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I'm trying to compute the Jordan canonical form for the matrix $$ \begin{pmatrix} 9 & 4 & 5 \\ -4 & 0 & -3 \\ -6 & -4 & -2 \end{pmatrix}. $$ My first step was to compute the characteristic polynomial and then, from that, find the minimal polynomial. But finding the minimal polynomial is proving to be somewhat of a challenge for some reason. Observe:

Let $A$ be the given matrix. Then, the characteristic polynomial $\chi_A(x)$ is given by the determinant of $xI - A$. Thus, $$ \chi_A(x) = x^3 - 7x^2 + 16x - 12 = (x-3)(x-2)^2. $$ So, the minimal polynomial $m_A(x)$ must be either $(x-3)(x-2)$ or $(x-3)(x-2)^2$. But, $$ (A-3I)(A-2I) = \begin{pmatrix} 42 & 16 & 25 \\ 16 & 6 & 9 \\ 36 & 16 & 20 \end{pmatrix} \ne \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. $$ And, similarly, $$ (A-3I)(A-2I)^2 = \begin{pmatrix} 294 & 64 & 125 \\ -64 & -12 & -27 \\ -216 & -64 & -80 \end{pmatrix} \ne \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. $$ Does this matrix not have a minimal polynomial, or something? Is such a thing even possible? What am I doing wrong here?

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    $\begingroup$ You should check your calculation of $(A-3I)(A-2I)$ and $(A-3I)(A-2I)^2$. $\endgroup$ – sharding4 Apr 15 '18 at 21:55
  • $\begingroup$ @sharding4 Hmm, I did, and I actually got something different. But it's still not the zero matrix. Not sure what's going on here. Are you telling me that this should, indeed, be the minimal polynomial? $\endgroup$ – thisisourconcerndude Apr 15 '18 at 22:04
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I get

\begin{align*} A-3I &= \pmatrix{ 6&4&5\cr -4&-3&-3\cr -6&-4&-5\cr } \\ A-2I &= \pmatrix{ 7&4&5\cr -4&-2&-3\cr -6&-4&-4\cr } \\ (A-3I)(A-2I) &= \pmatrix{ -4&-4&-2\cr 2&2&1\cr 4&4&2\cr } \\ (A-3I)(A-2I)^2 &= \pmatrix{ 0&0&0\cr 0&0&0\cr 0&0&0\cr } \end{align*}

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  • $\begingroup$ Ok, so I understand that matrix multiplication is not necessarily commutative, but the fact that squaring the $A-2I$ term and then (left) multiplying the result by $A-3I$ gives me the wrong answer seems a bit bizarre... $\endgroup$ – thisisourconcerndude Apr 15 '18 at 22:15
  • $\begingroup$ @thisisourconcerndude That shouldn't be a problem. $\pmatrix{ 6&4&5\cr -4&-3&-3\cr -6&-4&-5\cr }\pmatrix{ 3&0&3\cr -2&0&-2\cr -2&0&-2\cr } = \pmatrix{ 0&0&0\cr 0&0&0\cr 0&0&0\cr }$ but you should simply multiply $(A-3I)(A-2I)$ on the right by $A-2I$ $\endgroup$ – sharding4 Apr 15 '18 at 22:17
  • $\begingroup$ Oh, hell, I see where I went wrong. Just made a stupid arithmetic error, of course... Thanks. $\endgroup$ – thisisourconcerndude Apr 15 '18 at 22:19

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