Suppose $v_1,...,v_m$ is a linearly dependent list of vectors in V. Suppose also that $W\neq\{0\}$. Prove that there exist $w_1,...,w_m$ such that no $T\in L(V,W)$ satisfies $Tv_k=w_k$ for each k= $1,...,m$

My proof: Since $v_1,...,v_m$ is linearly dependent, this means that if $a_1v_1,...+a_mv_m=0$, then $a_1,...,a_m$ are not all zero.

Then apply a linear map $T$ to both side of $a_1v_1,...+a_mv_m=0$, and get $a_1Tv_1+...+a_mTv_m=0$

Then let $(w_1,...,w_m)$ be a linearly independent vector list in $W$

Suppose there exists a linear map $Tv_1=w_1,...,Tv_m=w_m$ under conditions above, then $a_1Tv_1+...+a_mTv_m=0$ this will be

$a_1w_1+...+a_mw_m=0$ and this implies $a_1=...=a_m=0$ and this contradicts with $a_1,...,a_m$ are not all zero.

Thus, when $(w_1,...,w_m)$ is linearly independent, there won't exist a linear map satisfying $Tv_k=w_k$ for each k= $1,...,m$

I'm kind of suspicious of my proof, and I feel there exists some logic problem. Any suggestions? Thanks in advance.

Edit: The above proof is flawed since I didn't consider if the dimension of W is less than m. I give another proof below.

Since $(v_1,...,v_m)$ is linearly dependent,for a vector $v$, this can be written in two ways:



In these two ways, there always exists $b_i\neq a_i$ for some $i$ since the vector list is linearly dependent.

Thus suppose there is a linearly map $T$ such that $Tv_1=w_1,...,Tv_m=w_m$

Then it is certain that $a_1Tv_1+...+a_mTv_m=b_1Tv_1+...+b_mTv_m$

This means that: $a_1w_1+...+a_mw_m=b_1w_1+...+b_mw_m\implies(a_1-b_1)w_1+...+(a_m-b_m)w_m=0$

Then if $a_1\neq b_1$, let $w_2=...=w_m=0$,but $w_1\neq 0$

Similarly, if $a_i\neq b_i$, just let $w_i\neq 0$ and other $w_j=0$

Then this kind of choice of $w_1,..,w_m$ will make a contradiction to the linear map $T$.

  • 2
    $\begingroup$ What if $\dim W<m$? $\endgroup$ Dec 14, 2021 at 3:52

1 Answer 1


The main problem with your proof is your understanding of linear dependence. Suppose that $$a_1v_1+\cdots+a_nv_n=0\ .$$ The statement that $v_1,\ldots,v_n$ are linearly dependent does not mean that the scalars are (definitely) not all zero, it means they are not necessarily all zero. Do you see the difference? To put it more formally, saying that the vectors are dependent means that there exist scalars, not all zero, for which the linear combination is zero.

Hint for a proof: if $v_1,\ldots,v_n$ are linearly dependent, then one of them can be written as a linear combination of the others, say $$v_1=a_1v_2+\cdots+a_nv_n\ .$$ Now the condition $$w_k=T(v_k)\quad\hbox{for all}\ k$$ (remember that you are trying to choose $w_k$ for which this is impossible) implies $$w_1=a_2w_2+\cdots+a_nw_n\ .$$ Can you choose $w_k$ such that this equation is definitely false?

  • $\begingroup$ In the example you gave, it seems that letting $v_2$ be zero(i.e. $w_2=...=w_n=0$), but keeping $w_1$ non zero will make a contradiction. I think I kind of had this idea, but I'm not sure. Also, based on this "vague“ idea I tried to rewrite my proof. Thus if you are interested, could you please look at my proof? $\endgroup$
    – M_k
    Dec 14, 2021 at 5:21
  • $\begingroup$ I think that's OK. I would suggest you try to write your argument as clearly as possible, remembering that it should be understandable to someone else, not just to yourself. This is always difficult and requires a lot of practice. $\endgroup$
    – David
    Dec 14, 2021 at 6:04

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