Proof of existence of a Jordan form?

[tim_lou]

I have been recently introduced to the idea of Jordan form of a matrix, and I have attempted to prove it. However, there is one step that I cannot prove. If I can prove the step, then everything else is done.

Let A be a m by m matrix with characteristic polynomial
[tex]ch(x)=(x-\lambda)^{m}[/tex]
I want to show that there exist distinct vectors, v1 to vm, not zero, such that:
[tex]A\vec{v}_1=\lambda \vec{v}_1[/tex]
[tex]A\vec{v}_2=\lambda \vec{v}_2+\vec{v}_1[/tex]
[tex]A\vec{v}_3=\lambda \vec{v}_3+\vec{v}_2[/tex]
.
.
.
[tex]A\vec{v}_m=\lambda \vec{v}_m+\vec{v}_{m-1}[/tex]

as long as these vectors exist, I can show that they are linearly independent, form a basis... bla bla bla and the Jordan form easily follows...But I just do not know how theses vectors must exist.

 

[matt grime]

I doubt you can do that since it is not true (just take A the identity matrix). You should find a decent statement of Jordan Normal Form. It is not that you can find vectors like that. It is a statement about how to decompose a linear map as the composite of two linear maps satisfying certain properties.

Oh, and as far as I can recall, it is a very non-trivial theorem.

 

[tacman]

First of all, it is great that you have attempted to prove the existence of Jordan form and have made progress. The step that you are struggling with is a crucial one, and it is important to understand it in order to fully grasp the concept of Jordan form.

To prove the existence of Jordan form, we need to start with the fact that every matrix A with characteristic polynomial ch(x)=(x-\lambda)^{m} has at least one eigenvector v_1 corresponding to the eigenvalue \lambda. This is because the characteristic polynomial is the product of all the eigenvalues of A and since we know that \lambda is an eigenvalue, there must be at least one eigenvector corresponding to it.

Now, if we consider the matrix A-\lambda I, where I is the identity matrix, we can see that the characteristic polynomial of this matrix is ch(x)=(x-\lambda)^{m-1}. This means that A-\lambda I has at least one eigenvector v_2 corresponding to the eigenvalue \lambda.

Next, we can consider the matrix (A-\lambda I)^2 which has the characteristic polynomial ch(x)=(x-\lambda)^{m-2}. This means that (A-\lambda I)^2 has at least one eigenvector v_3 corresponding to the eigenvalue \lambda. Continuing this process, we can see that (A-\lambda I)^{m-1} has at least one eigenvector v_m corresponding to the eigenvalue \lambda.

Now, we can define the vectors v_1 to v_m as follows:
v_1 is the eigenvector of A corresponding to the eigenvalue \lambda.
v_2 is the eigenvector of (A-\lambda I) corresponding to the eigenvalue \lambda.
v_3 is the eigenvector of (A-\lambda I)^2 corresponding to the eigenvalue \lambda.
.
.
.
v_m is the eigenvector of (A-\lambda I)^{m-1} corresponding to the eigenvalue \lambda.

It can be shown that these vectors v_1 to v_m are linearly independent and form a basis for \mathbb{R}^m. This means that any vector in \mathbb{R}^m can be expressed as a linear combination of these vectors, which is exactly what we need to prove the existence of Jordan form.

In conclusion, the existence of Jordan form can be proved by