# Proof of existence of a Jordan form?

1. Feb 24, 2007

### 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
$$ch(x)=(x-\lambda)^{m}$$
I want to show that there exist distinct vectors, v1 to vm, not zero, such that:
$$A\vec{v}_1=\lambda \vec{v}_1$$
$$A\vec{v}_2=\lambda \vec{v}_2+\vec{v}_1$$
$$A\vec{v}_3=\lambda \vec{v}_3+\vec{v}_2$$
.
.
.
$$A\vec{v}_m=\lambda \vec{v}_m+\vec{v}_{m-1}$$

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.

2. Feb 24, 2007

### 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.