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Let $T \in \mathcal{L} (M_{2 \times 2}(R))$ be given by $$X \mapsto \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}X.$$

How could I find the minimal polynomial of this operator? Unfortunately my math professor seems to want to do things his own way, and we haven't learned determinants or characteristic polynomials or any rules (or guidelines) for computation. All I know is that a minimal polynomial is a monic polynomial of smallest degree such that $p(T) = 0$. I tried setting

$$a_0\begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix} + a_1\begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}\begin{pmatrix} x_0 & x_1 \\ x_2 & x_3 \end{pmatrix} + a_2 \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}^2\begin{pmatrix} x_0 & x_1 \\ x_2 & x_3 \end{pmatrix} = 0.$$

But I don't really know where to go from here.

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  • $\begingroup$ That should go $$a_0\pmatrix{x_0&x_1\\x_2&x_3}+\cdots$$ surely? Is there any relation to the minimum polynomial of the matrix $\pmatrix{1&2\\4&3}$? $\endgroup$ – Lord Shark the Unknown Nov 29 '19 at 17:43
  • $\begingroup$ He's given us the hint that $a_0I + a_1 T + a_2 T^2 = 0$ $\endgroup$ – hellowold121 Nov 29 '19 at 17:46
  • $\begingroup$ Trace is $4$ and determinant is $-5$ so by Cayley-Hamilton $T^2-4T-5=0$ $\endgroup$ – J. W. Tanner Nov 29 '19 at 17:51
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This is not my usual method for finding minimal polynomial, but you were heading toward this solution. First, note that if we have $a_2T^2+a_1T+a_0I=0$, we can divide by $a_2$ to make the polynomial monic and get $T^2+a_1'T+a_0'I=0$.

Can you solve $\begin{pmatrix}9 & 8 \\ 16 & 17\end{pmatrix}+a_1'\begin{pmatrix}1 & 2 \\ 4 & 3\end{pmatrix}+a_0'\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}=\begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}$ for $a_1'$ and $a_0'$?

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  • $\begingroup$ @Lord Shark the Unknown: thanks for fixing the identity matrix $\endgroup$ – J. W. Tanner Nov 29 '19 at 17:50
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By Cayley-Hamilton theorem the characteristic polynomial $f(x) = x^2 - 4x - 5$ satisfy this polynomial. As $x^2 - 4x - 5 = (x+1)(x-5)$ and neither of the factors satisfy the operator, it must itself be the minimal polynomial.

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  • $\begingroup$ The domain of the operator is the space of $2\times2$ real matrices, not $\mathbb R^2$, so its characteristic polynomial is quartic, not quadratic. Your $f(x)$ does annihilate the fixed matrix in the operator’s definition, but not quite for the reason you’ve stated. $\endgroup$ – amd Nov 29 '19 at 18:27
  • $\begingroup$ ah I see... sorry for the confusion. $\endgroup$ – Siddhartha Nov 29 '19 at 23:59
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I think it’s worth expanding on Lord Shark the Unknown’s comment. Since $T:X\mapsto AX$ with a fixed matrix $A$, for any polynomial $p$ we have $p[T](X) = p[A]X$, so the minimal polynomials of $T$ and $A$ are identical. In particular, this means that the minimal polynomial is at most quadratic, hence the hint your professor gave you.

As described in hints and another answer, you can then find that the characteristic polynomial of $A$ is $\lambda^2-4\lambda-5=(\lambda-5)(\lambda+1)$. This is a product of distinct linear terms, so it is also the minimal polynomial. Interestingly, the characteristic polynomial of $T$ is the square of this polynomial.

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