Characteristic & minimal polynomial & geometric multiplicity

Question

Let $\mathbb{K}$ be a field and $1\leq n\in \mathbb{N}$.

Let $a\in M_n(\mathbb{K})$, such that $a_{ij}=0$ for all $i\leq j$.

It holds that $a^n=0$.

For $1\leq k\in \mathbb{N}$ we define \begin{equation*}J_k:=\begin{pmatrix}0 & 1 & 0 & \ldots & \ldots & 0 \\ 0 & 0 & 1 & 0 & \ldots & 0 \\ \vdots & \ddots & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 1 & 0 \\ \vdots & \ddots & \ddots & \ddots & \ddots & 1 \\ 0 & \ldots & \ldots & \ldots & \ldots & 0\end{pmatrix}\in M_k(\mathbb{K})\end{equation*}

The characteristic and the minimal polynomial is $m(\lambda )=\lambda^k$.

Let $1\leq k_1\leq k_2\leq \ldots \leq k_{\ell}\in \mathbb{N}$ with $n=\sum_{i=1}^{\ell}k_i$ and $$A=\begin{pmatrix}J_{k_1} & & \\ & \ddots & \\ & & J_{k_{\ell}}\end{pmatrix}+\lambda I_n\in M_n(\mathbb{K})$$ I want to determine the characteristic polynomial of $A$, the minimal polynomial of $A$ and the geometric multiplicity of $\lambda$.

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To calculate the chatacteristic polynomial do we do the following? $$\det (A-\lambda I_n)=\begin{vmatrix}J_{k_1} & & \\ & \ddots & \\ & & J_{k_{\ell}}\end{vmatrix}$$ Do we use here the above result for $J_k$ ?

Answer 1

Your matrix $a$ is lower triangular. For a triangular matrix, the diagonal entries are the eigenvalues, which means that in our case the characteristic polynomial must be $(x - 0)^n = x^n$. By the Cayley Hamilton, the desired result holds.

Alternatively, we could use an approach that uses the ideas from your previous question. In this case, let $U_k = \operatorname{span}\{e_n,e_{n-1},\dots,e_k\}$, where $e_1,\dots,e_n$ are the canonical basis vectors of $\Bbb K^n$. We find that $a(U_k) \subset U_{k+1}$ for $k = 1,\dots,n-1$, and $a(U_n) = \{0\}$. With this, we similarly conclude that $a^n(x) = 0$ for all $x \in \Bbb K^n$, which is to say that $a^n = 0$.

Answer 2

Here a is lower triangular matrix with all diagonal entries 0, clearly 0 is the only eigenvalue of a with algebraic multiplicity n. Also , dimension of eigenspace corresponding to eigenvalue 0 is 1 i.e, Geometric multiplicity of 0 is 1.
By your definition of Jᵢ are called Jordon blocks, and yes , we can use to find Jordon canonical form and through this one can find the minimal polynomial of a.
Here in this case 0 is the only eigenvalue of a , since G.M. of
0 is 1, there is possibility for only one Jordon block in the expression of Jordon canonical form, so hence this block is the Jordon canonical form here, Which is, For $1\leq k\in \mathbb{N}$ we define \begin{equation*}J_k:=\begin{pmatrix}b & 1 & 0 & \ldots & \ldots & 0 \\ 0 & b & 1 & 0 & \ldots & 0 \\ \vdots & \ddots & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 1 & 0 \\ \vdots & \ddots & \ddots & \ddots & \ddots & 1 \\ 0 & \ldots & \ldots & \ldots & \ldots & b\end{pmatrix}\in M_k(\mathbb{K})\end{equation*} and here k=n, I mean the order of a and b is eigenvalue of a and of course it's the case when b is eigenvalue of a , I am done that for better understanding of about Jordon block, and for your matrix a , all diagonal elements of Jₙ should 0, 0 is eigenvalue of the given a!
So, here characteristics polynomial and minimal polynomial are same for the case when G.M. of a eigenvalue is 1.
I think it will help now!