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Given a linear transformation $T: \mathbb{R^3} \rightarrow \mathbb{R^4}$ and take a standard basis $\{s = {(v_1,v_2,v_3})\}$ of $\mathbb{R^3}$.

If i know that $T(v_1),T(v_2),T(v_3)$ are not zero then these images are linearly independent?

My approach:

I think this is true.

Suppose that $T(v_1),T(v_2),T(v_3)$ are linearly dependent then there exist a linear combination of these images so that

$\lambda_1T(v_1)+\lambda_2T(v_2) + \lambda_3T(v_3) = 0$

Since $T$ is linear we can write it as: $T(\lambda_1v_1+\lambda_2v_2 + \lambda_3v_3) = 0$

Now i see that if a transformation sends everyvector to a zero vector then it is also linear and can't follow up?

Any hints ?

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    $\begingroup$ And what happens when at least one of $v_1,v_2,v_3$ is in the kernel of $T$. I guess we have to assume here, that $T$ must be injective. In general the above statement isn't true. $\endgroup$ – Fakemistake Jan 23 '18 at 21:19
  • $\begingroup$ @Fakemistake $v_1,v_2,v_3$ is basis of $\mathbb{R^3}$ and no this is the full question :( if it was injective then they are free but there was nothing given about $T$. $\endgroup$ – Khan Saab Jan 23 '18 at 21:26
  • $\begingroup$ I edited my answer below just right now because there was a huge mistake, so maybe you like it too. $\endgroup$ – Fakemistake Jan 24 '18 at 8:58
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My approach:

I think this is false.

Suppose that $T(v_1)=T(v_2)=T(v_3)=(1,1,1,1)^T$, which are linearly dependent, because the linear combination of these images gives

$T(v_1)+T(v_2) - 2T(v_3) = 0$.

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Here's a nice example:

Take the standard basis $e_1,e_2,e_3$ of $\mathbb{R}^3$ and $b_1,b_2,b_3,b_4$ of $\mathbb{R}^4$ and look at the matrix representation of $T$ which is given by \begin{align} \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3 \\ \end{pmatrix} \end{align} Then we have $f(e_1)=(1,1,1,1)^T,f(e_2)=(2,2,2,2)^T,f(e_3)=(3,3,3,3)^T$. All of them are different and not the zero vector, but they are linear dependent.

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