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I'm trying to prove that if a linear operator $f$ is diagonalisable then its minimal polynomial is the product of distinct linear factors. This is what I have so far:

Let $f$ be diagonalisable. So there exists a basis relative to which $f$ has a diagonal matrix, say $D$. So the characteristic polynomial of $f$ is given by $p_f(x)=(x-\lambda_1)(x-\lambda_2)\ldots(x-\lambda_s),$ where $\lambda_i$ are the diagonal entries of $D$.

I know that the minimal polynomial must divide the characteristic polynomial and have the same linear factors. Without loss of generality let the first $i$ linear factors be distinct. So I claim that the minimal polynomial $m_f(x)=\pm (x-\lambda_1)(x-\lambda_2)\ldots(x-\lambda_i).$

However, how can I now verify that $m_f(D)=0$ ?

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Denote $\lambda_1,\ldots,\lambda_s$ all distinct eigenvalues of $A$ and consider $$ \pi_A(\lambda)=\prod_{k=1}^s(\lambda-\lambda_k), $$ then if $A=SDS^{-1}$ we have $$ \pi_A(A)=\prod_{k=1}^s(A-\lambda_kI)=S\prod_{k=1}^s(D-\lambda_kI)S^{-1}=S\cdot 0\cdot S^{-1}=0 $$ since for every diagonal position there is always some $D-\lambda_kI$ in the product that has zero there. It means that this particular $\pi_A$ with simple roots annihilates $A$, hence, the minimal polynomilal must have simple roots too as it divides $\pi_A$.

P.S. Actually $\pi_A$ is the minimal polynomial, so you claim is, in fact, if and only if.

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  • $\begingroup$ All that really matters about this proof is "have zeros on different diagonal positions", and that is horribly vague. The diagonal matrices with diagonals $[0,1,1]$ and $[1,1,0]$ also have zeros on different diagonal positions, yet their product is not the zero matrix. $\endgroup$ – Marc van Leeuwen Nov 18 '16 at 7:24
  • $\begingroup$ @MarcvanLeeuwen Thanks for the comment, I see your point. Do you like the edit better? $\endgroup$ – A.Γ. Nov 18 '16 at 19:33
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This is very basic, and you do not need to use the characteristic polynomial or the fact that the minimal polynomial divides it at all. You just need to realise that "diagonalisable" means that the sum of the eigenspaces fills the whole space, so a linear operator is zero if (and obviously only if) it is zero on each of the eigenspaces.

Now on the eigenspace for an eigenvalue$~\lambda$, our $f$ acts by scalar multiplication by$~\lambda$. It easily follows that on this eigenspace any polynomial $P[f]$ acts by scalar multiplication by$~P[\lambda]$ (just check that $f^k$ acts by multiplication by $\lambda^k$, and then combine the monomials of the polynomial $P$ linearly). So by the above, $P[f]=0$ iff $P[\lambda]=0$ for every eigenvalue$~\lambda$. The minimal monic polynomial$~P$ with that property is the product of (just) one factor $X-\lambda$ for each distinct eigenvalue$~\lambda$ of$~f$; there are distinct linear factors.

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Without loss of generality, let the diagonal form $D$ have all the $i$ distinct eigenvalues in the top rows. For every eigenvalue $\lambda$, one of the top $i$ rows of the matrix $(D - \lambda I)$ will consist of zero entries. Hence, the determinant of the top left $i$-by-$i$ submatrix of $(D - I \lambda)$ is zero. But, this determinant is $m_{f}(\lambda)$.

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  • $\begingroup$ Nonsense, $m_f(D)$ is a matrix, so it cannot equal a determinant. $\endgroup$ – Marc van Leeuwen Nov 18 '16 at 7:26
  • $\begingroup$ @MarcvanLeeuwen Thank you for your courteous and constructive feedback. I've edited. $\endgroup$ – avs Nov 18 '16 at 19:37

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