Proof of existence of a Jordan form?

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SUMMARY

The discussion centers on the proof of the existence of a Jordan form for a matrix A, specifically when the characteristic polynomial is ch(x)=(x-\lambda)^{m}. The user seeks to demonstrate the existence of distinct non-zero vectors v1 to vm that satisfy a specific recurrence relation involving the matrix A and the eigenvalue λ. The conversation highlights the complexity of proving this step, emphasizing that the existence of such vectors is not guaranteed in all cases, particularly when considering matrices like the identity matrix.

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tim_lou
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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.
 
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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.
 

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