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Let $T: V \rightarrow V$ be a linear operator.

I need to demonstrate that if all nonzero vectors of $V$ are eigenvectors of $T$, then there is one specific $\lambda \in K$ such that $T(v) = \lambda v$, for all $v \in V$.

I understand that, if all nonzero vectors of $V$ are eigenvectors of $T$, then $T$ must be a scaling transformation. It just stretch or shrinks vectors, but doesn't change their directions.

So, the statement says that if it happens, then, there is a single $\lambda$ such that $T(v) = \lambda v$. In other words, if there is such transformation, then it scales all vectors by the same scalar $\lambda$.

Applying the transformation to our standard basis vectors, we have: $$ T(e_1) = \lambda_1 e_1 \\ T(e_2) = \lambda_2 e_2 \\ \vdots \\ T(e_n) = \lambda_n e_n $$

I understand I need to prove that $\lambda_1 = \lambda_2 = \dots = \lambda_n$, but I can't see how!

EDIT

$$ v = c_1e_1 + c_2e_2 + \dots + c_ne_n \\ T(v) = \mu v = \lambda_1c_1e_1 + \lambda_2c_2e_2 + \dots + \lambda_nc_ne_n \\ $$

Since what's multiplying $v$ coordinates is $\lambda_i$, then all of them must be $\mu$. I'm not sure how to 'mathematize' this. Is this idea correct?

EDIT 2 Extending the left hand side of EDIT 1, we have: $$ \mu v = \lambda_1c_1e_1 + \dots + \lambda_nc_ne_n \\ \mu(c_1e_1 + \dots + c_ne_n) = \lambda_1c_1e_1 + \dots + \lambda_nc_ne_n \\ \mu c_1e_1 + \dots + \mu c_ne_n = \lambda_1c_1e_1 + \dots + \lambda_nc_ne_n \\ $$

And since $e_i$ are linearly independent, $\mu = \lambda_1 = \lambda_2 = \dots = \lambda_n$. Is this proof correct?

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    $\begingroup$ Hint: consider $e_1+e_2$. $\endgroup$ May 17, 2012 at 18:23
  • $\begingroup$ I believe I've come to something. If ALL vectors of V are eigenvectors, then $T(v)$ is also an eigenvector, since T maps $V \rightarrow V$. Then it can be written as $\mu v$, where $\mu$ is another eigenvalue. Than $\mu = \lambda_1 = \lambda_2 = \dots = \lambda_n$. Is it correct? $\endgroup$ May 17, 2012 at 18:28
  • $\begingroup$ Your edit shows that $\mu = \lambda_i c_i$, $\forall i$, since the $e_i$ are linearly independent. $\endgroup$
    – copper.hat
    May 17, 2012 at 18:40
  • $\begingroup$ But $\mu$ is a scalar multiplying $v$, and $c_i$ are $v$ coordinates on standard basis, i.e., $c_ie_i$ are $v$ coordinates. If $\mu$ is multiplying $v$, why does $c_i$ comes to $\mu$ equation? $\endgroup$ May 17, 2012 at 18:44
  • $\begingroup$ You have now asked THREE different questions in the same post. What happens to the answers to your FIRST question, posted while this was the only one? $\endgroup$
    – Did
    May 17, 2012 at 19:13

3 Answers 3

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Since all $v\in V$ are eigenvectors, we can choose $e_i$, the $i$th unit vector. Then by assumption we have $T e_i = \lambda_i e_i$ for some $\lambda_i$. It follows that $T$ is diagonal, with elements $\lambda_1,...,\lambda_n$ on the diagonal.

Now choose $v=e_1+...+e_n$, again for some $\lambda$, we have $Tv=\lambda v$, so we have $$T v = T(e_1+...+e_n) = \lambda_1 e_n +... + \lambda_n e_n = \lambda (e_1+...+e_n).$$ Since the $e_i$ are linearly independent, it follows that $\lambda = \lambda_1 = ... = \lambda_n$. Hence $Tx = \lambda x$, $\forall x$.

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    $\begingroup$ When you write $v = e_1 + \dots + e_n$, you're considering it for unit vectors. Shouldn't you consider $v = c_1e_1 + \dots + c_ne_n$ with $c_i \in K$, in order to extend it to all vectors of $V$? $\endgroup$ May 17, 2012 at 18:51
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    $\begingroup$ What needs to be extended? If you know $T$ on a basis, you know $T$ everywhere, by linearity. First I show $T$ is diagonal, since $T(\sum x_i e_i) = \sum \lambda_i x_i e_i$, then I show that all diagonal elements are equal. $\endgroup$
    – copper.hat
    May 17, 2012 at 19:02
  • $\begingroup$ Sorry, I got confused mixing it with my try of demonstration. I'm still trying to assimilate your answer ;). $\endgroup$ May 17, 2012 at 19:11
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    $\begingroup$ This is a very nice proof. +1 $\endgroup$ May 17, 2012 at 19:43
  • $\begingroup$ @copper.hat $v$ is just a specific vector that sets $v = e_1+...+e_n$, you shouldn't show that it works for all $u \in V$? $\endgroup$
    – Avishay28
    Sep 25, 2017 at 7:46
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Assume for a contradiction that $v,w$ are eigenvectors for $\lambda\neq\mu$, respectively, (in particular they are nonzero) and that $T(v+w)=\nu(v+w)$ for some $\nu\in K$. Since any nonzero scalar multiple of an eigenvector is an eigenvector for the same eigenvalue, $v$ and $w$ cannot be linearly dependent. Then by this linear independence $\nu v+\nu w=T(v+w)=T(v)+T(w)=\lambda v+\mu w$ implies $(\nu,\nu)=(\lambda,\mu)$, which contradicts $\lambda\neq\mu$.

So if all nonzero vectors are eigenvectors, then all of them must be so for the same eigenvalue$~\lambda$, and one has $T=\lambda I$. (Pedantically, if $\dim V=0$ there are no eigenvectors at all, and one is free to choose$~\lambda$; "specific" in the question is not justified in this case.)

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Hint: Assume that $Tu=\lambda u$ and $Tv=\mu v$ for some nonzero vectors $u$ and $v$ and some $\lambda$ and $\mu$.

  • Show that $\{u,v\}$ is a linearly independent family.
  • Show that $\{u+v,au+bv\}$ is a linearly independent family, for every $a\ne b$.
  • Show that $\{T(u+v),u+v\}$ is not linearly independent.
  • Conclude that $\lambda=\mu$.
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  • $\begingroup$ What is a linearly free family? $\endgroup$
    – copper.hat
    May 17, 2012 at 18:38
  • $\begingroup$ @copper.hat: A Gallicism. $\endgroup$
    – Did
    May 17, 2012 at 18:44
  • $\begingroup$ A straight couple from the 60's maybe? $\endgroup$
    – copper.hat
    May 17, 2012 at 18:49
  • $\begingroup$ @copper.hat: Nice... :-) $\endgroup$
    – Did
    May 17, 2012 at 18:58

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