4
$\begingroup$

$V$ and $W$ are vector spaces with $dim(V)=n$. Prove that a linear transformation $T:V\rightarrow W$ is injective if and only if for a basis $B=(f_1,\ldots,f_n)$ of $V$, $T(f_1),\ldots,T(f_n)$ are linearly independent.

I'm first trying to prove the forward direction, by assuming that $T:V\rightarrow W$ is injective and showing that $T(f_1),\ldots,T(f_n)$ are linearly independent. So, by assumption, for any $r_1,r_2\in V$, $T(r_1)=T(r_2)\rightarrow r_1=r_2$. However, I don't see how I can use this to show that $T(f_1),\ldots,T(f_n)$ are linearly independent. I have to show that only the trivial relation holds in $c_1T(f_1)+\cdots+c_nT(f_n)$ for scalars $c_1,\ldots,c_n$, so $c_1=\cdots=c_n=0$, but how does this following from the transformation being injective alone?

$\endgroup$
  • $\begingroup$ Hint: if $T(f_1),\cdots, T(f_n)$ are not linearly independent, then they do not generate $V$. Thus $ker(T)$ is non trivial since $dim(ker(T))+dim (Rank(T))=dim V$... $\endgroup$ – Milly Oct 27 '14 at 5:45
  • $\begingroup$ @Milly $T(f_1),...,T(f_n)$ don't generate $V$, though. $f_1,...f_n$ do. $\endgroup$ – Hailey Oct 27 '14 at 5:49
  • $\begingroup$ Is this question not the same as math.stackexchange.com/questions/207798/… $\endgroup$ – Gerry Myerson Oct 27 '14 at 6:24
1
$\begingroup$

First assume that $T:V\rightarrow W$ is injective. So, we have to show that if $c_1T(f_1)+\cdots+c_nT(f_n)=0$ for scalars $c_1,\ldots,c_n$, then $c_1=\cdots=c_n=0$. By $c_1T(f_1)+\cdots+c_nT(f_n)=0$ we have $T(c_1(f_1)+\cdots+c_n(f_n))=0$, because $T$ is linear transformation. Since $T$ is injective we have $c_1(f_1)+\cdots+c_n(f_n)=0$ and since $f_i$'s are basis we have $c_i=0$ for $i=1,\ldots,n$.

For the converse, assume that $T$ carries independent subset onto independent subset. Let $\alpha\neq 0$ be a vector in $V$. Then the set $\{\alpha\}$ is independent. then the image of $\{\alpha\}$ is $\{T\alpha\}$ and by assumption it's independent. So $T\alpha\neq0$ and this shows that the null space of $T$ is the zero subspace, i.e., $T$ is injective.

$\endgroup$
  • $\begingroup$ Why do we have "since $T$ is injective...$c_1(f_1)+...+c_n(f_n) = 0$"? $\endgroup$ – Hailey Oct 27 '14 at 6:15
  • $\begingroup$ Because $T$ is $1:1$ iff $T$ is nonsingular, and $T$ is nonsingular if $T\alpha=0$ implies $\alpha=0$. $\endgroup$ – mathematics Oct 27 '14 at 6:19

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.