# Linear maps and Linear independence

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!

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.

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 $$\mathbb{R}^2$$.

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.