How to prove a power series of matrix is onto

In summary: If you don't know what it means, then you have no basis to discuss the problem.In summary, the conversation discusses how to prove that the map exp: M_2(R) --> M_2(R) is not onto. Various methods are suggested, including finding one element not in its image or using the fact that exp(A)exp(-A)=I. The conversation also addresses the assumption that one must know exp(A)=\sum1/n!(A)^n in order to prove the problem.
  • #1
happybear
19
0

Homework Statement


How do I prove a power series is onto? Since I cannot calculate directly, especially I haven't learned Jodarn Normal form.




Homework Equations





The Attempt at a Solution


By showing 1-1, I tried
∑(1/n!)[(M)^n-(N)^n]=0, what can I conclude from this step?
 
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  • #2
M is a square matrix, so is exp(M). Thus it is one-to-one if and only if it is onto which is if and only if it is invertible.
 
  • #3
Please don't send me private messages instead of answering in the thread.

From the pm, the question ought to have been written:

show that the map

exp: M_2(R) --> M_2(R)

is not onto.

This is a very different question from the one inferred from the first post.

What have you attempted.
 
  • #4
Sorry about that. How can I show it is not onto, since I cannot find a B in Y, and calculate its A in X? Is there any other to prove it instead of using Jordan Normal Form
 
  • #5
You are not aiming to calculate its image. Just to find one element not in its image, or some family of matrices not in its image.

Just think about the exp map for real numbers. It maps R onto the strictly positive real numbers. Now, negative matrices don't make much sense immediately, so what about the zero matrix? Can you show that is not the image.

If you don't like that, what else do you know? How about: what is exp(A)exp(-A)? Do you have any other results you can think of?
 
  • #6
I understand this part. But this is follow the assumption that we know
exp(A)=[tex]\sum[/tex]1/n!(A)^n

What if we don't know this theroem. All we know this is true for A is real number, not when A is matrix, then makes it hard to prove, right
 
  • #7
What? That's not an assumption so much as the definition of what the expression exp(A) means.
 

FAQ: How to prove a power series of matrix is onto

Can a power series of matrix be onto?

Yes, a power series of matrix can be onto if its range spans the entire codomain, meaning every element in the codomain can be reached through the power series.

How do you prove that a power series of matrix is onto?

To prove that a power series of matrix is onto, you need to show that for every element in the codomain, there exists at least one element in the domain that maps to it. This can be done through various methods such as using the invertibility of the matrix or showing that the determinant is non-zero.

What is the importance of proving that a power series of matrix is onto?

Proving that a power series of matrix is onto is important because it ensures that every element in the codomain has a corresponding element in the domain, making it possible to solve equations and perform calculations involving the matrix. It also guarantees that the matrix has an inverse, which is crucial in many applications.

Are there any specific criteria for a power series of matrix to be onto?

Yes, there are certain conditions that need to be met for a power series of matrix to be onto. For example, the matrix needs to be square and have a non-zero determinant. The power series must also be convergent for all elements in the domain.

Can a power series of matrix be onto without being one-to-one?

Yes, it is possible for a power series of matrix to be onto without being one-to-one. One-to-one means that every element in the domain maps to a unique element in the codomain, while onto means that every element in the codomain has at least one corresponding element in the domain. So, a power series can be onto without being one-to-one if multiple elements in the domain map to the same element in the codomain.

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