Original question : Suppose that $T\in\mathcal{L}(V)$, where $V$ is a finite-dimensional vector space, is such that every non-zero vector in $V$ is an eigenvector of $T$. Prove that $T$ is a scalar multiple of the identity operator.

My proof: Since every non-zero arbitrary vector $v_j\in V$ is an eigenvector of $T$, we have $$Tv_j=\lambda_j v_j$$with $v_j\neq0$.

Suppose $v_1,\cdots,v_n$ is basis of $V$ (Clearly, there can't have zero vector) and suppose an arbitrary vector $v\in V$ with $v\neq 0.$ Then $$Tv=T(a_1v_1+\cdots+a_nv_n)=a_1\lambda_1v_1+\cdots+a_n\lambda_nv_n$$ Since we know by theorem that if $a$ and $b$ are eigenvector of $T$, then $a+b$ is also eigenvector of $T$ corresponding to the same eigenvalues. Thus we have $\lambda_1=\cdots=\lambda_n$. Therefore $$Tv=\lambda_1(a_1v_1+\cdots+a_nv_n)=\lambda_1v.$$This means no matter what vector we put, we have the same eigenvalues for all eigenvectors. Therefore the eigenvalues is independent of the eigenvectors and we can conclude that $$Tv=\lambda_1 v=\lambda_1 Iv=(\lambda_1 I)v$$for which $T$ is a scalar multiple of identity operator.

Is my proof valid?

  • 1
    $\begingroup$ No this isn't valid. If $a$ and $b$ are eigenvectors corresponding to the same eigenvalue, then $a+b$ is an eigenvector corresponding to that eigenvalue. You're assuming what you want to prove. $\endgroup$ – saulspatz Jan 30 '18 at 17:03
  • $\begingroup$ @orole I don't follow this. $v_1+v_3 \mapsto v_1$ What is the eigenvalue? $\endgroup$ – saulspatz Jan 30 '18 at 17:12
  • $\begingroup$ @nevergraduating The statement is true, but the proof does not work. $\endgroup$ – InsideOut Jan 30 '18 at 17:32
  • $\begingroup$ @saulspatz, what do you mean by assuming what I want to prove? I want to prove $T$ is scalar multiple of identity operator isn't ? $\endgroup$ – Ling Min Hao Jan 30 '18 at 17:32
  • $\begingroup$ I meant that in the middle of the proof, you're trying to prove that all the basis vectors correspond to the same eigenvalue, but the theorem you quote is only valid if you already know that they're associated with the same eigenvalue. Sorry I wasn't clear. $\endgroup$ – saulspatz Jan 30 '18 at 17:44

Here's a hint. Let $e_1, e_2, ... e_n$ be a basis, $e_i \mapsto \lambda_i e_i, 1 \le i \le n.$ What can you say if $e_1 + e_2 \mapsto \lambda(e_1+e_2)?$

  • $\begingroup$ eigenvalues of $e_1,e_2$ and $e_1+e_2$ is $\lambda$ $\endgroup$ – Ling Min Hao Jan 30 '18 at 17:34
  • $\begingroup$ Right. Now generalize it for any pair of basis vectors. Doesn't that prove the theorem? $\endgroup$ – saulspatz Jan 30 '18 at 17:36
  • $\begingroup$ that's means the eigenvalue of $Tv=T(a_1v_1+\cdots+a_nv_n)=a_1Tv_1+\cdots+a_nTv_n=a_1\lambda v_1+\cdots+\lambda v_n=\lambda v$ is $\lambda$ ? $\endgroup$ – Ling Min Hao Jan 30 '18 at 17:45
  • $\begingroup$ Yes. (There's a small mistake. You wrote $\lambda v_n$ instead of $a_n \lambda v_n,$ but I'm sure that's just a typo.) $\endgroup$ – saulspatz Jan 30 '18 at 17:48

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