Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f \colon U \to V$ be a linear map and $\{u_1, u_2, \ldots, u_n\}$ be the set of linearly independent vectors in $U$.

Then the set $\{f(u_1), f(u_2),\ldots,f(u_n)\}$ is linearly independent iff

a. $f$ is one-one & onto

b. $f$ is one-one

c. $f$ is onto

d. $U = V$

I came to conclusion that $f$ has to be one-one. This is what I did.

We have $$a_1 u_1 + a_2 u_2+ \ldots + a_n u_n = 0.$$

For $\{f(u_1), f(u_2),\ldots,f(u_n)\}$ to be independent, we have to have $$a_1 f(u_1) + a_2 f(u_2) + \cdots + a_n f(u_n) = 0.$$

Simplifying we get $$f(a_1 u_1 + a_2 u_2 + \cdots + a_n u_n) = 0$$ i.e. $f(0)=0$.

This implies that $\ker f = \{0\}$ i.e. $f$ has to be one-one.

Is my logic correct? Also what's the correct answer and why?

share|cite|improve this question
$u_i$ are assumed to be linearly independent. You can therefore not have $a_i$ with $a_1 u_1 + \dots + a_n u_n = 0$. – goobie Feb 12 '13 at 9:23
All four answers are wrong. The second set is linearly independent if $f$ is one-one, but it's not an "only if". Try to find an example of a linear map that isn't one-one but still preserves linear independence of some (not every!) set. – Gerry Myerson Feb 12 '13 at 9:23
$d$ is false since if $f =0$ is the constant zero map, $f$ is linear but $f(u_i)$ are not linearly independent. – goobie Feb 12 '13 at 9:34

In the light of @Gerry's comment, take the following linear transformation: $$T:\mathbb R^3\to\mathbb R^2$$ $$T(a_1,a_2,a_3)=(a_1-a_2,2a_3)$$ $N(T)=\{(a,a,0)\mid a\in\mathbb R\}$ and $R(T)=\mathbb R^2$. So a,b,d is wrong. Search for another for c.

share|cite|improve this answer
Here, I consider $\{(1,0,0),(0,0,1)\}$ which is a part of standard basis or $R^3$. $T$ takes it to $\{(1,0),(0,2)\}$ which contains two independent vectors. – Babak S. Feb 12 '13 at 9:56
So which is the correct answer? b or c? – John Feb 12 '13 at 14:30
Helpful, like usual! +1 – amWhy Feb 12 '13 at 15:56
John, I already said all four answers are wrong. Have I ever lied to you? – Gerry Myerson Feb 13 '13 at 5:46
@Gerry :-) This question was asked in one entrance exam. Anyway, so I take it that it's printing mistake (iff instead of if) and so the correct answer is one-one. – John Feb 13 '13 at 6:08

$T:V\rightarrow W$ be a linear map such that $KerT=\{0\}$. Then $T$ maps any basis of $V$ to a basis of $W$. $V,W$ be finite dimensional vector spaces over the field $F$, $\dim W\le \dim V$

share|cite|improve this answer
Not true if the dimension of $W$ exceeds that of $V$. Also not true in infinite dimensions. – Gerry Myerson Feb 12 '13 at 11:21
If $\dim W\lt\dim V\lt\infty$ then you can't have a trivial kernel. Also, there's nothing in the original problem about a basis. – Gerry Myerson Feb 12 '13 at 11:38

OP has clarified that one wants a proof that if $f$ is linear and one-one then the image of a linearly independent set is linearly independent.

So, assume $u_1,\dots,u_n$ linearly independent, assume $f$ linear and one-one, and assume $$a_1f(u_1)+\cdots+a_nf(u_n)=0\tag1$$ By linearity of $f$, we have $$f(a_1u_1+\cdots+a_nu_n)=0\tag2$$ Every linear map $f$ satisfies $$f(0)=0\tag3$$ Since $f$ is assumed one-one, (2) and (3) imply $$a_1u_1+\cdots+a_nu_n=0\tag4$$ Since $u_1,\dots,u_n$ are assumed linearly independent, (4) implies $$a_1=\cdots=a_n=0\tag5$$ So we have proved that under the hypotheses, (1) implies (5). But that's precisely the definition of linear independence of $f(u_1),\dots,f(u_n)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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