A matrix \(P\in\mathbb{R}^{3\times3}\) satisfies \(P^3-4P^2+5P-2I=0\), where \(I\) is a \(3\times3\) identity matrix. The set of all possible eigenvalues of matrix \(P\) is ____.
Show Hint
If a matrix satisfies a polynomial equation, every eigenvalue of the matrix must satisfy that same scalar polynomial equation.
Step 1: Turn the matrix equation into a scalar equation.
The matrix $P$ obeys $P^3-4P^2+5P-2I=0$. Take any eigenvalue $\lambda$ of $P$ with eigenvector $v$, so $Pv=\lambda v$. Applying the whole polynomial to $v$ gives $(P^3-4P^2+5P-2I)v=(\lambda^3-4\lambda^2+5\lambda-2)v=0$, and since $v\neq 0$ we get the scalar condition
\[ \lambda^3-4\lambda^2+5\lambda-2=0 \]
Step 2: Find the roots by trial division.
Trying small integers, $\lambda=1$ gives $1-4+5-2=0$, so $(\lambda-1)$ is a factor. Dividing out,
\[ \lambda^3-4\lambda^2+5\lambda-2=(\lambda-1)(\lambda^2-3\lambda+2) \]
The quadratic factors as $(\lambda-1)(\lambda-2)$, since $1\times2=2$ and $1+2=3$ match its middle and last terms. So
\[ \lambda^3-4\lambda^2+5\lambda-2=(\lambda-1)^2(\lambda-2) \]
Step 3: Match the roots to the matrix size.
$P$ is $3\times3$, so it carries exactly $3$ eigenvalues counted with repeats, and each one must come from the set $\{1,2\}$ found above. Nothing in the problem fixes exactly how many times $1$ or $2$ repeats, only that no eigenvalue can be anything other than $1$ or $2$.
Step 4: Reject the wrong sets.
$\{1,2,3\}$ fails since $\lambda=3$ does not satisfy the cubic. $\{1\}$ and $\{2\}$ each drop one of the two genuine roots. Only $\{1,2\}$ lists exactly the values that can actually occur.
\[ \boxed{\{1,2\}} \]
