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If $A=\begin{bmatrix} 0 & 0 & 1 \\ 3 & 1 & 0 \\ -2&1&4\end{bmatrix}$, find $A^5 - 27A^3 + 65A^2$

$$A=\begin{bmatrix} 0 & 0 & 1 \\ 3 & 1 & 0 \\ -2&1&4\end{bmatrix}$$

Let $\lambda$ be its eigenvalue, then

$$(A-\lambda I) = \begin{bmatrix} 0-\lambda & 0 & 1 \\ 3 & 1-\lambda & 0 \\ -2&1&4-\lambda\end{bmatrix}$$

$$|A-\lambda I| = -(\lambda)^3 + 5(\lambda)^2 - 6(\lambda) +5$$

Using Cayley-Hamilton theorem

$$A^3-5^2+6A-5=0$$

How do I use this find $A^5 - 27A^3 + 65A^2$?

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    $\begingroup$ By Cayley-Hamilton, it is actually $A^3-5A^2+6A-5I=O$. $\endgroup$ Sep 8, 2018 at 15:45

4 Answers 4

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To add to the above answer, using $A^3 = 5A^2 - 6A + 5$ to simplify further $A^2(5A^2 -33A + 70) = A(5(5A^2 - 6A + 5) - 33A^2 + 70A) = A(-8A^2 + 40A + 25) = -8(5A^2 -33A + 70) + 40A^2 + 25A = 289A - 560$

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  • $\begingroup$ small typo at the end: you must plug in $A^3$. $\endgroup$
    – farruhota
    Sep 8, 2018 at 16:48
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I don't see a way to find the answer directly but you can certainly simplify your calculations; since $A$ commutes with itself we may factor the polynomial however we want. Write $$ A^5-27A^3+65A^2=A^2(A^3-27A+65)=A^2\left[(A^3-5A^2+6A-5)+(5A^2-33A+70)\right]. $$ Now by Cayley Hamilton you only need to compute $A^2(5A^2-33A+70)$, which cuts down on the powers of $A$ that you need to calculate directly and reduces the problem to a bunch of addition and two matrix multiplications.

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Doing long division: $$A^5-27A^3+65A^2=(A^2+5A-8)(\underbrace{A^3-5A^2+6A-5}_{=0})+73A-40.$$ Hence it is: $$A^5-27A^3+65A^2=73A-40I=\begin{pmatrix}-40&0&73\\ 219&33&0\\-146&73&252\end{pmatrix}.$$

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Let

$p(x) =x^5 -27x^3 + 65x^2; \tag 1$

one can always compute $p(A)$ directly, but perhaps the following requires less arithmetical work:

Let

$\chi(x) = x^3 - 5x^2 +6x - 5 \tag 2$

be the characteristic polynomial of $A$.

By the division algorithm for polynomials, there exist unique

$q(x), r(x) \in \Bbb Q[x] \tag 3$

such that

$p(x) = q(x)\chi(x) + r(x), \tag 4$

with either $r(x) = 0$ or $\deg r(x) < 3 = \deg \chi(x)$; thus, since

$\chi(A) = 0, \tag 5$

we have from (4)

$r(A) = p(A) - \chi(A)q(A) = p(A) - 0 \cdot q(A) = p(A); \tag 6$

thus, we compute $r(x)$ by synthetic (long) division of polynomials; it's not too much work; we find

$q(x) = x^2 + 5x - 8, \tag 7$

and

$r(x) = 73x - 40; \tag 8$

the reader may easily check that

$p(x) = x^5 - 27x^3 + 65x^2$ $= (x^2 + 5x - 8)(x^3 - 5x^2 +6x - 5) + 73 x - 40 = q(x)\chi(x) + r(x); \tag 9$

thus

$p(A) = r(A) = 73A - 40I = 73 \begin{bmatrix} 0 & 0 & 1 \\ 3 & 1 & 0 \\ -2 & 1 & 4 \end{bmatrix} - 40I = \begin{bmatrix} -40 & 0 & 73 \\ 219 & 33 & 0 \\ -146 & 73 & 252 \end{bmatrix}. \tag{10}$

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