Let $f$ and $g$ be two linear operators, show that eigenvalues of $g$ are of the form $a, a-1, a-2, \ldots, a - (n - 1)$?

Question

Let $f$ and $g$ be two linear operators with $\dim(L) = n$, where $L$ is a vector space over a field $F$ with characteristic zero. Assume that $f^{n} = 0$, $\dim \ker(f) = 1$, and $gf-fg = f$. How it can be shown that the eigenvalues of $g$ have the form $a$, $a-1$, $a-2$, $\ldots$, $a - (n - 1)$ for some $a$ from $F$?

Answer 1

I assume that $f$ and $g$ are linear operators on $L$ (you did not specify, but it is clear enough). Let $U_j$ denote $\ker f^j$. Clearly, $\{0\}=U^0\subseteq U^1\subseteq U^2\subseteq \ldots \subseteq U^{n-1}\subseteq U^n=L$. Also, by induction, $$k\ f^k=gf^k-f^kg$$ for all $k=0,1,2,3,\ldots$.

We claim that $U^j$ is an invariant subspace with respect to $g$. For $j=0$, this is trivial. For $j=1$, we have $fu=0$ for all $u\in U^1$. Therefore, $$0=fu=(gf-fg)u=g(fu)-f(gu)=-f(gu).$$ Hence, $gu\in U^1$. We now suppose that each of $U^0,U^1,\ldots,U^{k-1}$ is an invariant subspace wrt $g$. Let $u\in U^k$. Then, $$0=k\ f^ku=(gf^k-f^kg)u=g(f^ku)-f^k(gu)=0-f^k(gu).$$ Consequently, $gu\in U^k$.

Now, since $\dim \ker f=1$, $U^1$ is $1$-dimensional. So, $g|_{U^1}:U^1\to U^1$ must be a scalar multiplication by some $a\in F$. For each $j=1,2,\ldots, n$, note that $f|_{U^j}:U^{j}\to U^{j-1}$ is a surjective map with $1$-dimensional kernel (this part requires the assumption that $f^n=0$). So we have $\dim U^j=j$ for $j=0,1,2,\ldots,n$.

Let $u_1\in U^1\setminus U^0=U^1\setminus\{0\}$ be arbitrary, so $U^1=\operatorname{span} \{u_1\}$. Note that $gu_1=au_1$. There exists $v_2\in U^2\setminus U^1$ such that $fv_2=u_1$. Note that $U^2=\operatorname{span}\{u_1,v_2\}$. We have $$f(gv_2)=\big(gf-(gf-fg)\big)v_2=g(fv_2)-fv_2=gu_1-u_1=(a-1)u_1=f\big((a-1)v_2\big).$$ That is, $$f\big(gv_2-(a-1)v_2\big)=0,$$ so $gv_2=(a-1)v_2+t_2u_1$ for some $t_2\in F$ since $\ker f=U^1=\operatorname{span}\{u_1\}$. That is, $$g\big(v_2-t_2u_1\big)=(a-1)\big(v_2-t_2u_1\big).$$ Setting $u_2=v_2-t_2u_1$, we have $U^2=\operatorname{span}\{u_1,u_2\}$ with $gu_2=(a-1)u_2$. (As I wrote, I realized that the step to find $u_2$ can be combined with the paragraph below, but well, since it had already been written, I am leaving this here.)

Suppose, for some integer $k$, $1\leq k\leq n$, we have $u_1,u_2,\ldots,u_{k-1} \in L$ such that $gu_j=(a+1-j)u_j$ and $\operatorname{span}\{u_1,u_2,\ldots,u_j\}=U^j$ for $j=1,2,\ldots,k-1$. As before, there exists $v_k\in U^k\setminus U^{k-1}$ such that $fv_k=u_{k-1}$. We have \begin{align}f(gv_k)&=\big(gf-(gf-fg)\big)v_k=g(fv_k)-fv_k=gu_{k-1}-u_{k-1}\\&=\big(a+1-(k-1)\big)u_{k-1}-u_{k-1}=(a+1-k)u_{k-1}.\end{align} That is, $$f\big(gv_k-(a+1-k)v_k\big)=0,$$ so $gv_k=(a+1-k)v_k+t_ku_1$ for some $t_k\in F$ since $\ker f=U^1=\operatorname{span}\{u_1\}$. That is, $$g\big(v_k-t_ku_1\big)=(a-1)\big(v_k-t_ku_1\big).$$ Setting $u_k=v_k-t_ku_1$, we have $U^k=\operatorname{span}\{u_1,u_2,\ldots,u_k\}$ with $gu_k=(a+1-k)u_k$.

Finally, because $U^n=L$, we have found a basis $\{u_1,u_2,\ldots,u_n\}$ of $L$, for which $gu_j=(a+1-j)u_j$ for all $j=1,2,\ldots,n$. Thus, the eigenvalues of $g$ are $a$, $a-1$, $a-2$, $\ldots$, $a-(n-1)$.