If a two vectors $\bf u$ and $\bf v$ in linearly independent, can their images be linearly dependent given they were transformed by a linear transformation $T$?

I think not.

My reasoning

If the images are linearly dependent, then $c_1T({\bf u})+c_2T({\bf v})={\bf 0}$ for scalars $c_1$ and $c_2$, not both $0$. Then

$c_1T({\bf u})+c_2T({\bf v})={\bf 0}$

$T(c_1{\bf u}+c_2{\bf v})=T({\bf 0})$

$c_1{\bf u}+c_2{\bf v}={\bf 0}$

Therefore, ${\bf u}$ and $\bf v$ are linearly dependent.

My reservation

A homework problem seems to contradict the above. Here's the problem:

Let $\mathbb R^n\rightarrow\mathbb R^m$ be a linear transformation. Suppose $\{{\bf u},{\bf v}\}$ is a linearly independent set, but $\{T({\bf u}),T({\bf v}\})$ is a linearly dependent set. Show that $T({\bf x})={\bf 0}$ has a nontrivial solution.

  • $\begingroup$ If $T:\mathbb{R}^n\to\mathbb{R}^m$, what happens when $m=1$? $\endgroup$ – Todd Wilcox Sep 23 '15 at 4:52
  • $\begingroup$ If those scalars exist it's because the images are linearly dependent, not independent. You got the definition mixed up. $\endgroup$ – JKEG Sep 23 '15 at 4:53
  • $\begingroup$ Good catch. I meant to say "dependent." $\endgroup$ – Joel Christophel Sep 23 '15 at 5:01

It is possible that they can be linearly dependent. Consider the map $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ which maps $(x,y)$ to $(x,0)$. Then $(1,0)$ and $(1,1)$ are linearly independent, but $T(1,0) = T(1,1) = (1,0)$.

The problem with your argument is that $T(c_1 u + c_2 v) = T(0) = 0$ does not imply that $c_1 u + c_2 v = 0$. It just means that $c_1 u + c_ 2 v$ is a solution of $T(x) = 0$. When $u$ and $v$ are linearly independent, it must be the case that $c_1 u + c_2 v$ is nonzero, so this solution is non-trivial.

  • $\begingroup$ Good explanation. Am I correct in thinking that your last sentence only holds when ${\bf u}$ and ${\bf v}$ are linearly independent because otherwise, they could both be the zero vector? $\endgroup$ – Joel Christophel Sep 23 '15 at 5:28
  • $\begingroup$ I edited the answer to address this - linear independence is necessary to justify the word "non-trivial". $\endgroup$ – Dorebell Sep 23 '15 at 10:34

Hint :

Can we have a linear transformation from a n-dimensional space to a m-dimensional space where $n \gt m$ ? . If yes, then definitely for some $u$ and $v$, the images will be linearly dependent.

Projection is such a linear transformation. That should answer your question in a more intuitive way. @Dorebell has already provided a detailed answer.


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