# Linear Algebra - Dimension, Basis

1. Apr 1, 2004

### discoverer02

Linear Algebra -- Dimension, Basis

Had a horrible lecture this morning and the following was presented at a lightning quick pace. I'm not sure whether I messed up my notes in my haste to get them all down or if I'm completely not understanding something.

I have two sets of vectors: W = {w1,w2,w3,.....,wm} and V = {v1,v2,...,vn} V is linearly independent.
If the elements of W are linear combinations of the elements in V and there are more elements in W than V, m>n, W is linearly dependent or independent?

I wrote down W is linearly independent, but I'm thinking this is wrong.

Clarification would be VERY GREATLY APPRECIATED.

Thanks.

Last edited: Apr 1, 2004
2. Apr 1, 2004

Linearly independent means that the vectors do not depend on each other.

Let's say that the first n vectors in W are constant multiples of the first n vectors in V, i.e.

$$w_n = a_n v_n$$

Soyou'll agree that W will be linearly independent if it comprised only the n vectors, as V is. In fact, you'll probably agree that W is almost identical to V except in scaling.

Except now we have extra vectors in W, which are linear combinations of the vectors in V. We've run out of independent vectors in V, so we have to start adding two or more V vectors together to get a vector in W, i.e.

$$w_m = b^nv_n$$
for m>n

So the question is, are the extra vectors in W independent of the first n vectors in W? Clearly, they cannot be, since all of the linearly independent vectors have already been used!

Therefore, W is not linearly independent.

3. Apr 1, 2004

### HallsofIvy

Staff Emeritus
Since V is linearly independent, its "span" (the set of a linear combinations of vectors in V) is a subspace- a vector space in its own right- which has dimension n. A set of vectors, such as W, in that space with more than n members CANNOT be independent. W is dependent.

4. Apr 1, 2004

### discoverer02

Thanks very much for the help. I can now try to make sense of the rest of my notes and hopefully sleep soundly tonight.