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?

  • 1
    $\begingroup$ 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


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.

  • $\begingroup$ 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

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 $T(0)=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.