0
$\begingroup$

Question: let $T$ be linear operator on finite dimensional Vector space $V$ then $T$ is diagonalizable if and only if $V$ is direct sum of one dimensional $T$-invariant subspaces.

My attempt: if $T$ is diagonalizable linear operator on $n$ dimensional Vector space $V$ then $V$ has basis say $\beta =\{v_1,...,v_n\}$ consisting of an eigenvectors of $T$.

Taking $W_i=span\{v_i\}$ where $i=1,..,n$ then

$W_i ∩Wj=\{0\}$ for $i≠j$ and $V=W_1+...+W_n$.

Hence $V=W_1⊕...⊕W_n$ i.e. $V$ is direct sum of one dimensional $T$ invariant subspaces.

I had skip to prove $W_i$ are $T$ invariant but I think it's obvious by definiition of $W_i$.

Is above proof is valid? In fact I saw the above question/ theorem first time. Is the statement of above theorem is correct?

Please help me.

$\endgroup$
  • 1
    $\begingroup$ No, you are asked to prove an equivalence and you attemped only one direction. The implication you started with is correct though. $W_i$ is T-invariant because these spaces are generated by eigenvectors. So, include your attempt for the converse. $\endgroup$ – EpsilonDelta Jun 8 at 13:08
  • $\begingroup$ @EpsilonDelta Thanks. :-) $\endgroup$ – Akash Patalwanshi Jun 9 at 7:37

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.