Answer: (a) The characteristic equation is \(\lambda^{3}-12\lambda^{2}+12\lambda+80=0\).
The eigenvalues are \(\lambda=-2,\ 4,\ 10\).
(b) \(\mathbf{A}^{4}=132\mathbf{A}^{2}-224\mathbf{A}-960\mathbf{I}\), so \(p=132\), \(q=-224\), \(r=-960\).
(c) One suitable choice is
\(\mathbf{P}=\begin{pmatrix}2&0&0\\1&3&3\\-1&-1&1\end{pmatrix}\), \(\mathbf{D}=\begin{pmatrix}625&0&0\\0&1&0\\0&0&2401\end{pmatrix}\).
Then \((\mathbf{A}-3\mathbf{I})^{4}=\mathbf{PDP}^{-1}\).
(a) The characteristic equation comes from
\(\det(\mathbf{A}-\lambda\mathbf{I})=0\).
So
\(\det\begin{pmatrix}-2-\lambda&0&0\\0&7-\lambda&9\\4&1&7-\lambda\end{pmatrix}=0\).
Expanding along the first row,
\((-2-\lambda)\det\begin{pmatrix}7-\lambda&9\\1&7-\lambda\end{pmatrix}=0\).
Hence
\((-2-\lambda)\big((7-\lambda)^2-9\big)=0\).
Now
\((7-\lambda)^2-9=49-14\lambda+\lambda^2-9=\lambda^2-14\lambda+40\).
Therefore
\((-2-\lambda)(\lambda^2-14\lambda+40)=0\).
Expanding,
\(-\lambda^3+12\lambda^2-12\lambda-80=0\),
so equivalently
\(\lambda^3-12\lambda^2+12\lambda+80=0\).
Also,
\(\lambda^3-12\lambda^2+12\lambda+80=(\lambda+2)(\lambda^2-14\lambda+40)=(\lambda+2)(\lambda-4)(\lambda-10)\).
So the eigenvalues are
\(\lambda=-2,\ 4,\ 10\).
(b) Since \(\mathbf{A}\) satisfies its characteristic equation,
\(\mathbf{A}^3-12\mathbf{A}^2+12\mathbf{A}+80\mathbf{I}=\mathbf{0}\).
Multiplying by \(\mathbf{A}\),
\(\mathbf{A}^4-12\mathbf{A}^3+12\mathbf{A}^2+80\mathbf{A}=\mathbf{0}\).
So
\(\mathbf{A}^4=12\mathbf{A}^3-12\mathbf{A}^2-80\mathbf{A}\).
From the cubic relation,
\(\mathbf{A}^3=12\mathbf{A}^2-12\mathbf{A}-80\mathbf{I}\).
Substitute this into the expression for \(\mathbf{A}^4\):
\(\mathbf{A}^4=12(12\mathbf{A}^2-12\mathbf{A}-80\mathbf{I})-12\mathbf{A}^2-80\mathbf{A}\).
Simplifying gives
\(\mathbf{A}^4=132\mathbf{A}^2-224\mathbf{A}-960\mathbf{I}\).
Hence \(p=132\), \(q=-224\), \(r=-960\).
(c) We find eigenvectors corresponding to the eigenvalues found in part (a).
For \(\lambda=-2\):
\(\mathbf{A}+2\mathbf{I}=\begin{pmatrix}0&0&0\\0&9&9\\4&1&9\end{pmatrix}\).
Let the eigenvector be \(\begin{pmatrix}x\\y\\z\end{pmatrix}\). Then
\(9y+9z=0\Rightarrow y=-z\),
\(4x+y+9z=0\Rightarrow 4x+8z=0\Rightarrow x=-2z\).
Taking \(z=-1\), one eigenvector is
\(\mathbf{v}_{-2}=\begin{pmatrix}2\\1\\-1\end{pmatrix}\).
For \(\lambda=4\):
\(\mathbf{A}-4\mathbf{I}=\begin{pmatrix}-6&0&0\\0&3&9\\4&1&3\end{pmatrix}\).
Then
\(-6x=0\Rightarrow x=0\),
\(3y+9z=0\Rightarrow y=-3z\).
Taking \(z=-1\), one eigenvector is
\(\mathbf{v}_{4}=\begin{pmatrix}0\\3\\-1\end{pmatrix}\).
For \(\lambda=10\):
\(\mathbf{A}-10\mathbf{I}=\begin{pmatrix}-12&0&0\\0&-3&9\\4&1&-3\end{pmatrix}\).
Then
\(-12x=0\Rightarrow x=0\),
\(-3y+9z=0\Rightarrow y=3z\).
Taking \(z=1\), one eigenvector is
\(\mathbf{v}_{10}=\begin{pmatrix}0\\3\\1\end{pmatrix}\).
Thus we may take
\(\mathbf{P}=\begin{pmatrix}2&0&0\\1&3&3\\-1&-1&1\end{pmatrix}\),
with columns \(\mathbf{v}_{-2},\mathbf{v}_4,\mathbf{v}_{10}\).
Then
\(\mathbf{A}=\mathbf{P}\begin{pmatrix}-2&0&0\\0&4&0\\0&0&10\end{pmatrix}\mathbf{P}^{-1}\).
So
\(\mathbf{A}-3\mathbf{I}=\mathbf{P}\begin{pmatrix}-5&0&0\\0&1&0\\0&0&7\end{pmatrix}\mathbf{P}^{-1}\).
Raising to the fourth power,
\((\mathbf{A}-3\mathbf{I})^4=\mathbf{P}\begin{pmatrix}(-5)^4&0&0\\0&1^4&0\\0&0&7^4\end{pmatrix}\mathbf{P}^{-1}\).
Hence
\(\mathbf{D}=\begin{pmatrix}625&0&0\\0&1&0\\0&0&2401\end{pmatrix}\),
and therefore
\((\mathbf{A}-3\mathbf{I})^4=\mathbf{PDP}^{-1}\).
