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Let T be a linear operator on a finite-dimensional vector space V. Prove that if the characteristic polynomial of T splits, then so does the characteristic polynomial of the restriction of T to any T-invariant subspace of V.

Theorem: Let T be a linear operator on a finite-dimensional vector space V, and let W be a T-invariant subspace of V. Then the characteristic polynomial of $T_W$ divides the characteristic polynomial of T.

Can I use this theorem to argue since $T_W$ is a factor of the polynomial of T, so it splits?

Let $T$ be a linear operator on a finite-dimensional vector space $V$.

Deduce that if the characteristic polynomial of $T$ splits, then any non-trivial $T$-invariant subspace of $V$ contains an eigenvector of $T.$

Let $W$ be a $T$-Invariant subspace. $W\neq\{0\}$($\because$ Given that $W$ is non-trivial). The characteristic polynomial of $T$ restricted to $W$ divides the characteristic polynomial of $T$. Then since nontrivial, there exists an eigenvalue for $det(W_1-tI)=0$ for every $W_1 \in T_{|W}$, hence it has at least one eigenvector.

Is this reasoning correct?

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  • $\begingroup$ Hi! With $p(x) $ "splits" do you mean that splits in linear factor (i. e. $p(x) = (x-a_1)...(x-a_k)$)? $\endgroup$
    – Menezio
    Commented Apr 14, 2020 at 21:59
  • $\begingroup$ @Menezio yes exactly $\endgroup$
    – spruce
    Commented Apr 14, 2020 at 22:53

1 Answer 1

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Yes, you did it right. Formally:

Named $p(t)$ the characteristic polynomial of $T$ and $p_W(t)$ che characteristic polynomial of $T_{W}$. By hypotesis we have $p(t)=(t-a_1)...(t-a_n)$ (eventually with $a_i=a_j$ for some $i,j$). Thanks to the theorem you mentioned above we have $$ p_W(t) \ | \ p(t) $$ And this implies that $p_W(t)$ is just a product of some factors of $p(t)$; hence $p_W(t)=(t-a_{i_1})...(t-a_{i_k})$ for some indices $i_1,...,i_k$ where $k=\dim W$.

The second part follows directly: since $k>0$, we have an eingenvalue $a_{i_1}$ of $T_W$ and so an eingenvector relative to it.

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