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Let $T:V\to W$ be an injective linear transformation. Given that $(v_1,\dots,v_n)$ is a basis for $V$, I want to show that $(T(v_1),\dots,T(v_n))$ is linearly independent. I've tried proving it by contradiction but have not made any significant progress. Any tips?

Arnaud D.
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Michael Lodger
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  • I've changed the formulation to make your question more clear, can you confirm that I understood your question correctly and didn't change the meaning? If you are not satisfied you can rollback to the previous version. – Arnaud D. Mar 30 '17 at 14:32
  • Or alternatively http://math.stackexchange.com/questions/923237/prove-that-if-t-is-injective-and-v-i-is-linearly-independent-then-tv-i?rq=1, http://math.stackexchange.com/questions/207798/prove-transformation-is-one-to-one-iff-it-carries-linearly-independent-subsets-o?noredirect=1&lq=1 – Arnaud D. Mar 30 '17 at 14:37

2 Answers2

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I assume you are asking to show that $Tv_1, \dots, Tv_n$ are linearly independent.

Suppose $Tv_1, \dots, Tv_n$ were not linearly independent. Then there are scalars $\lambda_1, \dots, \lambda_n$, not all of them zero, such that $$T(\lambda_1 v_1 + \dots + \lambda_n v_n) = \lambda_1 Tv_1 + \dots + \lambda_n Tv_n = 0. $$ Since $T$ is injective, it follows that $\lambda_1 v_1 + \dots + \lambda_n v_n = 0$. But this is a contradiction, since $(v_1, \dots, v_n)$ was assumend to be a basis and hence linearly independent.

Ken Duna
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agb
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    Note that you don't really need contradiction here. If you assume that you have some linear combination equal to zero, in the end you get that those scalars are all zero. Thus the $Tv_i$ are linearly independent.

    There is nothing wrong with doing a proof by contradiction, per se. But many people I talk to agree that it is better to do a direct proof when possible (just from a stylistic standpoint).

     – Ken Duna Mar 30 '17 at 14:19
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    I know, but since the question suggested a proof by contradiction I gave one :-) In general I prefer direct proofs as well. – agb Mar 30 '17 at 17:44
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This is similar to agb's answer but it is a direct proof, instead of one by contradiction.

Suppose we have $\lambda_1,\dots,\lambda_n$ such that $$ T(\lambda_1 v_1 + \dots + \lambda_nv_n) = \lambda_1T(v_1) + \dots + \lambda_nT(v_n) = 0 $$ Since $T$ is injective and $T(0) = 0$ we have that $\lambda_1 v_1 + \dots + \lambda_nv_n = 0$. Since $\{v_1,\dots,v_n\}$ is linearly independent $\lambda_1 = \dots = \lambda_n = 0$ and so $\{T(v_1),\dots,T(v_n)\}$ is linearly independent.

Kai Rüsch
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