I apologize for asking a trivial question, probably. However, I am learning matrix polynomial and appreciate your pointers.
I have come across the following question (as I am preparing for an exam) and not sure how to proceed.
If a matrix $\mathbf{A} \in M_3\left(\mathbb{R}\right)$ has eigenvalues $\lambda_1 =1, \lambda_2=1$, and $\lambda_3 = 3$, then how to find a matrix polynomial of the form $r(\mathbf{A}) = r_2\mathbf{A}^2 + r_1 \mathbf{A} + r_0 \mathbf{I}$ such that $r(\mathbf{A}) = \mathbf{A}^4 + \mathbf{A}^{-1}$?
Thank you so much in advance
I'm not so sure this is trivial. The easiest way to solve it (IMO) is to take your last equation (which I'll write as $R=A^4 + A^{-1}$), multiply both sides by $A$, then substituting $R=r_2 A^2 + r_1 A + r_0 I$. This produces $$\eqalign{ AR &= A^5 + I\cr A(r_2 A^2 + r_1 A + r_0 I) &= A^5 + I\cr 0 &= A^5 - r_2 A^3 - r_1 A^2 - r_0 A + I.\cr }$$ Now use the eigenvalues you're given; the characteristic polynomial of $A$ is $(1-\lambda)^2(3-\lambda)$. The Cayley-Hamilton Theorem states that $$0 = (I - A)^2 (3-A).$$ That means that (I think) you're looking for a polynomial $P=-A^2+bA+{1\over3}\,I$ such that $$(I-A)^2(3I-A) \cdot \left(-A^2+bA+{1\over3}\,I\right) = A^5 - r_2 A^3 - r_1 A^2 - r_0 A + I.$$ (The coefficient of $A^2$ in $P$ has to be $-1$, to make the coefficient of $A^5$ on the LHS equal to $1$. Similarly, the coefficient of $I$ in $P$ must be ${1\over3}$ to make the coefficient of $I$ on the LHS equal to $1$.)
You're now choosing $b$ so that the coefficient of $A^4$ on the LHS is $0$. Using Maple, it is found that the coefficient of $A^4$ is $-5-b$, so $b=-5$. Now, multiply out the left-hand side to get the coefficients $r_2$, $r_1$, and $r_0$. (I get $r_2={55\over3}$, $r_1=-{101\over3}$, and $r_0={52\over3}$.)