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Here's the statement that I am to prove/disprove:

If the minimum polynomial of $A$ is $A^5 = 5 A^2$, then $A$ is not diagonalizable.

First, I'm confused about how this can actually be a minimum polynomial. From my understanding, the minimum polynomial is a specific polynomial of least degree which $A$ satisfies. But, couldn't I just multiply both sides of that equation by $A^{-2}$? Then the minimum polynomial would be $A^3 - 5I = 0$, which is of lesser degree. That's really confusing for me!!

Anyway, I continue with this problem focusing on the given polynomial. I know that $A$ is at least 5x5, because the minimum polynomial is of degree 5. However, there are only two distinct roots of the minimum polynomial:

$$A^5 - 5A^2 = 0 \Longleftrightarrow A^2 (A^3 - 5I) = 0 \Longrightarrow$$ eigenvalues: $0,\,5^{1/3}$.

Is that correct? If there were 5 distinct roots, and I knew $A$ were exactly 5x5, then this question would be cake, and it's definitely diagonalizable. Instead, all I know is that there are only 2 eigenvalues, which does not necessarily mean that these are no more than 2 eigenvectors, so $A$ may or may not be diagonalizable (e.g., when looking at roots of $I$).

Any help is appreciated.


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Are you using A as an unknown for polynomials? Or is it a matrix? – DonAntonio Nov 27 '12 at 4:48
I think we can assume that what's meant is that the minimum polynomial of $A$ is $t^5-5t^2$. – Gerry Myerson Nov 27 '12 at 4:51
Isn't it as simple as this: if a matrix is diagonalizable, then its minimal polynomial is simply $\prod(x-\lambda)$ over the distinct eigenvalues, so it can't have repeated roots? – Gerry Myerson Nov 27 '12 at 4:57
You can't multiply by $A^{-2}$ unless $A$ is actually invertible, which in this case it's not. – Greg Martin Nov 27 '12 at 7:11
@GerryMyerson - Buy, an identity matrix of size n has fewer than n roots and is diagonalizable (already diagonalized, obviously). – Clark Henry Nov 27 '12 at 9:44
up vote 1 down vote accepted

If you meant (a guess) that $\,m_A(x):=x^5-5x^2\,$ is the minimal polynomial of a matrix $\,A\,$, then since $\,m_A(x)=x^2(x^3-5)\,\,,\,\,A$ is not diagonalizable since it is not the product of different linear factors (never mind whether the field of definition of $\,A\,$ contains or not the roots of the factor $\,x^3-5\,$).

Diagonalization does not depend on the number of different eigenvalues a matrix has but whether there are as many linearly independent eigenvectors of the matrix as its order, or not.

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