Let p(t)=t^5-6t^4+6t^3-6t^2+5t.a) Factorise p(t) as a product of degree 1 polynomials.

The given polynomial is
${\displaystyle p\left(t\right)=t^5-6t^4+6t^3-6t^2+5t}$
We have to find the roots of this polynomial to write it as product of degree 1 polynomials.
a) Finding the roots
${\displaystyle p\left(t\right)=t^5-6t^4+6t^3-6t^2+5t}$
${\displaystyle p\left(t\right)=t(t^4-6t^3+6t^2-6t+5)}$
As t=1 also satisfies the polynomial so dividing the rest of the polynomial by t-1, we get
${\displaystyle p\left(t\right)=t(t-1)(t^3-5t^2+t-5)}$
Now t=5 satisfies the rest of the polynomial so dividing the cubic polynomial by t-5, we get
${\displaystyle p\left(t\right)=t(t-1)(t-5)(t^2+1)}$
Now as we know that
${\displaystyle t^2+1=0}$
${\displaystyle \Rightarrow t=i,-i}$
Hence, the given polynomial can be written as
${\displaystyle p\left(t\right)=t(t-1)(t-5)(t+i)(t-i)}$
b) We have to find a matrix A such that characteristic of matrix A and the given polynomial p(t) are equal.
That is,
${\displaystyle P_A\left(t\right)=p\left(t\right)}$
As we know the roots of characteristic polynomial are eigen values of the matrix .
Hence the required matrix A must have eigen values 0,1,5 ,i and −i.
Also we know that eigen values of diagonal matrix are diagonal entries it self.
Therefore, the required matrix A can be given as
$A=\left[\begin{array}{ccccc}0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0\\ 0& 0& 5& 0& 0\\ 0& 0& 0& i& 0\\ 0& 0& 0& 0& -i\end{array}\right]$
c) We have to give an example of 3 degree polynomial with real coefficients and two imaginary roots.
The polynomial q(t) can be taken as
${\displaystyle q\left(t\right)=(t^3-5t^2+t-5)}$
It has three roots such that
${\displaystyle q\left(t\right)=(t-5)(t+1)(t-i)}$
where two of the roots are imaginary.
Now the required matrix whose characteristic polynomial is equal to the polynomial q(t) can be given as
$B=\left[\begin{array}{ccc}5& 0& 0\\ 0& i& 0\\ 0& 0& -i\end{array}\right]$
Clearly, ${\displaystyle P_B\left(t\right)=q\left(t\right)}$