}/K_{\lambda_i}^{n_i-2},\ldots, K_{\lambda_i}^{1}/0$ will give a canonical basis for the component $\ker(T-\lambda_iI)^{n_i}$.

Question Do anyone knows a standard or elementary textbook reference for this method? I couldn't find any intuition/ idea behind the proof. It will be better if one suggests some simple example to illustrate this procedure.

I know the only following link:

$\endgroup$
  • $\begingroup$ Peter Lax's Linear Algebra and Its Applications has this construction in the appendix I think. $\endgroup$ – Mark Apr 1 '15 at 9:26
  • $\begingroup$ Note that "canonical basis" in point 5) is completely out of place. While the form of the matrix is canonical, the basis is very far from canonical. While there is discussion about what "canonical" means in general, there must be something that puts a canonical thing apart from arbitrary similar things in its class. Here there is just nothing of that. $\endgroup$ – Marc van Leeuwen Apr 2 '15 at 9:33
0
$\begingroup$

Check Halmos "Finite Dimensional Vector Spaces" classic (1958 or 1974 reprint), section 58. It may be phrased differently than your approach, but it is essentially the same method.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.