Yes, there is a basis for and it would be the number 1. The problem I see is with your . You didn't use the that you defined as the basis for . It should be
.
for every
Find a basis for kerT.
Attempt:
First, I chose a basis for : W = ((1,...,0),(0,1,...,0),...,(0,0...1))
and T(W) = ((1),(2),(3),...,(k),...,(n))
so basically, I don't know what I'm doing...is there such thing as a basis for R?
is T(W) really (1+2+3+...+k+...+n)??
and how do I come up with the representation matrix?
I took it away, because it may not be a matrix. I thought about it more and decided that it was better not to post that. Unfortunately, I don't have my book near me so that I can look this up. I will see what I can find to be sure how to answer you.
My suggestion is to think of what happens when you multiply a matrix and a vector together, you get a linear combination of the rows of the matrix. Well, if we are multiplying on the right. We could multiply on the left and then we get linear combinations of the columns. This is why I decided to not post what I had. It wasn't completely clear which side we were acting on. Let me read some and get back to you.
Okay, it looks like I had the right idea, but was wrong in my process.
By your definition of the linear transformation, given a vector the linear transformation is given by
.
So we are transforming a vector in to a scalar in . So what does this transform look like? First, we must realize that we are multiplying on the left. Thus, must be since is and we need a scalar.
Now remember in my previous post that when we multiply vectors and matrices we get linear combinations. In this case our matrix is actually a vector and the linear combination is . Thus, is given by
.
Now you can see why I delete my previous post, it was wrong.
Oh yeah, sorry. Now we want to find a basis for the space of vectors that are sent to by , i.e., we want . If were even then the easiest way I could think of doing this would be easy, all we need is one vector and then split it into a linear combination of vectors. One such vector would be
and so one such basis is
If is odd we could just require one of the indices to be zero and then do everything else in almost he same way.
I had a dream about this problem and realize the basis for the kernel that I gave you was wrong.
If a vector is to be in the kernel, the transformation T must send it to 0. However, the individual vectors I gave you are not sent to zero.
The following vectors are sent to zero
.
I am unsure if this is the complete basis though. It seems like I might be missing something. Anyway, this is the idea behind it.
A variation on Ivleph's solution (because I don't like fractions!):
so .
That means we can write as
That should make the basis obvious.
The dimension of is, of course, n and we have put one condition on the kernel: so the kernel has dimension n- 1.
We also know that has to be true because of the "rank-nullity theorem". T maps all of into R and so has rank 1. The nullity must be n-1 so that "1+ (n-1)= n".
I love how you had a dream about this problem. That is so funny...and a bit nerdyI had a dream about this problem and realize the basis for the kernel that I gave you was wrong.
You guys are awesome. Thanks for your help. I'll review this more when I have time.