# Finding ch. polynomial and Jordan normal form of $f$ knowing $\dim\ker f=2$ and there are $a,b$ not in $\ker f$ such that $f^2(a)=0, f(b)=b$

Given a vector space V of dimension $$4$$, let $$f$$ be an endomorphism such that $$\dim(\ker f)=2$$. Assuming there exist $$a,b\in V\setminus \ker f$$ such that $$f^2(a)=0, f(b)=b$$ I should find $$\chi_f$$ and the Jordan normal form of $$f$$.

I would like feedback on my approach, which I came up with trying to follow the answer to a related question of mine : $$f(f(a))=0, f(b)=b$$ and we know both $$f(a)$$ and $$b$$ are nonzero, which means $$f(a)$$ is an eigenvector of $$f$$, associated with the eigenvalue $$\lambda_0=0$$ and $$b$$ is an eigenvector of $$f$$, associated with the eigenvalue $$\lambda_1=1$$.

In particular the geometric multiplicity of $$\lambda_0$$ is $$\dim\ker (f-\lambda_0I)=\dim(\ker f)=2$$, i.e. there are $$2$$ Jordan blocks relative to $$\lambda_0=0$$. Now, $$f(a)$$ and $$b$$ form a basis for a $$2\times2$$ Jordan block, therefore we can conclude $$J=\begin{bmatrix} 1&0&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&0&0&0\end{bmatrix} \ \text{and} \ \chi_f(\lambda)=\lambda^3(\lambda-1).$$Is this correct?

Your answer is correct and your reasoning is mostly correct. However I don't understand how you can conclude that $$f(a), b$$ form a basis for a $$2 \times 2$$ Jordan block? For starters they have different eigenvalues so they can't possibly be in the same Jordan block. Want you want to say is the following.

There are at least two distinct $$0$$-Jordan blocks, corresponding to $$\operatorname{ker}(f)$$, and $$\operatorname{ker}(f^{2})$$. From the data you have got it you should be able to see that $$\operatorname{dim}\operatorname{ker}(f) = 2$$, and that $$\operatorname{dim}\operatorname{ker}(f^{2}) > \operatorname{dim}\operatorname{ker}(f) = 2.$$ But since $$\operatorname{dim}\operatorname{ker}(f - \operatorname{Id}) \geq 1,$$ we have that $$3 \leq \operatorname{dim}\operatorname{ker}(f^{2}) \leq 4 - 1 = 3,$$ and so $$\operatorname{dim}\operatorname{ker}(f^{2}) = 3$$. Thus we have

$$\operatorname{dim}\operatorname{ker}(f) = 2, \ \operatorname{dim}\operatorname{ker}(f^{2}) = 3, \ \operatorname{dim}\operatorname{ker}(f - \operatorname{Id}) = 1$$

Then our Jordan decomposition is

$$\operatorname{ker}(f) \oplus \left(\operatorname{ker}(f^{2}) / \operatorname{ker}(f)\right) \oplus \operatorname{ker}(f - \operatorname{Id})$$

With $$\operatorname{dim}\operatorname{ker}(f) = 2, \operatorname{dim}\left( \operatorname{ker}(f^{2}) / \operatorname{ker}(f) \right) = 1$$, and $$\operatorname{dim}\operatorname{ker}(f - \operatorname{Id}) = 1$$.

Thus we have a $$(2\times 2)$$-Jordan block for eigenvalue zero, a $$(1 \times 1)$$-Jordan block for eigenvalue zero, and a $$(1 \times 1)$$-Jordan block for eigenvalue $$1$$. So we are done.

In case there is any confusion, the bases for the Jordan blocks are:

$$(2 \times 2)$$-$$0$$-Jordan block: $$\{f(a), a\}$$,

$$(1 \times 1)$$-$$0$$-Jordan block: $$\{u \}$$ where $$u \in \ker{f} \backslash \operatorname{Span}\{ f(a) \}$$,

$$(1 \times 1)$$-$$1$$-Jordan block: $$\{ b \}$$

• Thank you for your answer and your good explanation! I just have one concern: how do you know $\dim\ker(f^2)$ is strictly larger than $\dim\ker(f)$? Switching to a representative matrix $A$, it's equivalent to stating $\operatorname{rk}(A^2)<\operatorname{rk}(A)$, but in general we may only say $\operatorname{rk}(AB)\le\min\{\operatorname{rk}(A), \operatorname{rk}(B)\}$ - how do we exclude $\operatorname{rk}(A^2)=\operatorname{rk}(A)$ here? – Learner Jan 22 at 12:52
• Because $a$ is something that is not in $\ker{f}$ but $f(a)$ is in $\ker{f}$ – Adam Higgins Jan 22 at 14:43
• Ok sorry for not following, but how does that affect the cardinalities of the bases of $\ker f$ and $\ker(f^2)$? Could you please make it explicit for me? – Learner Jan 22 at 14:56
• @Learner The cardinality of a finite dimensional vector space is the maximum size of a linearly independent set. If $U$ is a proper vector subspace of $V$, with $v \not\in U$, and $\mathcal{B}$ a basis of $U$, then if $\left| \mathcal{B}\right| =n$, then $\mathcal{B} \cup \{v\}$ is a linearly independent subset of $V$. Hence the dimension of $V$ is no less than $n+1$. – Adam Higgins Jan 22 at 15:03
• OH right, right! I was mixing up $a$ and $f(a)$, thank you again! – Learner Jan 22 at 15:07