1
$\begingroup$

Let $T$ be a linear operator on finite dimensional vector space $V$ such that the characteristic polynomial $P_T(t)$ splits to linear factors, and $\lambda_1 ... \lambda_k$ are the eigenvalues of $T$.

Given $S(x) = \sum\lambda_iv_i$ for $1\leq i \leq k$

Where $v_1, ...,v_k$ are the unique vectors such that $v_i \in K_{\lambda_i}$ and $x=\sum v_i$, where $K_\lambda$ is the generalized eigenspace of $\lambda$

How do I show:

$(a)$ S is diagonalizable, and

$(b)$ (T-S) is nilpotent, and commutes with $S$

My thoughts:

$(a)$ S obviously has eigenvalues $\lambda_1, ... , \lambda_k$, since the $v_i$s are linearly independent, but I'm not sure where to take the proof. Is it enough to show that for any $x$, you can take the basis $\{v_1, ..., v_k\}$ such that $[S]_\beta$ is a diagonal matrix of eigenvalues?

$\endgroup$
  • $\begingroup$ Your first two lines in the questions are unfinished, I think: "...if ...splits and....are eigenvalues of T..." .... what ? Also, what is $\; K_{\lambda_i}\;$ ? Eigenspaces? The $\;v_i\;$ are eigenvalues of $\;T\;$ which are a basis? I think you should re-edit your question. $\endgroup$ – DonAntonio Apr 6 '16 at 22:15
  • $\begingroup$ Initially worded it differently, fixed the first couple lines. Will fix it. $\endgroup$ – George Apr 6 '16 at 22:19
  • $\begingroup$ Also, what is $\;P_T(t)\;$ ? By the splitting thing I presume it is the minimal polynomial of $\;T\;$ ? And it splits into different linear factors? $\endgroup$ – DonAntonio Apr 6 '16 at 22:20
  • $\begingroup$ I've updated the question again. $P_T(t)$ is the characteristic polynomial of $T$, and the linear factors are not necessarily unique, that is, its eigenvalues do not necessarily have multiplicity 1. $\endgroup$ – George Apr 6 '16 at 22:24
  • 1
    $\begingroup$ You can use schur's theorem, if the characteristic polynomial of T splits, then there exists an orthonormal basis $ \beta$ s.t $[T]_\beta$ is upper triangular, S is also upper triangular, so when you subtract them, the diagonal will only have zeros, and a strictly upper triangular matrix is nilpotent, moreover, diagonal matrices commute. $\endgroup$ – ಠ_ಠ Apr 7 '16 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.