The given polynomial is

\(\displaystyle{p}{\left({t}\right)}={t}^{{5}}-{6}{t}^{{4}}+{6}{t}^{{3}}-{6}{t}^{{2}}+{5}{t}\)

We have to find the roots of this polynomial to write it as product of degree 1 polynomials.

a) Finding the roots

\(\displaystyle{p}{\left({t}\right)}={t}^{{5}}-{6}{t}^{{4}}+{6}{t}^{{3}}-{6}{t}^{{2}}+{5}{t}\)

\(\displaystyle{p}{\left({t}\right)}={t}{\left({t}^{{4}}-{6}{t}^{{3}}+{6}{t}^{{2}}-{6}{t}+{5}\right)}\)

As t=1 also satisfies the polynomial so dividing the rest of the polynomial by t-1, we get

\(\displaystyle{p}{\left({t}\right)}={t}{\left({t}-{1}\right)}{\left({t}^{{3}}-{5}{t}^{{2}}+{t}-{5}\right)}\)

Now t=5 satisfies the rest of the polynomial so dividing the cubic polynomial by t-5, we get

\(\displaystyle{p}{\left({t}\right)}={t}{\left({t}-{1}\right)}{\left({t}-{5}\right)}{\left({t}^{{2}}+{1}\right)}\)

Now as we know that

\(\displaystyle{t}^{{2}}+{1}={0}\)

\(\displaystyle\Rightarrow{t}={i},-{i}\)

Hence, the given polynomial can be written as

\(\displaystyle{p}{\left({t}\right)}={t}{\left({t}-{1}\right)}{\left({t}-{5}\right)}{\left({t}+{i}\right)}{\left({t}-{i}\right)}\)

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

\(A=\begin{bmatrix}0&0&0&0&0\\0&1&0&0&0\\0&0&5&0&0\\0&0&0&i&0\\0&0&0&0&-i\end{bmatrix}\)

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

\(\displaystyle{q}{\left({t}\right)}={\left({t}^{{3}}-{5}{t}^{{2}}+{t}-{5}\right)}\)

It has three roots such that

\(\displaystyle{q}{\left({t}\right)}={\left({t}-{5}\right)}{\left({t}+{1}\right)}{\left({t}-{i}\right)}\)

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

\(B=\begin{bmatrix}5&0&0\\0&i&0\\0&0&-i\end{bmatrix}\)

Clearly, \(\displaystyle{P}_{{B}}{\left({t}\right)}={q}{\left({t}\right)}\)