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
Best Answer
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.