Look at post #5 again. Hint: Solve for u.
-Dan
Prove or disprove:
{v1...vk} is a set of linearly independent vectors.
Vector u cannot be written as a linear combination of the above set.
Then {v1...vk, u} is linearly independent.
Like Steven G wrote:
I would look at the contrapositive of the theorem.
We are given that {v1...vk} is a set of linearly independent vectors.
Suppose {v1...vk, u} is a linearly dependent set.
Case 1: u=0.
Then u = 0v1 + ... +0vk.
This means that u can be written as a linear combination of {v1...vk} -- contradiction.
Case 2: u=/= 0
Since {v1...vk} is a set of linearly independent vectors, we know that the only linear combination that sums to the zero vector is the trivial one.
Since we assume that {v1...vk, u} is a linearly dependent set, we know that there exists c1, ..., ck, b, where c1,...,b are not all 0, such that
c1v1 + c2v2 +...+ ckvk + bu = 0.
Let b=/=0, since we know that not all scalars are equal to zero.
Then u= (-c1/b)v1+...+(-ck/b)vk
which means that u can be written as a linear combination of {v1...vk}, which means that there is a contradiction, hence {v1...vk, u} is a linearly independent set?