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Is the following Proof Correct?

Given that $T\in\mathcal{L}(V,W)$ and $v_1,v_2,...,v_m$ is a list of vectors in $V$ such that $Tv_1,Tv_2,...,Tv_m$ is a linearly independent list of vectors in $W$ then $v_1,v_2,...,v_m$ is linearly independent.

Proof. Let $a_1,a_2,...,a_m$ be arbitrary scalars in $\mathbf{F}$ and assume that $$0=a_1v_1+a_2v_2+\ .\ .\ . +a_mv_m\tag{1}$$ Applying $T$ to to above linear combination we have $$0=T(0)=T(a_1v_1+a_2v_2+\ .\ .\ . +a_mv_m)\tag{2}$$ the additivity of $T$ implies that $$0=T(a_1v)_1+T(a_2v_2)+\ .\ .\ . +T(a_mv_m)\tag{3}$$ and the homogeneity of $T$ necessitates that $$0=a_1Tv_1+a_2Tv_2+\ .\ .\ . +a_mTv_m\tag{4}$$ but $Tv_1,Tv_2,...,Tv_m$ is a linearly independent list in $W$ consequently $a_1=a_2=...\ a_m=0$ therefore $v_1,v_2,...,v_m$ is a linearly independent list in $V$.

$\blacksquare$

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  • $\begingroup$ In short, this says that if $S \subset W$ is a linearly independent set, then $T^{-1}(S) \subset V$ is a linearly independent set. $\endgroup$ – Sean Roberson Aug 1 '17 at 16:59
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Yes, the proof looks fine to me.

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