Linear Algebra Jordan form.

Question

Suppose $f$ $\in$ to $L(C^{10},C^{10}) $ linear functions and has only one eigenvalue say λ .

Suppose $g=f-λid $. And

$$\dim(\text{Ker}(g))=4$$ $$\dim(\text{Ker}(g^2))=7$$ $$\dim(\text{Ker}(g^3))=9$$ $$\dim(\text{Ker}(g^4))=10$$.

Find the Jordan matrix of the $f$ subject to λ.!!

Help if it is possible .I know that the $\text{Ker}(g)$ has all the eigenvectors of $f$ .So how do i find the information i need from the given dimensions for my Jordan matrix.

Answer 1

Hints (justifications omitted):

• There is only one eigenvalue $\lambda$ so the matrix's diagonal elements are all $\lambda$. The question is how large the Jordan blocks are (i.e., where to put the $1$s above the diagonal).
• $\dim(\ker(g))=4$ implies that there are four Jordan blocks.
• more generally, $\dim(\ker(g^n))-\dim(\ker(g^{n-1}))$ (for $n>1$) tells you how many of the Jordan blocks have size $\ge n$. [Recall that $\ker(g^n)$ contains the generalized eigenvalues of order $\le n$.]

This should allow you to answer the question, but you need to make sure you understand the theorem for the Jordan decomposition and why these statements are true. It is unclear how much you know; if you ask more specific questions people can help you better.

Answer 2

I'll try to sketch why $\dim(\ker g^i)-\dim(\ker g^{i-1})=r_i$, the number of Jordan blocks of size $\ge i$. In particular, the number of Jordan blocks is $r=r_1$. We'll suppose we work in $\mathbf C^n$.

First note the matrix of $g$ is block-diagonal matrix of the form: $$J_l=\begin{pmatrix}0&1&0&\dots&0\\ 0&0&1&\dots&0\\[-1ex] \vdots&\vdots&\vdots&\ddots&\vdots\\[-1ex] 0&0&0&\dots&1\\0&0&0&\dots&0\end{pmatrix}$$ Each such block clearly has rank $l-1$, hence the sum of the ranks of these blocks is $\;\displaystyle\sum_l (l-1)=\sum_l l- r=n-r$, which proves $\;\dim(\ker g)=r$.

Next, as we have: $$J_l^2=\begin{pmatrix}0&0&1&0&\dots&0\\ 0&0&0&1&\dots&0\\[-3ex] \vdots&\vdots&\vdots&&\ddots&\vdots\\[-1.5ex] 0&0&0&0&\dots&1\\0&0&0&0&\dots&0\\0&0&0&0&\dots&0 \end{pmatrix}$$ we see that $$\operatorname{rk}(J_l^2)=\begin{cases}l-2=\operatorname{rk}(J_l)-1&\text{ if}\enspace l\ge 2\\0&\text{ if}\enspace l=1\end{cases},\enspace\text{hence}\enspace \dim(\ker g^2)=\dim(\ker g)+r_2.$$ And so on: on a Jordan block, when the exponent steps from $k$ to $k+1$, the overdiagonal of 1s moves one step further north-east, hence the rank diminishes by 1 and dimension of the kernel increases by $1$ for those $l\ge k+1$.

Application:

The sequence of dimensions (counting $\dim g^0=0$): $\; 0,4,7,9,10$. Thus there are:

• $4$ Jordan blocks (of dimension $\ge 1$, by definition);
• $3$ blocks of size $\ge 2$;
• $2$ blocks of size $\ge 3$;
• $1$ block of size $\ge 4$.

All this implies there is $1$ block of size $1$, $1$ block of size $2$ and $1$ block of size $3$. Hence the block of size $\ge 4$ has size $10-(1+2+3)=4$.