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?

  • $\begingroup$ $u_i$ are assumed to be linearly independent. You can therefore not have $a_i$ with $a_1 u_1 + \dots + a_n u_n = 0$. $\endgroup$ – goobie Feb 12 '13 at 9:23
  • 2
    $\begingroup$ 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. $\endgroup$ – Gerry Myerson Feb 12 '13 at 9:23
  • $\begingroup$ $d$ is false since if $f =0$ is the constant zero map, $f$ is linear but $f(u_i)$ are not linearly independent. $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – mrs Feb 12 '13 at 9:56
  • $\begingroup$ So which is the correct answer? b or c? $\endgroup$ – John Feb 12 '13 at 14:30
  • $\begingroup$ Helpful, like usual! +1 $\endgroup$ – Namaste Feb 12 '13 at 15:56
  • $\begingroup$ John, I already said all four answers are wrong. Have I ever lied to you? $\endgroup$ – Gerry Myerson Feb 13 '13 at 5:46
  • $\begingroup$ @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. $\endgroup$ – 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 $F$, $\dim W\le \dim V$

  • $\begingroup$ Not true if the dimension of $W$ exceeds that of $V$. Also not true in infinite dimensions. $\endgroup$ – Gerry Myerson Feb 12 '13 at 11:21
  • $\begingroup$ 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. $\endgroup$ – 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 $f(u_1),\dots,f(u_n)$.


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