It is well-known if $\lambda$ is an eigen value of a square matrix $A$ of order $n\times n$ then $\lambda^k$ will be eigen value of $A^k$ for every positive integer $k$. Also, if $f(x)$ be a polynomial in $x$, then $f(\lambda)$ is an eigen value of the matrix $f(A)$.
Suppose the characteristic equation of $A$ is $$\lambda^3+3\lambda^2+2\lambda+1=0.............(1)$$ I was willing to find the characteristic equation of (i) $A^2$, (ii) $A^3,$ and (iii) $A^3+A^2+A+I$.
I managed to find the results for (i) and (ii) by using the transformation $y=x^2, x^3$ respectively in Eq(1). However I got stuck in (iii) to find any useful method. What I tried is the following:
Let $f(t)=t^3+t^2+t+1$. Then eigen value of $A^3+A^2+A+I_3=f(A)$ are $f(\lambda)$ if $\lambda$ be the eigen values of $A$. Here $AX=\lambda X$ for some non-null column matrix $X$. However from here, I am getting no direction. Is there any adequate technique to obtain the solution? Please help.
Let $\ B=A^3+A^2+A+I\ $, and for each natural number $\ n\ $ let $\ r_n(x)\ $ be the remainder of dividing $\ (x^3+x^2+x+1)^n\ $ by $\ x^3+3x^2+2x+1\ $. Since $\ A^3+3A^2+2A+I=0\ $, then $$ B^n=r_n(A) $$ for all $\ n\ $. Since each $\ r_n(x)\ $ has degree at most $\ 2\ $, the four polynomials $ 1,r_1(x),r_2(x), r_3(x)\ $ span a space of dimension at most $\ 3\ $ over the rational numbers. Therefore, there exist four constants, $\ b_0, b_1,b_2,b_3\ $, not all zero, such that $$ b_3r_3(x)+b_2r_2(x)+b_1r_1(x)+b_0=0\ , $$ which you can find by equating the coefficients of $\ x^0,x\ $ and $\ x^2\ $ in the polynomial on the left of the above equation to zero and solving the resulting linear equations. As it happens, the solution space of these equations has dimension $\ 1\ $, and is spanned by a vector with $\ b_3\ne0\ $. You then have $$ B^3+\frac{b_2}{b_3}B^2+\frac{b_1}{b_3}B+\frac{b_1}{b_3}I=0\ , $$ and so $$ x^3+\frac{b_2}{b_3}x^2+\frac{b_1}{b_3}x+\frac{b_1}{b_3} $$ must be the characteristic polynomial of $\ B\ $.