Linear Algebra, help?

OK. All problems that ask you to show that a set is linearly independent start the same way: you write:

c1s1 + c2s2 + ....cksk = 0.
Now we need to find out something about the c's.
If the only way that the above equation will work is if all the
 c's = 0  then it is independent (by definition). If there are nontrivial solutions for the c's (meaning you can get the equation to work with at least one of the c's not=0 ), then it's dependent.

This problem requires a little trick. This is a very common trick that is used in Linear Algebra and once you see it once, you'll know to use it to do many similar problems.

The only information we have in this problem is that all the s vectors are orthogonal. You translate that as meaning that their dot products = 0. So lets take the dot product of both sides of the equation with s1:

s1 * ( c1s1 + c2s2 + c3s3 +....+ cksk) = s1*0
(I'm using * to indicate the dot product)
Distribute the dot product:

s1*c1s1 + s1*c2s2 + s1*c3s3 + ....+ s1*cksk = s1*0 = 0
Now notice what happens on the left: since all the s vectors are orthogonal, the dot product of any 2 different ones =0. So we get that  s1*c2s2 = c2s1*s2 = 0. Similarly, s1*c3s3 = c3s1*s3 = 0 and this continues up to cks1*sk = 0. The only term that does not equal zero on the left side is the first where we dot s1 with itself:  s1*c1s1 = c1s1*s1 = c1 ||s1||^2  (I've used the fact that when you dot a vector with itself you get the square of its length.). THe right side of the equation = 0. So we are left with:   c1||s1||^2 = 0. Now none of the s vectors =0, so the length of an s vector can't be zero, so it must be that c1 = 0. So we've found that c1 = 0.

Now, we dot both sides of the equation with s2 and go through the same business. We will be left with c2||s2||^2 =0, so c2=0. Guess what you do next? Yes, dot both sides with s3...and you get that c3=0. Continue this way and you find that all the c's =0. But that means that the set of s vectors is linearly independent ! DONE

If you take a vector and multiply it by its transpose, you get the same result as taking the dot product of the vector with itself-namely you get the square of its length.
Example: if v = [ v1, v2], then vTv = v1v1 + v2v2 = v1^2 + v2^2 = ||v||^2.

Don't get upset about these problems. There are many groups of problems in Linear Algebra that turn out to be easy once you learn the little trick that is needed. Dotting both sides of a vector equation is often done when you are dealing with orthogonal vectors.  The solution to any problem that requires proving independence will start with the equation
c1v1 + c2v2 + ...+ cnvn = 0. There are many such little rules of thumb. Once someone shows you the few little tricks that come up, you'll do fine.You might even grow to enjoy the subject. (I did !).

Let me know if you need more details here.