Consider the linear transformation $\ltdefn{Q}{P_4}{P_4}$ defined by \begin{align*} &\lt{Q}{a+bx+cx^2+dx^3+ex^4}\\ &=(-46a-22b+13c+5d+e)+(117a+57b-32c-15d-4e) x+\\ &\quad\quad (-69a-29b+21c-7e)x^2+(159a+73b-44c-13d+2e)x^3+\\ &\quad\quad (-195a-87b+55c+10d-13e)x^4 \end{align*}
Choose a simple basis to compute with, say \begin{equation*} B=\set{1,\,x,\,x^2,\,x^3,\,x^4} \end{equation*}
Then it should be apparent that the matrix representation of $Q$ relative to $B$ is \begin{equation*} M=\matrixrep{Q}{B}{B}= \begin{bmatrix} -46 & -22 & 13 & 5 & 1 \\ 117 & 57 & -32 & -15 & -4 \\ -69 & -29 & 21 & 0 & -7 \\ 159 & 73 & -44 & -13 & 2 \\ -195 & -87 & 55 & 10 & -13 \end{bmatrix} \end{equation*}
Compute the characteristic polynomial, eigenvalues and eigenvectors according to the techniques of Section EE, \begin{align*} \charpoly{Q}{x} &=-x^5+6 x^4-x^3-88 x^2+252 x-208\\ &=-(x-2)^2 (x+4) \left(x^2-6x+13\right)\\ &=-(x-2)^2 (x+4) \left(x-(3+2i)\right) \left(x-(3-2i)\right)\\ \end{align*} \begin{align*} \algmult{Q}{2}&=2 & \algmult{Q}{-4}&=1 & \algmult{Q}{3+2i}&=1 & \algmult{Q}{3-2i}&=1 \end{align*} \begin{align*} \lambda&=2\\ M-(2)I_5&= \begin{bmatrix} -48 & -22 & 13 & 5 & 1 \\ 117 & 55 & -32 & -15 & -4 \\ -69 & -29 & 19 & 0 & -7 \\ 159 & 73 & -44 & -15 & 2 \\ -195 & -87 & 55 & 10 & -15 \end{bmatrix} \rref \begin{bmatrix} 1 & 0 & 0 & \frac{1}{2} & -\frac{1}{2} \\ 0 & 1 & 0 & -\frac{5}{2} & -\frac{5}{2} \\ 0 & 0 & 1 & -2 & -6 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\\ \eigenspace{M}{2}&=\nsp{M-(2)I_5} =\spn{\set{ \colvector{-\frac{1}{2}\\\frac{5}{2}\\2\\1\\0},\, \colvector{\frac{1}{2}\\\frac{5}{2}\\6\\0\\1} }} =\spn{\set{ \colvector{-1\\5\\4\\2\\0},\, \colvector{1\\5\\12\\0\\2} }} \end{align*} \begin{align*} \lambda&=-4\\ M-(-4)I_5&= \begin{bmatrix} -42 & -22 & 13 & 5 & 1 \\ 117 & 61 & -32 & -15 & -4 \\ -69 & -29 & 25 & 0 & -7 \\ 159 & 73 & -44 & -9 & 2 \\ -195 & -87 & 55 & 10 & -9 \end{bmatrix} \rref \begin{bmatrix} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & -3 \\ 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\\ \eigenspace{M}{-4}&=\nsp{M-(-4)I_5} =\spn{\set{\colvector{-1\\3\\1\\2\\1}}} \end{align*} \begin{align*} \lambda&=3+2i\\ M-(3+2i)I_5&= \begin{bmatrix} -49-2 i & -22 & 13 & 5 & 1 \\ 117 & 54-2 i & -32 & -15 & -4\\ -69 & -29 & 18-2 i & 0 & -7 \\ 159 & 73 & -44 & -16-2 i & 2 \\ -195 & -87 & 55 & 