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I know there are different ways to approach the JNF, one popular of them being expanding the Eigenspace by taking a potence of $\ker (f-\lambda id)$ where $f$ is an Endomorphism and $\lambda$ an Eigenvalue of the respective Eigenspace, where I will just care about one (the one of $\lambda$ in the following).

After expanding to the generalized eigenspace and showing the cyclic decomposition of the vector space one considers the nilpotent endomorphism $(f-\lambda id)$ and finds a decomposition again.

I know that $f\ker(f-\lambda id)^k=\lambda\ker(f-\lambda id)^k$.

But if we build the JNF matrix, we have something like this

$f\ker(f-\lambda id)^k=\ker(f-\lambda id)^{k+1}+\lambda\ker(f-\lambda id)^k$.

otherwise the JNF wouldn't have the form it does. My problem is, although I get the decompositions, I don' t understand how the basis is obtained, or why the construction of the matrix has too look like above.

1 Answers 1

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Suppose there's only one eigenvalue, $\lambda$. Let $r$ be the index of nilpotency of $A-\lambda I$ and consider the sequence: $$\ker(A-\lambda I)\varsubsetneq\bigl(\ker(A-\lambda I)\bigr)^2\varsubsetneq\dots\varsubsetneq\bigl(\ker(A-\lambda I)\bigr)^r=\bigl(\ker(A-\lambda I)\bigr)^{r+1}=\dotsm $$ The strategy is the following: find the maximal number of linearly independent vectors in $\;\bigl(\ker(A-\lambda I)\bigr)^r\smallsetminus \bigl(\ker(A-\lambda I)\bigr)^{r-1}$ (i.e. if you've heard of quotient vector spaces lift a basis of the quotient space $\;\bigl(\ker(A-\lambda I)\bigr)^r/\bigl(\ker(A-\lambda I)\bigr)^{r-1}$).

$A-\lambda I$ maps this basis to a system of linearly independent vectors in $\;\bigl(\ker(A-\lambda I)\bigr)^{r-1}\smallsetminus \bigl(\ker(A-\lambda I)\bigr)^{r-2}$. Complete to a maximal system of linearly independent vectors in $\;\bigl(\ker(A-\lambda I)\bigr)^{r-1}\smallsetminus \bigl(\ker(A-\lambda I)\bigr)^{r-2}$, and map it by $A-\lambda I$ to a set of linearly independent vectors in $\bigl(\ker(A-\lambda I)\bigr)^{r-2}$.

ascading this way, you end up in a set of linearly independent vectors in the eigenspace $\ker(A-\lambda I)$, which you complete in a basis of the eigenspace. This basis is by construction a Jordan basis

Note:

Setting $d_k=\dim\bigl(\ker(A-\lambda I)\bigr)^k$ and $d_0=0$, the sequence $(d_k-d_{k-1})$ is decreasing and its value is the number of Jordan blocks of size $\ge k$. In particular $d_1$ (the dimension of the eigenspace) is the total number of Jordan blocks.

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    Thanks for the answer. I wanted to avoid using quotient spaces. I know this way, there must be another one without it. – 2017-01-18
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    The quotient spaces are here just as a comment aside to explain what happens really behind the scene. All one has to retain is the maximal number of linearly independent vectors in each kernel, but not in the previous one, and their mapping by $A-\lambda I$. – 2017-01-18
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    But it is not clear how to obtain this basis of the kernel, and thats the point. How do I know, that the basis in the kernel looks like this? – 2017-01-18
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    Which kernel? There are plenty of kernels around here? – 2017-01-18
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    The kernel $\ker(A-\lambda id)^r$. – 2017-01-18
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    Well, it's a linear system to solve, as for any linear system, writing the matrix in reduced row echelon form. This yields a basis of the space of solutions, of which you eliminate those which belong to the previous kernel. – 2017-01-18