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Consider the matrix $$A = \frac{1}{4} \begin{bmatrix} 17 & -6 & -9 & -12 \\ 15 & 26 & 69 & 36 \\ 3 & 6 & 29 & 12 \\ -12 & -12 & -48 & -16 \end{bmatrix}. $$ I am given that $(A-I)(A-2I)(A-5I)=0$. I have to explain why this matrix is diagonalizable and then find the minimal polynomial of this matrix.

For the diagonalizability, I was thinking that since the matrix has 3 distinct eigenvalues, then the minimal polynomial would have 3 distinct linear factors, making this matrix diagonalizable. However I'm not sure how to show this and I'm also not sure if this is necessarily correct.

As for finding the minimal polynomial, I was given a hint that says to use the trace but I don't see the relationship between the two.

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  • $\begingroup$ How do you know the matrix has three distinct eigenvalues? As soon as one of the factors is $0$, it doesn't matter what the other factors are. $\endgroup$ – saulspatz Aug 13 '20 at 21:07
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    $\begingroup$ Suppose the minimal polynomial is $(x-2)(x-5)$. Then won't $A$ also satisfy $(x-a)(x-2)(x-5)$ for any value of $a$? $\endgroup$ – saulspatz Aug 13 '20 at 21:13
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    $\begingroup$ One possible way: every factor of the expression is linear and the minimal polynomial divides that expression hence its factors are linear. $\endgroup$ – cgss Aug 13 '20 at 21:23
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    $\begingroup$ The matrix satisfies the polynomial $(x-1)(x-2)(x-5)$; in particular, the minimal polynomial divides $(x-1)(x-2)(x-5)$. That means the minimal polynomial splits and has no repeated roots, hence the matrix is diagonalizable. The minimal polynomial is one of the factors of this polynomial. It is clearly none of $x-1$, $x-2$, or $x-5$. You can check to see if it is any of the degree 2 factors; otherwise, it is the degree 3 polynomial. $\endgroup$ – Arturo Magidin Aug 13 '20 at 21:30
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    $\begingroup$ Alternatively, the trace is the sum of the eigenvalues, and every eigenvalue must show up in the minimal. If the minimal were, say, $(x-1)(x-2)$, then the eigenvalues would have to be $1,1,1,2$; or $1,1,2,2$; or $1,2,2,2$, so the trace would have to be $5$, $6$, or $7$. Etc. $\endgroup$ – Arturo Magidin Aug 13 '20 at 21:32
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Since $(A-I)(A-2I)(A-5I)=0$ the minimal polynomial $m(A)$ is a divisor of $(x-1)(x-2)(x-5)$, hence it has only simple roots, which is a necessary and sufficient condition of diagonalisability.

Edit: concerning the minimal polynomial, we know the eigenvalues are among $\{1,2,5\}$. To determine it, you only have to determine which, among $A-I, A-2I, A-5I$ has a rank less than $4$.

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  • $\begingroup$ You've omitted the part about the minimal polynomial. $\endgroup$ – saulspatz Aug 13 '20 at 21:40
  • $\begingroup$ @saulspatz: Oh yes! I'll complete it in a moment $\endgroup$ – Bernard Aug 13 '20 at 21:46

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