10 & -16-2 i \end{bmatrix}\\ &\quad\quad\rref \begin{bmatrix} 1 & 0 & 0 & 0 & -\frac{3}{4}+\frac{i}{4} \\ 0 & 1 & 0 & 0 & \frac{7}{4}-\frac{i}{4} \\ 0 & 0 & 1 & 0 & -\frac{1}{2}+\frac{i}{2} \\ 0 & 0 & 0 & 1 & \frac{7}{4}-\frac{i}{4} \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\\ \eigenspace{M}{3+2i}&=\nsp{M-(3+2i)I_5} =\spn{\set{\colvector{\frac{3}{4}-\frac{i}{4} \\ -\frac{7}{4}+\frac{i}{4} \\ \frac{1}{2}-\frac{i}{2} \\ -\frac{7}{4}+\frac{i}{4} \\ 1}}} =\spn{\set{\colvector{3-i\\-7+i\\2-2i\\-7+i\\4}}} \end{align*} \begin{align*} \lambda&=3-2i\\ M-(3-2i)I_5&= \begin{bmatrix} -49+2 i & -22 & 13 & 5 & 1 \\ 117 & 54+2 i & -32 & -15 & -4 \\ -69 & -29 & 18+2 i & 0 & -7 \\ 159 & 73 & -44 & -16+2 i & 2 \\ -195 & -87 & 55 & 10 & -16+2 i \end{bmatrix}\\ &\quad\quad\rref \begin{bmatrix} 1 & 0 & 0 & 0 & -\frac{3}{4}-\frac{i}{4} \\ 0 & 1 & 0 & 0 & \frac{7}{4}+\frac{i}{4} \\ 0 & 0 & 1 & 0 & -\frac{1}{2}-\frac{i}{2} \\ 0 & 0 & 0 & 1 & \frac{7}{4}+\frac{i}{4} \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\\ \eigenspace{M}{3-2i}&=\nsp{M-(3-2i)I_5} =\spn{\set{\colvector{\frac{3}{4}+\frac{i}{4} \\ -\frac{7}{4}-\frac{i}{4} \\ \frac{1}{2}+\frac{i}{2} \\ -\frac{7}{4}-\frac{i}{4} \\ 1}}} =\spn{\set{\colvector{3+i\\-7-i\\2+2i\\-7-i\\4}}} \end{align*}
It is straightforward to convert each of these basis vectors for eigenspaces of $M$ back to elements of $P_4$ by applying the isomorphism $\vectrepinvname{B}$, \begin{align*} \vectrepinv{B}{\colvector{-1\\5\\4\\2\\0}}&=-1+5x+4x^2+2x^3\\ \vectrepinv{B}{\colvector{1\\5\\12\\0\\2}}&=1+5x+12x^2+2x^4\\ \vectrepinv{B}{\colvector{-1\\3\\1\\2\\1}}&=-1+3x+x^2+2x^3+x^4\\ \vectrepinv{B}{\colvector{3-i\\-7+i\\2-2i\\-7+i\\4}}&=(3-i)+(-7+i)x+(2-2i)x^2+(-7+i)x^3+4x^4\\ \vectrepinv{B}{\colvector{3+i\\-7-i\\2+2i\\-7-i\\4}}&=(3+i)+(-7-i)x+(2+2i)x^2+(-7-i)x^3+4x^4\\ \end{align*}
So we apply Theorem EER and the Coordinatization Principle to get the eigenspaces for $Q$, \begin{align*} \eigenspace{Q}{2}&=\spn{\set{-1+5x+4x^2+2x^3,\,1+5x+12x^2+2x^4}}\\ \eigenspace{Q}{-4}&=\spn{\set{-1+3x+x^2+2x^3+x^4}}\\ \eigenspace{Q}{3+2i}&=\spn{\set{(3-i)+(-7+i)x+(2-2i)x^2+(-7+i)x^3+4x^4}}\\ \eigenspace{Q}{3-2i}&=\spn{\set{(3+i)+(-7-i)x+(2+2i)x^2+(-7-i)x^3+4x^4}} \end{align*} with geometric multiplicities \begin{align*} \geomult{Q}{2}&=2 & \geomult{Q}{-4}&=1 & \geomult{Q}{3+2i}&=1 & \geomult{Q}{3-2i}&=1 \end{align*}