2
$\begingroup$

Let $V , W$ be two vector spaces and $f : V \rightarrow W$ a linear map. Let $w_1,...,w_n$ be elements of $W$ that are linearly independent and let $v_1,...,v_n$ be elements of V such that $f(v_i) = w_i$ for $i=1,...,n.$ Prove that $v_1,...,v_n$ are linearly independent.

This is the actual proof, but i'm not sure if its complete:

I will prove this by contradiction.

Suppose $V_a,V_b\in V $ such that they are linearly dependent vectors. then $F(V_a)=W_b$ $V_b=\alpha V_a$ then: $f(V_b)=f(\alpha V_a)=\alpha f(V_a)=\alpha W_a$

Since $W$ has independent vectors, this results as False. So by contradiction we know that $V$ has independent vectors

$\endgroup$
3
$\begingroup$

You are on the right track, but you have to generalize from two vectors to $n$. But the same idea can be used: Suppose, that $v_1, \ldots, v_n$ are linearly dependent, then for some $\lambda_i\in k$ ($k$ the ground field), not all $\lambda_i = 0$, we have $\sum_i \lambda_i v_i = 0$. Applying $f$, we have by linearity $$ 0 = f(0) = f\left(\sum_i \lambda_i v_i\right) = \sum_i \lambda_i f(v_i) = \sum_i \lambda_i w_i $$ As not all $\lambda_i$ are 0, this contradicts the idependence of the $w_i$.

$\endgroup$
0
$\begingroup$

$f(v_i)=w_i, \;i=1,...,n.$ Choose any $v=\sum_{i=1}^{n} \alpha_i v_i \in{V}, f(v)=\sum_{i=1}^{n} \alpha_i w_i \in{W}.$

Now, $ f(v)=0 \;(v \in{ker(f)})\;$ if and only if $\alpha_i=0 \;\forall{1 \le i \le n}. \;$ But $ v=\sum_{i=1}^{n} \alpha_i v_i=0,$ hence $ker(f)=\{0\},$ and $v=0$ if and only if $\alpha_i=0 \;\forall{1 \le i \le n}.$ Therefore, $i=1,...,n, v_i$ are linearly independent.

$\endgroup$
0
$\begingroup$

Consider $T(a_1v_1+a_2v_2....+a_nv_n)=0$ Then by linearity you get $a_1w_1+.....+a_nw_n=0$. As $w_1,....,w_n$ are linearly independent, you get $a_1=a_2....=a_n=0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.