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Does any know how to go about proving the following statement?

Let $v_1, v_2, \dots, v_n \in V$ be a linearly independent vectors. Furthermore, let $T \in \mathcal{L}(V)$ be an invertible linear transformation. Prove that $Tv_1, Tv_2, \dots, Tv_n$ are linearly independent.

Clearly we need to show that the equation

$$ c_1Tv_1 + c_2Tv_2 + \cdots + c_nTv_n = 0 $$

has only the trivial solution. We know that the equation

$$ a_1v_1 + a_2v_2 + \cdots + a_nv_n = 0 $$

has only the trivial solution. Can we use this, with the linearly of $T$, to prove the claim?

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Hint $T$ is Invertible $\Rightarrow T $ is Injective

as $T$ is linear transformation so

$c_1 T(v_1)+\dots+c_nT(v_n)=0\Rightarrow T(c_1v_1+\dots+c_nv_n)=0\Rightarrow c_1v_1+\dots+c_nv_n=0$ as $T$ is injective.

so $c_i=0\forall i$ as $\{v_i\}_{i=1}^{n}$ are linearly indipendent.

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By the linearity of $T$, we have

$$ c_1Tv_1 + c_2Tv_2 + \cdots c_nTv_n = 0 $$

$$ T(c_1v_1 + c_2v_2 + \cdots c_nv_n) = 0 $$

Since $T$ is invertible, it must be injective. Since it is injective, $T(u) = 0$ if and only if $u = 0$. By the linearity of $v_1, v_2, \dots, v_n$, we have that $c_1 = c_2 = \cdots = c_n = 0$. Thus, $Tv_1, Tv_2, \dots, Tv_n$ are linearly independent.

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Let $T(\sum c_kv_k)=0$ then since $T$ is invertible $\sum c_kv_k=0$. Then the result follows because $v_k$'s are independent.

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  • $\begingroup$ Any of the other answers are more explicitly. $\endgroup$ – Valerin Nov 28 '13 at 3:05
  • $\begingroup$ Another form to do this is as follow: Since $T$ is bijective then $T^{-1}$ exist and is also (of course) bijective. Then $$\sum c_kTv_k=0\Rightarrow T^{-1}(\sum c_kTv_k)=0$$ but this is precisely $$\sum c_kv_k=0\Rightarrow c_k=0, \forall k.$$ I have used the fact that $T^{-1}Tv_k=Iv_k=v_k$. $\endgroup$ – Valerin Nov 28 '13 at 3:12

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