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Let $V$ be a finite-dimensional vector space and $T\colon V\to V$ be a linear operator over the field $F$, and let $$f = (x-c_1)^{d_1}\cdots(x-c_k)^{d_k}$$ $$p = (x-c_1)^{r_1}\cdots(x-c_k)^{r_k}$$ be the characteristic polynomial and the minimal polynomial for $T$, respectively. If $W_i$ is the null space of $(T-c_i)^{r_i}$, then I am familiar with the fact that $\dim W_i = d_i.$

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What can we say about the dimension of the null space of $(T-c_i)^{d_i}$?

When $d_i = r_i$, it is clear, but when $d_i>r_i$ do we have enough information to determine the dimension, or what more would we need to know?

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Yes. If $d_i > r_i$, then $\ker(T - c_i)^{r_i} = \ker(T - c_i)^{d_i}$. To show this, you can argue as following:

  1. Assume $T \colon V \rightarrow V$ is nilpotent of index $r < d = \dim V$. The characteristic polynomial of $T$ is $x^d$, the minimal polynomial of $T$ is $x^r$ and since $T^r = 0$ we have $\{ 0 \} = \ker(T^r) = \ker(T^{r+1}) = \dots = T^{d} = \ker(T^{r + k})$ for all $k \geq 0$. This handles the case where $k = 1$, $c_1 = 0$.
  2. Split $V$ into a direct sum of $T$-invariant subspaces $V = \oplus_{i=1}^k W_i$ and apply the previous item to each $(T - c_i I)|_{W_i}$. Then note that $\ker((T - c_i I)|_{W}^m) = \ker((T - c_i I)^m)$ because $T - c_i I$ is invertible on $\oplus_{j \neq i} W_j$.
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