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Question: Use Cayley-Hamilton theorem to express $A^5-4A^4-7A^3+11A^2-A-10I$, where $$A= \begin{pmatrix}1&4\\2&3\end{pmatrix}$$

Attempt: I only know how to get the characteristic polynomial equation which is $\lambda^2-4\lambda -5$ for that matrix. By definition $A^2-4A-5=0$. But where does that 5 degree polynomial come from?

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Note: It's not "by definition" that $A^2-4A-5I=0$; it's as a consequence of the Cayley-Hamilton Theorem. – Arturo Magidin Feb 19 '12 at 23:46
up vote 4 down vote accepted

So, the characteristic polynomial of $A$ is $p(t) = t^2 - 4t -5$.

You are interested in the polynomial $P(t) = t^5 -4t^4 -7t^3+11t^2 - t - 10$, evaluated at $A$.

We can divide $P(t)$ by $p(t)$ with remainder. We get: $$P(t) = (t^3-2t+3)p(t) + (t+5).$$ Now, because $A$ commutes with itself and with all scalars, we can evaluate both sides at $A$ and retain equality. So $$P(A) = (A^3-2A+3I)p(A) + (A+5I).$$ But $p(A)=0$ by the Cayley-Hamilton Theorem.

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Cayley-Hamilton lets you express $A^2$ as $rA+s$ for some numbers $r$ and $s$. Then it lets you express $A^3,A^4,\dots$ in the same form. So it lets you express any polynomial in $A$ in that form.

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Continuing your analysis, $$A^2 - 4A -5I=0\implies A^2 = 4A+5I$$ Thus, $$\begin{array}{cc} A = A\\ A^2 = 4A + 5I\\ A^3 = 4A^2 + 5A &= 4(4A+5I) + 5A = 21A +20I\\ \vdots \end{array}$$

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