Let $W$ be a vector space over $\mathbb R$ and let $T:\mathbb R^6 \to W$ be a linear transformation such that $S = \{Te_2, Te_4, Te_6\}$ spans $W$. Wich one of the following must be true?

  • (A) $S$ is a basis of $W$
  • (B) $T(\mathbb R^6) \ne W$
  • (C) $\{Te_1, Te_3, Te_5\}$ spans $W$
  • (D) $\ker T$ contains more than one element

I'm having trouble starting this problem.

Here are my findings so far:

I tried Dimension theorem and found that $\ker(T)$ contains more than three elements so (D) is incorrect.
And taking $\dim(T(\mathbb R^6))=\dim(W)$ we get $T(\mathbb R^6)=W$ so (B) is incorrect.
I think (C) option is correct but still not able to prove it.

  • 2
    $\begingroup$ Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. $\endgroup$ – AlexR Jan 28 '15 at 17:05
  • $\begingroup$ What have you done so far and where exactly are you stuck? $\endgroup$ – Timbuc Jan 28 '15 at 17:05
  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – AlexR Jan 28 '15 at 17:05
  • $\begingroup$ @AlexR Thanks for suggestions.I’ll make sure I don’t repeat these mistakes in future. $\endgroup$ – ap21 Jan 29 '15 at 12:28
  • $\begingroup$ @ap21 The question was now reopened. See the edit made to your question to see how you should formulate it in the future. Is the problem solved now or do you need any further assistance? $\endgroup$ – AlexR Jan 30 '15 at 15:27

Such a problem is most effectively tackled by looking at each point and asking oneself How could this become wrong?. Let me get you started:

Since $W$ is spanned by a set of three vectors, it has dimension at most $3$. Let's thus assume $W = \mathbb R^d$ with $d\in\{0,1,2,3\}$ for the tasks (this will actually be enough).

For (A) to go wrong, $S$ must be linearly dependent. This is the case if our $d$ is smaller than $3$, wich is certainly possible. A concrete example for $d = 2$ would be $$T = \pmatrix{1&2&3&0&0&0\\0&0&0&1&2&3}\\ S = \{Te_2, Te_4, Te_6\} = \{(2,0)^T, (0,1)^T, (0,3)^T\}$$

Is there a way to make (B) wrong?

and so on...

I will be happy to provide further help given that you actually show some work. Consider this a "kick-off" ;)

  • $\begingroup$ Thanks for help.I tried Dimension theorem and found that Ker(T) contains more than three elements so (d) is incorrect.And taking dim(T($\mathbb{R}^6$))=dim(W) we get T($\mathbb{R}^6$)=W so (b) is incorrect.I think (c) option is correct but still not able to prove it.Please help! $\endgroup$ – ap21 Jan 29 '15 at 13:34
  • $\begingroup$ @ap21 Very good. Everything is correct. For a counter-example to (C) chose the columns of $T$ such that $e_1, e_3$ and $e_5$ are linearly dependent. $\endgroup$ – AlexR Jan 29 '15 at 14:45
  • $\begingroup$ @ap21 If $ker(T)$ contains more than $3$ elements, then it certainly contains more than $1$ element, so shouldn't option d) be correct? $\endgroup$ – Diya Aug 27 '15 at 18:28
  • 1
    $\begingroup$ @Diya You're right, option (D) is true (since $\dim \ker T > 0$). I don't expect the answerer to be around anymore though looking at his rep and last seen... $\endgroup$ – AlexR Aug 27 '15 at 18:39
  • 1
    $\begingroup$ @AlexR thanks a lot! I was doing this problem and needed the help :) :) $\endgroup$ – Diya Aug 27 '15 at 18:41

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