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If we have a set of linearly independent vectors, can they be mapped to a set of linearly dependent vectors? and vice versa. for example,

Given T : V → W is a linear map where $T(v_1) = w_1,··· ,T(v_n) = w_n$ for some vectors $w_1,··· ,w_n ∈ W$.

I think if the vectors {$w_1,...,w_n$} are linearly independent then it doesn't necessarily mean that the vectors {$v_1,...,v_n$} must also be linearly independent. If we let $v_1= c_1v_2+...+c_{n-1}v_n$ , for some scalars $c$, and similarly write the other vectors in $V$ as linear combinations of each other, then {$v_1,...,v_n$} would be dependent but would still be sent to linearly independent vectors in $W$.

Am I on the right track or is there a flaw in my logic/math. Thanks!

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To adress your first question: Yes, linearly independent vectors may be mapped to linearly dependent vectors, consider for example the constant $0$ map. Nevertheless, if you require that the map is injective, linearly independent vectors are mapped to linearly independent vectors which is easy to check.

To adress the second question: Whenever you have linearly dependent vectors their images will be linearly dependent, too.

Proof: Let $v_1,\dots,v_n$ be linearly dependent, i.e. we can find $\lambda_1,\dots,\lambda_n$ with not all $\lambda_i=0$ such that $\lambda_1 v_1+\dots + \lambda_n v_n=0$. Then for the images we have $$ \lambda_1 T(v_1)+\dots + \lambda_n T(v_n)=T(\lambda_1 v_1+\dots + \lambda_n v_n)=T(0)=0. $$ Hence, $T(v_1),\dots,T(v_n)$ are linearly dependent.

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Yes, we can map a set of linearly independent vectors to a set of linearly dependent vectors. Take for instance the linear map represented by the matrix $$\left[\matrix{1 & 0\\ 0 & 1\\ 1& 1}\right]$$ The (linearly independent) standard basis of $\mathbb{R}^3$, consisting of $[1,0,0]$, $[0,1,0]$, $[0,0,1]$, gets mapped to the vectors $[1,0]$, $[0,1]$ and $[1,1]$, which obviously form a linearly dependent set of vectors in $\mathbb{R}^2$.

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Suppose that $c_1w_1+...+c_nw_n=0$.

ie suppose that $T(c_1v_1+...+c_1v_n)=0$ (by linearity and $T(v_i)=w_i$.

Now if T is injective, you have that $c_1v_1+...+c_1v_n=0$, which in turn implies that $c_i=0$ for all i.

so {$w_1,...w_n$} is linearly independent too.

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