Image and kernel of iterated linear transformation intersect trivially

In summary, it is shown that for a linear transformation f:V -> V on a finite-dimensional vector space V, there exists a positive integer m such that the image and kernel of f^m intersect trivially. This is proved by observing that the images and kernels of successive iterations of the transformation form increasing and decreasing chains of subspaces, which must eventually stabilize due to the finite-dimensionality of V.
  • #1
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Homework Statement


Given a linear transformation f:V -> V on a finite-dimensional vector space V, show that there is a postive integer m such that im(f^m) and ker(f^m) intersect trivially.


Homework Equations





The Attempt at a Solution


Observe that the image and kernel of a linear transformation f:V -> V are each subspaces of V.
The kernels of the successive iterations of the transformation form an increasing chain of subspaces.
The images of the successive iterations of the transformation form a decreasing chain of subspaces.
Since V is finite-dimensional, the chains of the kernels and images must eventually stabilize, say at m1 and m2 respectively. Let m be max{m1,m2}. So the kernel and image of the transformation are stable after m iterations.
Now let v be some element contained in both the image and the kernel of f^m.
This means that some element w in V with f^m(w) = v, and that f^m(v) = 0.

I need to obtain a contradiction, but I have been unsuccessful so far.

Thanks for your help.
 
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  • #2
If the image is stabilized then f^m is a 1-1 map of Im(f^m)->Im(f^m). f^m(f^m(w))=f^m(v)=0. Seems like there is a contradiction there to me if you assume v is nonzero. What is it?
 
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  • #3
Thank you, Dick. I am kind of dumb, so it took me a while to see what you meant.
 

1. What is the definition of an image and kernel of an iterated linear transformation?

The image of an iterated linear transformation is the set of all possible outputs that can be obtained by applying the transformation to every possible input. The kernel, on the other hand, is the set of all inputs that produce a zero output when the transformation is applied.

2. How do the image and kernel of an iterated linear transformation intersect?

The image and kernel of an iterated linear transformation intersect if and only if the only input that produces a zero output when the transformation is applied is the zero vector. In other words, the only common element between the image and kernel is the zero vector.

3. What is the significance of the image and kernel intersecting trivially?

When the image and kernel of an iterated linear transformation intersect trivially, it means that the transformation is both injective and surjective. This is significant because it implies that the transformation is an isomorphism, meaning it preserves both the linear structure and dimensionality of the original vector space.

4. Can the image and kernel of an iterated linear transformation intersect non-trivially?

Yes, it is possible for the image and kernel of an iterated linear transformation to intersect non-trivially. This can happen when the transformation is not injective (meaning there are multiple inputs that produce the same output) or not surjective (meaning there are some outputs that are not produced by any input).

5. How can we determine if the image and kernel of an iterated linear transformation intersect trivially?

To determine if the image and kernel of an iterated linear transformation intersect trivially, we can use the rank-nullity theorem. This theorem states that the dimension of the image plus the dimension of the kernel is equal to the dimension of the original vector space. If the dimension of the kernel is zero, then the image and kernel intersect trivially.

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