- #1
thejinx0r
- 27
- 0
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).
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).