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Let be $V$ a vector space over a field $F$ with finit dimension. Let be $T$ a lineal operator in $V$. Supose that the characteristic polynomial of $T$, $p(x)$, is of the form $p(x)=(x-c)^{k}g(x)$ with $k > \in \mathbb{N}^{+}$, $c \in F$ and $g(c) \neq 0$, and consider $W$ the space of the eigenvectors associated with $c$.

Prove that:

  1. $Dim(W) \leq k$
  2. If $Dim(W)<k$, then $T$ is not diagonizable

I'm not sure of how to solve the problem. How can I prove it?

I would really appreciate your help!

luisegf
  • 715

1 Answers1

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Hint

  1. Take a basis $(e_1, \dots, e_m)$ of the eigenspace associated to the eigenvalue $c$ and complete it into a basis $(e_1, \dots, e_m, e_{m+1}, \dots, e_n)$ of the space $V$. Now compute the characteristic polynomial of $T$ in that basis.
  2. By contraposition if $T$ was diagonalizable, $T$ would be similar to a diagonal matrix having $k^\prime$ $c$ and the diagonal. Computing the characteristic polynomial leads to $k=k^\prime$ meaning that the eigenspace associated to $c$ has its dimension equal to $k$.
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