Comparing subspaces of a linear operator

In summary, for a linear operator S: U->U on a finite dimensional vector space U, we can prove that Ker(S) = Ker(S^2) if and only if Im(S) = Im(S^2). This can be shown by using the concepts of the kernel and image of a linear operator and the fact that for any vector v in U, some of its basis elements are in Ker(S) and the remaining elements are in Im(S). By this reasoning, we can conclude that the dimensions of Ker(S) and Ker(S^2) are equal, as well as the dimensions of Im(S) and Im(S^2).
  • #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).
 
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  • #2
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.
 

Related to Comparing subspaces of a linear operator

1. What is a linear operator?

A linear operator is a mathematical function that maps one vector space to another, while preserving the structure of the vector space. It is a special type of linear transformation that operates on vector spaces.

2. How do you compare subspaces of a linear operator?

To compare subspaces of a linear operator, you can look at their dimensions, basis vectors, and how they are related to each other. You can also use the rank-nullity theorem to compare the dimensions of the null space and range of the linear operator.

3. What is the importance of comparing subspaces of a linear operator?

Comparing subspaces of a linear operator allows for a better understanding of the properties and behavior of the operator. It can also help in solving problems involving the operator, such as finding eigenvalues and eigenvectors.

4. Can subspaces of a linear operator be equal?

Yes, subspaces of a linear operator can be equal if they have the same basis vectors and dimensions. This means that they represent the same vector space, and any vector in one subspace can be expressed as a linear combination of the basis vectors in the other subspace.

5. Can you compare subspaces of different linear operators?

Yes, subspaces of different linear operators can be compared if they operate on the same vector space. However, the comparison may be more complex as the operators may have different properties and behaviors.

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