I want to know how to tackle this exercise:

Prove that if a vector is uniquely expressed as a linear combination of the vectors {$v_1,v_2,...,v_n$} then this vectors are linearly independent. Prove also the contrapositive.

Actually this idea came up to me when writing the question check for mistakes and contradictions please.

They don't allow me to embed images so they've added a link,check it clicking this phrase

If the link does not work. What I did was to write $v$ as a unique linear combination of {$v_1,v_2,...,v_n$} like this: $v=c_1v_1+...+c_nv_n$

Then assuming that {$v_1,v_2,...,v_n$} are linearly dependent in search for a contradiction. $0=k_1v_1+...+k_nv_n$ where $k_1,...,k_n$ are not all equals to $0$. let's say $k_n$ different from $0$

$v_n=(k_1v_1/-k_n)+...+(k_{n-1}v_{n-1}/-k_n)$ but when we substitute this in the linear combination of $v$ is a contradiction because we have assumed that it was unique.

So {$v_1,v_2,...,v_n$} are linearly independent.

So if this is correct I just need a hint on proving the contrapositive

  • $\begingroup$ There is a logical equivalence between $A \Rightarrow B$ and $\neg B \Rightarrow \neg A$. That seems to be the contrapositive. Alas your proof by contradiction seems to be that proof to me. $\endgroup$ – mvw Mar 30 '16 at 3:59
  • $\begingroup$ I think you have the right idea but $v_n=(c_1v_1/-k_1)+...+(c_nv_n/-k_1)$ is not correct; it should be $v_n=-(k_1v_1/k_n)-\ldots-(k_{n-1}v_{n-1}/k_n)$. It might also help if, after you do the substitution, you show just why it leads to an expression for $v$ which is different from $v=c_1v_1+...+c_nv_n$. $\endgroup$ – ForgotALot Mar 30 '16 at 4:04
  • $\begingroup$ @ForgotALot Your are right, however I don't know how to put the subscripts n-1 in the right way, can you fix it? $\endgroup$ – César Rosendo Mar 30 '16 at 4:14
  • $\begingroup$ @ForgotALot Thanks btw $\endgroup$ – César Rosendo Mar 30 '16 at 4:24
  • $\begingroup$ Just in case, if you see a formula and you want to know how it was written, click your right mouse button and select "Show Math As," then "TeX Commands." $\endgroup$ – ForgotALot Mar 30 '16 at 4:24

Assume first that $v = \sum c_i v_i$ is the unique expression of $v$ as a linear combination of the $v_i$. Then adding $v$ to both sides of the linear dependence relation $0 = \sum a_i v_i$ yields $$ \sum c_i v_i = v = \sum (a_i + c_i)v_i.$$

Therefore by uniqueness we must have $a_i + c_i = c_i$ for all $i$, i.e., $a_i = 0$ for all $i$. This proves linear independence.

For the converse, assume linear independence of the $v_i$ and write $v = \sum a_i v_i = \sum b_i v_i$ in any two ways. Then $0 = \sum (a_i - b_i)v_i$, so by linear independence we have $a_i - b_i = 0$ for all $i$, i.e., $a_i = b_i$. Therefore the expression of $v$ in terms of $v_i$ is unique.

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That is the approach I would take. Assume that $v_1...v_n$ are dependent. Then some non trivial combination = 0. Therefore, any vector cannot be expressed as a $unique$ combination.

However, this proof really is a proof of the contrapostion. If the vectors are not linearly independent, then there is no unique combination.

To prove it the other way...$\mathbf v$ is a unique combination of vectors $\mathbf v_1,\mathbf v_2,...\mathbf v_n$.
$\mathbf v= c_1\mathbf v_1+c_2\mathbf v_2+...c_n\mathbf v_n \iff c_1...c_n$ equal whatever it is that they equal, then there exist no $k_1...k_n$ such that $k_1\mathbf v_1+k_2\mathbf v_2+...k_n\mathbf v_n = 0$ therfore $\mathbf v_1,\mathbf v_2,...\mathbf v_n$ are independent.

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