How can I guarantee that all the number of generalized eigenvectors are equal to the algebraic multiplicity?

Question

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.

Answer 1

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Answer 2

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.