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Prove if T : R n → R n is a 1−1 linear operator, and {v1, v2, . . . , vn} is a basis of R n, then so is {T (v1), T (v2), . . . , T (vn)}.

I am not sure how I can show that. Any help or tips?

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If you suppose $T(v_1) , \ldots , T(v_n)$ are not linearly independent, can you then construct a non-trivial element of the kernel? – ah11950 Mar 21 '14 at 0:48
up vote 5 down vote accepted

Well, you have $n$ vectors, so it suffices to check that $\{T(v_1),\dots,T(v_n)\}$ is a collection of linearly independent vectors. So suppose $$ a_1T(v_1)+\dots+a_nT(v_n)=0 $$ for some real numbers $a_1,\dots,a_n$. Then, since $T$ is a linear operator, $$ T(a_1v_1+\dots+a_nv_n)=0. $$ But $T$ is $1$-to-$1$, and we already know that $T(0)=0$ for any linear operator, so it follows that $$ a_1v_1+\dots+a_nv_n=0. $$ But $\{v_1,\dots,v_n\}$ is a basis for $\Bbb{R}^n$. What does this tell us about the values of $a_1,\dots,a_n$?

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very interesting approach I will have to think about it. – makakas Mar 21 '14 at 0:53

Here is another approach.

Note that if $u=Tv$ then $u=T(\alpha_1v_1+\ldots+\alpha_nv_n)=\alpha_1T(v_1)+\ldots\alpha_nT(v_n)$. Therefore $\{T(v_1),\ldots,T(v_n)\}$ spans the image of $T$.

Use the rank-nullity theorem ($\dim\ker T+\dim\operatorname{im} T=\dim\Bbb R^n$) to conclude that the above set must be a basis.

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