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Can someone please point me in the right direction on how to solve the matrix equation below.

If $$A = \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 2 & -1\end{bmatrix}$$ find constants $p$, $q$ and $r$ such that $$A^3 = p A^2 + q A +r I_3$$ where $I_3$ is the $3 \times 3$ identity matrix.

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Let $B \in Mat(n, \mathbb{C})$ with $n \in \mathbb{N}$. Note that there is a polynomial $P(X) \in \mathbb{C} [X]$ such that $\deg (P(X)) \leq n$ and $P(B) = 0$. Moreover, the characteristic polynomial $f_B(X)$ of $B$ satisfies the equation $f_B(B) = O$.

Here, $A \in Mat(3, \mathbb{C})$, then we can find its characteristic polynomial which is $ - A^3 - A^2 + 2A + I.$ Thus $A^3 = -A^2 + 2A +I$.

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We get $$\det(\lambda I-A) = \begin{vmatrix} \lambda & -1 & 0 \\ 0 & \lambda & -1 \\ -1 & -2 & \lambda+1 \end{vmatrix}=\lambda^3+\lambda^2-2\lambda-1.$$ Thus, using Cayley-Hamilton theorem, we have that

$$A^3+A^2-2A-I=0.$$ That is

$$A^3=-A^2+2A+I.$$

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I think finding the characteristic equation $\det (A-\lambda I)$ is most direct.

However this will work.

Choose a vector that seems easy to work with.. how about $v = \pmatrix{1\\0\\0}$

Find $Av,A^2v, A^3v.$ Since you are working in $\mathbb R^3$ and you now have 4 vectors, There is some non-trivial linear combination $c_0 v+ c_1 Av+ c_2 A^2v+c_3A^3v = 0$

and $(c_0 I+ c_1 A+ c_2 A^2+c_3A^3)v = 0$

If $v$ is a (generalized) eigenvector, you may short-circuit. If not, $c_0 I+ c_1 A+ c_2 A^2+c_3A^3 = 0$

$Av = \pmatrix{0\\0\\1}, A^2v = \pmatrix{0\\1\\-1},A^3v = \pmatrix{1\\-1\\3}$

$A^3v + A^2v-2Av - Iv = 0$

$A^3 + A^2-2A - I = 0$

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$A=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 2 & -1 \end{pmatrix}\quad \det(A)=1\times \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}=1\quad \operatorname{tr}(A)=0+0-1=-1$

$A^2=\begin{pmatrix} 0 & 0 & 1 \\ 1 & 2 & -1 \\ -1 & -1 & 3 \end{pmatrix} \quad\operatorname{tr}(A^2)=0+2+3=5$


$\chi_A(X)=X^3-\operatorname{tr}(A)X^2-\frac 12\left(\operatorname{tr}(A^2)-\operatorname{tr}(A)^2\right)X-\det(A)=X^3+X^2-2X-1$

By Cayley-Hamilton $\chi_A(A)=0\iff A^3=-A^2+2A+I$

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  • $\begingroup$ Note for the lazy: calculating only the diagonal elements of $A^2$ is sufficient. $\endgroup$ – zwim Jul 12 '17 at 0:04

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