I have the following question in my exam preparation:
I think I have to prove that the eigenvalues of A are 2,2,1, but I can't understand how.
It's pretty clear that the Jordan Normal Form is:
But can I prove it?
Thanks,
Alan
2026-03-30 22:57:33.1774911453
Jordan Normal Form of A, Given A Cubed
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Jordan normal form is usually specified with two matrices $\bf P$ and $\bf J$ such that $\bf P^{-1}JP = A$. I assume you mean that $\bf J$ is the matrix you mention. $\bf A^3$ will then be $\bf P^{-1}JPP^{-1}JPP^{-1}JP$, where you see that $PP^{-1}$ will cancel and leave a $\bf P^{-1}J^3P$. Then you can use the block property of matrix multiplication. The jordan blocks are all 1x1 and the block sizes are preserved under multiplication. That means that $\bf J$ must have the same block sizes as $\bf J^3$. So unless is allowed to be complex, the only solutions can be positive real roots to 8 = x^3 and 8 = y^3 and 1 = z^3.