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Let $V$ and $W$ be vector spaces and let $T: V\to W$ be a linear transformation. Let $\{v_1,v_2,\ldots,v_p\}$ be a linearly dependent set of vectors in $V$. Show that $\{Tv_1,Tv_2,\ldots,Tv_p\}$ is also linearly dependent.

Here's what I have, I need someone to tell me if they think this works.

Since $\{v_1,\ldots,v_p\}$ is linearly dependent, we know that there are scalars that are not all zero (lets this scalar be an) that make $a_1v_1+\ldots+a_nv_n = 0$

Once we apply the transformation we get $T(a_1v_1+\ldots+a_nv_n)=T(0)$

Because are told that this thing is a linear transformation, we know its closed under addition and scalar multiplication. We can use this to change the set to.

$T(a_1v_1)+\ldots+T(a_nv_n) = T(0)$

and we can also pull those scalars out since its dependent.

$a_1T(v_1)+...+a_nT(v_n) = T(0)$

the scalars are still not all zero since we just factored them out of the transformation. Therefore, we have a set here that has scalars, not all zero that gives us $T(0)$ which is $0$. This set with the transformation applied is linearly dependent which is what we needed to show.

The only question I have here is about the end where I said that $T(0) = 0$. Can I make that assumption or am I missing something?

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    $\begingroup$ No. For all linear maps, $T(0)=0$. Just a remark: it's not because the vectors are linearly dependent that you can pull out the scalars. It's just because $T$ is a linear map. $\endgroup$
    – Bernard
    Nov 22, 2017 at 19:35
  • $\begingroup$ Please familiarize yourself with MathJax. $\endgroup$
    – user499203
    Nov 22, 2017 at 19:38
  • $\begingroup$ $\{v_j\}_{j=1}^{n}$ is linearly depended implies that that there are constants $\{a_j\}$ such that $v_1 = \sum_{j=2}^{n} a_j v_j$. Applying $T$ to this and exploiting linearity, we have $$ Tv_1 = T\left( \sum_{j=2}^{n} a_j v_j \right) = a_j\sum_{j=2}^{n} Tv_j,$$ which shows dependence. $\endgroup$
    – Xander Henderson
    Nov 22, 2017 at 19:40
  • $\begingroup$ Yes, your proof is fine. The one that always causes students difficulties is the other direction. Is it true that if $v_1,\dots, v_p$ are linearly independent, then $T(v_1),\dots,T(v_p)$ are likewise? If not, give an example and give a sufficient hypothesis to make it be true. $\endgroup$ Mar 9, 2021 at 1:37

2 Answers 2

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As noted in the comments, $T(0) = 0$ holds for any linear map:

$$T(0) = T(0 + 0) = T(0) + T(0) = 2T(0) \implies T(0) = 0$$


A bit easier is to show the contrapositive of your statement: if $\{Tv_1, \ldots, Tv_p\}$ is linearly independent, then $\{v_1, \ldots, v_p\}$ is also linearly independent.

Take scalars $\alpha_1, \ldots, \alpha_p$ such that $0 = \alpha_1 v_1 + \cdots + \alpha_p v_p$.

Apply $T$ on both sides:

$$0 = T(0) = T(\alpha_1 v_1 + \cdots + \alpha_p v_p) = \alpha_1 Tv_1 + \cdots + \alpha_p Tv_p$$

Therefore, $\alpha_1 = \ldots = \alpha_p = 0$, which shows that $\{v_1, \ldots, v_p\}$ is linearly independent.

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  • $\begingroup$ Yes, I agree that that would be easier but the teacher wanted us to avoid that so we could really see what was going on in the transformation. Does what I have make sense at least? $\endgroup$
    – Bret Hisey
    Nov 22, 2017 at 20:45
  • $\begingroup$ @BretHisey Yes, your proof is perfectly valid. $\endgroup$ Nov 22, 2017 at 20:46
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By linearity of $T$ you have

$T(\alpha u+\beta v)=\alpha T(u)+\beta T(v)$ for scalars $\alpha,\beta$ belonging to underlying field and $u,v\in V$.

Put $\alpha=\beta=0$ to see $T(0)=0$.

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