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I have found many questions in the Stack but, I couldn't find what I want... Since my major is physics, I'm not good at the terms in mathematics. So it was hard to reach understanding Jordan normal form.. If you can answer with an easy approach, I'll express my gratitude in advance.

For the sake of asking in detail, I take an example that the characteritic of a $5\times 5$ matirx $A$ be $p(t)=(t-\lambda)^5$. (the algebraic multiplicity is 5.)

In this case, I can find the linearly independent eigenvectors by finding how many linearly independent rows or columns is in $A-\lambda I$. Let the number of independent rows or columns in $A-\lambda I$ is 3. Then, The number of Jordan block is 3. Following this manner up to $(A-\lambda I)^5$, I can find the size of each block.

Here is my question. Can I guarantee that there are totally 5 generalized eigenvectors? If I assume that there is the excess or lack of generalized eigenvectors, what happens? I think that the contradictions arise definitely.

Jae Hoon Jeong
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    one simple thing: if you calculate the minimal polynomial instead, and the eigenvalue $\lambda$ occurs as $(x- \lambda)^k,$ then the largest Jordan block for $\lambda$ is size $k$ by $k.$ Anyway, – Will Jagy Mar 23 '22 at 02:48
  • I have found how to derive the minimal polynomial... However I couldn't understand... is there any good doccument for the derivation of minimal polynomial? – Jae Hoon Jeong Mar 23 '22 at 02:50
  • My advice is to do many problems for which all the matrix entries and eigenvalues are integers. The practice that will help is taking such a matrix $A,$ finding a matrix $P$ of all integers such that $P^{-1} A P = J$ is in Jordan form. I've answered a bunch of such questions, I guess I can put links to them as an answer – Will Jagy Mar 23 '22 at 03:51
  • Every $n\times n$ matrix with (real or) complex entries has $n$ generalized eigenvectors (over the complex numbers). If $A$ has Jordan form $J$, then there's a matrix $P$ with $P^{-1}AP=J$, and the columns of $P$ are the (generalized) eigenvectors. – Gerry Myerson Mar 23 '22 at 04:23
  • To Gerry Myerson. As far as I know, any matrix has Jordan forms. My question is about whether every $n\times n$ matrix has indeed $n$ generalized eigenvectors. – Jae Hoon Jeong Mar 23 '22 at 05:22
  • Yes, every $n\times n$ matrix $A$ has a Jordan form $J$. And the relation between $A$ and $J$ is $P^{-1}AP=J$ for some invertible matrix $P$. And the columns of $P$, of which there are $n$, are the generalized eigenvectors of $A$. So $A$ has $n$ generalized eigenvectors. – Gerry Myerson Mar 23 '22 at 11:40
  • When I wrote, "If $A$ has Jordan form $J$ ...," that was to be interpreted as "If we let $J$ be the Jordan form of $A$ ...." It wasn't meant to suggest that there are matrices that don't have a Jordan form. – Gerry Myerson Mar 23 '22 at 11:43
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    To Gerry Myerson. Oh! we can think like that! I like the logic because it is very simple! thank you so much!! – Jae Hoon Jeong Mar 24 '22 at 07:50
  • The number of Jordan blocks corresponding to eigenvalue $\lambda$ is the nullity of (A-$\lambda$I) and not its rank. – Lawrence Mano Mar 25 '22 at 19:22

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The characteristic polynomial of the following matrix is $p(x)=x^5$, and its minimal polynomial is the same: $$ A=\left[\begin{array}{ccccc} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right] $$ Indeed, it is easy to verify that $A^5=0$ and $A^4\ne 0$ by applying powers of $A$ to the column vector with all zeroes, except for a $1$ in the bottom place: $$ \left[\begin{array}{c}0 \\ 0 \\ 0 \\ 0 \\ 1\end{array}\right]\mapsto\left[\begin{array}{c}0 \\ 0 \\ 0 \\ 1 \\ 0\end{array}\right]\mapsto\left[\begin{array}{c}0 \\ 0 \\ 1 \\ 0 \\ 0\end{array}\right]\mapsto\left[\begin{array}{c}0 \\ 1 \\ 0 \\ 0 \\ 0\end{array}\right]\mapsto\left[\begin{array}{c}1 \\ 0 \\ 0 \\ 0 \\ 0\end{array}\right]\mapsto\left[\begin{array}{c}0 \\ 0 \\ 0 \\ 0 \\ 0\end{array}\right]. $$ So the minimal polynomial for $A$ is the same as its characteristic polynomial $p(x)=x^5$. The Jordan canonical form is designed to deal with this type of scenario.

Disintegrating By Parts
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Is this a correct basis for the Jordan normal form?

Will Jagy
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  • Thank you so much! I'll read all of them! If I get several questions though, I'll ask you again. – Jae Hoon Jeong Mar 23 '22 at 05:24
  • I have read several sites you recommended. Unfortunately, finding generalized eigenvectors for such specific examples is not what I want to know... What I want to know exactly is whether we can find always $n$ generalized eigenvectors in a $n\times n$ matrix. – Jae Hoon Jeong Mar 23 '22 at 05:34