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$V$ 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?

mathematics
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Hailey
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  • 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 $dim(ker(T))+dim (Rank(T))=dim V$... – Milly Oct 27 '14 at 05:45
  • @Milly $T(f_1),...,T(f_n)$ don't generate $V$, though. $f_1,...f_n$ do. – Hailey Oct 27 '14 at 05:49
  • Is this question not the same as http://math.stackexchange.com/questions/207798/prove-transformation-is-one-to-one-iff-it-carries-linearly-independent-subsets-o – Gerry Myerson Oct 27 '14 at 06:24

1 Answers1

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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.

mathematics
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