# Linearly independent in different vector spaces

If a set of vectors are linearly independent in $\mathbb{R}^n$, are they also linearly independent in the vector space $V$?

Edit: Here is the full question: Let $B = \{v_1,...,v_n\}$ be a basis for a vector space $V$ and let $u_1,..., u_k \in V$. If $\{[u_1]_B,...,[u_k]_B\}$ is linearly independent in $\mathbb{R^n}$, then $\{u_1,...,u_k\}$ is linearly independent in $V$.

• Well, what is $V$? – Casteels Mar 11 '15 at 14:46
• They wouldn't even be in V unless V is in $\mathbb{R}^n$. – Paul Sundheim Mar 11 '15 at 14:48
• V is a vector space with the basis B. The vectors U1, U2, ... Uk are all part of V. If [U1]B, [U2]B, ... [Uk]B are linearly independent in Rn, are they also linearly independent in V? – user222771 Mar 11 '15 at 14:51
• The full question doesn't make it better. The same problem prevails, it doesn't make sense unless $V=\mathbb R^n$. – Git Gud Mar 11 '15 at 14:53
• I suggest you ask the instructor. He knows what your book means by [U1]B or whatever it is. Perhaps a column vector consisting of the coefficients of vector U1 in the basis B. – GEdgar Mar 11 '15 at 14:57

I guess we should also assume $V$ is a vector space over $\mathbb{R}$.

Suppose we have scalars $a_1,\ldots, a_k$ such that $$a_1u_1+\cdots +a_ku_k=0.$$

Now we know that for each $i$, there are scalars $b_{i,1},b_{i,2},\ldots$ such that $$u_i=b_{i,1}v_1+b_{i,2}v_2+\cdots+b_{i,n}v_n.$$ Hence,

$$(\sum_{j=1}^k a_j b_{j,1})v_1 + (\sum_{j=1}^k a_j b_{j,2})v_2+\cdots + (\sum_{j=1}^k a_j b_{j,n})v_n = 0.$$

Now: Why must each coefficient of $v_i$ here be equal to $0$? Why, then, must each $a_j$ be equal to zero? What, then, can you say about $u_1,\ldots,u_k$?

Assuming you mean, would the image of a set of independent vectors in $\mathbb{R}^n$ under a linear map $A:\mathbb{R}^n\rightarrow V$ stay independent, then if $A$ is injective, then yes.

For a full explanation, see my answer here: