# Comparing subspaces of a linear operator

1. Jan 27, 2009

### thejinx0r

Statement
Let S be a linear operator S: U-> U on a finite dimensional vector space U.
Prove that Ker(S) = Ker(S^2) if and only if Im(S) = Im(S^2)

So, I'm really not sure about how to prove this properly. I have a few ideas, but this one seemed to make sens intuitively to me. So, I'm just going to write out the first part. And if it is correct, then the converse should be similar.

Everything up to the *** is mathematically correct.
After that, I'm on shaky grounds.

Proof
Suppose Im(S)=Im(S^2)
Then
U \ {Im(S) \ 0 } =U \ {Im(S^2) \ 0}
***
Then, for all u in U \ {Im(S) \ 0 } except for u=0_vector
S(u) does not belong to Im(S)
Therefore, u must belong to Ker(S) and Ker(S^2).
This implies that for all u in U such that S(u) does not belong to Im(S) belongs to both Ker(S) and the kernel (S^2).

2. Jan 28, 2009

### Staff: Mentor

There are some concepts about the kernel and image of a linear operator that you're not using that might be helpful.

You can assume that dim(U) = n, which means that any basis for U has n linearly independent vectors.

In general, dim(Ker(S)) + dim(Im(S)) = n, and for this problem dim(Ker(S^2)) + dim(Im(S^2)) = n.

For any vector v in U, with v = c_1*x_1 + c_2*x_2 + ... + c_n*x_n, part of this vector is mapped by S to the zero vector in U, and the rest of this vector is mapped to some nonzero vector. IOW, some of the x_i's in the basis for U are in Ker(S), say m of them, with m<=n, and the other n - m of them are in Im(S).

That's how I would approach it.