$n\times n$ Matrix with Fewer than n Eigenvalues

Question

I have a problem which asks for all JCF of a 7x7 matrix with characteristic polynomial $t^3(t-1)^2$. Shouldn't the algebraic multiplicity of all Eigenvalues add up to 7? Why not? Are there any examples that come to mind?

Answer 1

Theorem: Let $A$ be an $n \times n$ matrix and $f(\lambda)=\det(A-\lambda I)$ be its characteristic polynomial. Then $f(\lambda)$ is a polynomial of degree $n$.

${}$

Minimal polynomial: The monic polynomial $p(x)$ of least degree such that $p(A) = 0$ for a square matrix $A$ called the minimal polynomial of $A$ .

---

Your claimed characteristic polynomial $t^3(t-1)^2$ is of degree $5$, so must not be characteristic polynomial of $A=(a_{ij})_{7 \times 7}$ but minimal polynomial of $A$.

So the possible characteristic polynomial are $t^5(t-1)^2$ or, $t^3(t-1)^4$ or, $t^4(t-1)^3$.

Case I: If we take characteristic polynomial of $A$ as $t^5(t-1)^2$, then the algebraic multiplicity of the eigenvalue $0$ and $1$ is $5$ and $2$ respectively.

Since the matrix is unknown, the geometric multiplicity of the eigenvalues are not to be determined. So we can choose the geometric multiplicity of the eigenvalue $0$ as $1$, or $2$, or $3$, or $4$, or $5$ and the geometric multiplicity of the eigenvalue $1$ as $1$, or $2$.

So the possible combinations of geometric multiplicities of the eigenvalue $0$ and $1$ are respectively $\{1,1\}, \{1,2\}, \{1,3\}, \{1,4\}, \{1,5\}, \{2,1\}, \{2,2\}, \{2,3\}, \{2,4\}, \{2,5\}$.

then follow the rules to find the Jordan Canonical form of $A$.

${}$

$1.$ Geometric multiplicity of an eigenvalue $=$ number of Jordan Block associated with that eigenvalue.

$2.$ If algebraic multiplicity of the eigenvalue $\lambda$ $=$ geometric multiplicity $=k$(say), then $[\lambda]$ will be $k$ identical Jordan blocks.

$3.$ If algebraic multiplicity of the eigenvalue $\lambda$ $(= k)$ $\neq$ geometric multiplicity $(= m\quad \text{and}\quad k \gt m)$ , then $J_1=\begin{pmatrix} \lambda & 1 & 0 & . . . & 0 & 0 \\ 0 & \lambda & 1 & . . . & 0 & 0\\ 0 & 0 & \lambda & . . . & 0 & 0\\ .\\ .\\ .\\ 0 & 0 & 0 & . . . & \lambda & 1\\ 0 & 0 & 0 & . . . & 0 & \lambda \end{pmatrix}_{k \times k}$ is the only one Jordan blocks associated with the eigen value $\lambda$.

Other two cases are exactly same as Case I.