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I am wondering about the theorem

If $T: V \to U$ is a linear non-singular transformation and $\{v_1,..,v_k\}$ is a linearly independent subset of V then the images of T are also independent.

I know the standard proof is something along the lines of

$A_1v_1+...+A_kv_k=0$ so $T(A_1v_1+...+A_kv_k)=0$

And $T$ is linear so we have $A_1T(v_1)+...+A_kT(v_k)=0$

because $T$ is non singular $\ker T=\{0\}$ so and the V are independent thus zero .

I am wondering of there is an alternate way to to do this along the lines of we know $dimV=\dim IM F$ ie $V=im F$ and use the Theorem that there exists a unique linear mapping $G: V\to U$ such that $G(v_1)=u_1$ for $i= 1,2,...$

So we could have wrote it as

$A_1u_1+...+A_ku_k=0$ and said because the u form a basis for Image, as they span the images as well, they must be independent?

Thanks all for comments

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    $\begingroup$ please edit your question in Latex... $\endgroup$ Commented Feb 19, 2015 at 21:42
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    $\begingroup$ Sorry about that I had wrote it originally on my phone but it should be fixed now :) $\endgroup$ Commented Feb 19, 2015 at 21:54

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Let $W=\text{Sp}(B)=\text{Sp}\{x_1,\dotsc,x_n\}$ be the subspace of $V$ which is of dimension $n$ and $B$ acts as a basis for $W$. Consider the restriction map of $T$ to $W$, then $T:W\to U$ also defines a non-singular linear map as you can see. Clearly $T(W)=\text{Sp}\{Tx_1,\dotsc,Tx_n\}=\text{Sp}T(B)$. Use the dimension theorem to conclude that $\dim T(W)=n$. Since a linear map is uniquely determined by its action on the basis $B$, so $T(B)$ must contain $n$-linearly independent vectors i.e. $T(B)$ is linearly independent (otherwise if $Tx_i$ and $Tx_j$ are linearly dependent then $T(W)$ has dimension less than $n$ which shows a contradiction).

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  • $\begingroup$ We have not covered the topics of restriction maps. Is what I wrote make any sort of sense or at least a good possible start? $\endgroup$ Commented Feb 20, 2015 at 0:40

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