The columns of $P$ are the coordinates of the vectors $\mathbf f_j$, so they form a basis exactly when
$$
\boxed{\det P\ne0}.
$$
For a <linear transformation> $\alpha$, define its <matrix representation> in the standard basis by
$$
[\alpha(\mathbf v)]_{\mathbf e}=A[\mathbf v]_{\mathbf e},
\qquad A_{ij}=[\alpha(\mathbf e_j)]_i.
$$
The <change-of-basis matrix> satisfies $[\mathbf v]_{\mathbf e}=P[\mathbf v]_{\mathbf f}$. Therefore
$$
P[\alpha(\mathbf v)]_{\mathbf f}
=A P[\mathbf v]_{\mathbf f},
$$
and multiplication by $P^{-1}$ gives the <similar matrix> relation
$$
\boxed{\widetilde A=P^{-1}AP}.
$$

The characteristic polynomial of the displayed $A$ is
$$
(\lambda-1)(\lambda-2)^2.
$$
Eigenvectors $(2,-1,2)^T$, $(1,0,0)^T$, and $(0,-1,3)^T$ give
$$
P=\begin{pmatrix}2&1&0\\-1&0&-1\\2&0&3\end{pmatrix},
\qquad
B=\begin{pmatrix}1&0&0\\0&2&0\\0&0&2\end{pmatrix}.
$$
Indeed, $\det P=1$ and
$$
AP=\begin{pmatrix}2&2&0\\-1&0&-2\\2&0&6\end{pmatrix}=PB,
$$
so $P^{-1}AP=B$. Finally, the <diagonalization of a matrix> gives
$$
A^nP=PB^n
=\boxed{\begin{pmatrix}
2&2^n&0\\
-1&0&-2^n\\
2&0&3\,2^n
\end{pmatrix}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 1 / 6B / Solution

A diagonalization is a factorization $A=PDP^{-1}$ in which $D$ is diagonal and the columns of $P$ form an eigenvector basis.


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