Solution

We want to find a non-trivial linear cominination of which is equal to 0. If these are the columns of a matrix we know how to do this (find the nullspace) so we reduce the problem to this form by writing

[ ] = [ ]

write this as VA = U

If we can solve AX= , i.e

of we can find a vector X = such that =

Then since VAX = UX (matrix multiplication) we see that = 0 (the zero vector) which is what we wanted to do.

The nullspace of has for a basis so any multiple of this will work. In particular the vector is a valid solution! Note that as shown here, when a multiple of a vector can be used it usually simplifies hand calculations to clear the denominator.

There is a conceptually easy way to understand this reduction to a matrix problem. It goes thus:

Create a special vector whose entries are the basis vectors . Make a vector . Then it is not hard to see that the coefficient matrix satisfies the equation . To say that the vectors in are dependent is the same thing as saying for some nonzero vector .

But this means . Since has independent entries, the vector must be zero!

Thus we simply have to find the null space of and can use the answer for a valid .

Thus, remember the criterion for independence/dependence when we take a vector of vectors . These are dependent iff there is a nonzero vector of scalars such that .

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