The matrix \(\mathbf{M}\) is upper triangular, so its eigenvalues are the diagonal entries \(2\), \(m\) and \(1\). Since \(m\neq 0,1,2\), these eigenvalues are distinct, so \(\mathbf{M}\) is diagonalisable.
We find eigenvectors for each eigenvalue.
For \(\lambda=2\):
\[
\mathbf{M}-2\mathbf{I}
=
\begin{pmatrix}
0 & m & 1\\
0 & m-2 & 7\\
0 & 0 & -1
\end{pmatrix}.
\]
Let \(\mathbf{x}=(x,y,z)^T\). Then \((\mathbf{M}-2\mathbf{I})\mathbf{x}=\mathbf{0}\) gives \(z=0\), then \((m-2)y+7z=0\), so \(y=0\). Hence \(x\) is free, and an eigenvector is
\[
\mathbf{v}_1=
\begin{pmatrix}
1\\
0\\
0
\end{pmatrix}.
\]
For \(\lambda=m\):
\[
\mathbf{M}-m\mathbf{I}
=
\begin{pmatrix}
2-m & m & 1\\
0 & 0 & 7\\
0 & 0 & 1-m
\end{pmatrix}.
\]
From the last row, \((1-m)z=0\). Since \(m\neq 1\), we get \(z=0\). The first row becomes
\[
(2-m)x+my=0.
\]
A convenient choice is \(x=m\), \(y=m-2\). Thus an eigenvector is
\[
\mathbf{v}_2=
\begin{pmatrix}
m\\
m-2\\
0
\end{pmatrix}.
\]
For \(\lambda=1\):
\[
\mathbf{M}-\mathbf{I}
=
\begin{pmatrix}
1 & m & 1\\
0 & m-1 & 7\\
0 & 0 & 0
\end{pmatrix}.
\]
From \((m-1)y+7z=0\), choose \(z=m-1\). Then \(y=-7\). The first row gives
\[
x+m(-7)+(m-1)=0,
\]
so
\[
x=6m+1.
\]
Hence an eigenvector is
\[
\mathbf{v}_3=
\begin{pmatrix}
6m+1\\
-7\\
m-1
\end{pmatrix}.
\]
Therefore, we may take
\[
\mathbf{P}
=
\begin{pmatrix}
1 & m & 6m+1\\
0 & m-2 & -7\\
0 & 0 & m-1
\end{pmatrix},
\qquad
\mathbf{D}
=
\begin{pmatrix}
2 & 0 & 0\\
0 & m & 0\\
0 & 0 & 1
\end{pmatrix}.
\]
Then
\[
\mathbf{M}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}.
\]
Now
\[
\mathbf{M}^7
=
\left(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\right)^7
=
\mathbf{P}\mathbf{D}^7\mathbf{P}^{-1}.
\]
Multiplying by \(\mathbf{P}\) on the right gives
\[
\mathbf{M}^7\mathbf{P}
=
\mathbf{P}\mathbf{D}^7.
\]
Since
\[
\mathbf{D}^7
=
\operatorname{diag}(2^7,m^7,1^7)
=
\operatorname{diag}(128,m^7,1),
\]
we get
\[
\mathbf{M}^7\mathbf{P}
=
\mathbf{P}
\begin{pmatrix}
128 & 0 & 0\\
0 & m^7 & 0\\
0 & 0 & 1
\end{pmatrix}.
\]
This means that the first column of \(\mathbf{P}\) is multiplied by \(128\), the second column by \(m^7\), and the third column is unchanged.
