# Characteristic & minimal polynomial & geometric multiplicity

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 $$\lambda$$.



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 $$J_k$$ ?

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 $$a^n = 0$$.

• I got stuck right now. The characteristic polynomial of $a$ is equal to the determinant $\det (a-\lambda I_n)=\begin{vmatrix}J_{k_1} & & \\ & \ddots & \\ & & J_{k_{\ell}}\end{vmatrix}$ or not? And the diagonal elements of that triangular matrix are $0$, or not? Isn't the determinant equal to $0$ ? Commented Jul 11, 2020 at 9:01
• @MaryStar Yes, it is true that $\det(A - \lambda I_n)$ is the characteristic polynomial. I don't know where you're getting $$\begin{vmatrix}J_{k_1} & & \\ & \ddots & \\ & & J_{k_{\ell}}\end{vmatrix}$$ from, but this is obviously no equal to the characteristic polynomial since it does not depend on $\lambda$. Commented Jul 11, 2020 at 9:21
• @MaryStar In your question, you say that $$a=\begin{pmatrix}J_{k_1} & & \\ & \ddots & \\ & & J_{k_{\ell}}\end{pmatrix}+\lambda u_n\in M_n(\mathbb{K}),$$ but this is not necessarily true. We can say that if $a$ has only a single eigenvalue, then it is similar to a matrix that has the form on the right Commented Jul 11, 2020 at 9:22
• @MaryStar To the second question of your comment: I'm not sure what you're asking. We know that $a$ is a lower triangular matrix with zeros on the diagonal, and the matrix on the right is an upper triangular matrix with zeros on the diagonal. Both of these matrices have $0$ as their only eigenvalue. Commented Jul 11, 2020 at 9:24
• @Mary Star what you are given \begin{vmatrix}J_{k_1} & & \\ & \ddots & \\ & & J_{k_{\ell}}\end{vmatrix}, this is possible but just you have take only one block Jᵢ, because number of blocks depends on the geometric multiplicity of eigenvalue! Commented Jul 11, 2020 at 9:44

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!

• I got stuck right now. At $A$ the diagonal elements are $\lambda$, or not? Commented Jul 11, 2020 at 10:43
• I am taking about Jordon blocks, of course it have eigenvalue as it's diagonal entries! Commented Jul 11, 2020 at 12:00
• Sorry , I can't type this kind matrix and that's why I just copy , paste for the matrix typo part from your question! That's why there can be some similar part of your question part, sorry! Commented Jul 11, 2020 at 12:14
• @Mary Star is this answer still unacceptable? Please you can tell me if I did miss something! Commented Jul 12, 2020 at 4:25