Question

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 ____.

• A. \(\{1,2,3\}\)
• B. \(\{1\}\)
• C. \(\{1,2\}\)
• D. \(\{2\}\)

Hint:

If a matrix satisfies a polynomial equation, every eigenvalue of the matrix must satisfy that same scalar polynomial equation.

Answer 1

The Correct Option is C

Step 1: Set up the eigenvalue equation.
If \(P\) satisfies a polynomial equation \(p(P)=0\), then every eigenvalue \(\lambda\) of \(P\) must satisfy the same polynomial equation \(p(\lambda)=0\). This is because if \(Pv=\lambda v\) for an eigenvector \(v\), then \(P^k v=\lambda^k v\) for every power \(k\), so applying the whole polynomial to \(P\) and acting on \(v\) gives \(p(P)v=p(\lambda)v=0\), and since \(v\neq 0\), we need \(p(\lambda)=0\).
Here the given equation is
\[ P^3-4P^2+5P-2I=0 \] so every eigenvalue of \(P\) satisfies
\[ \lambda^3-4\lambda^2+5\lambda-2=0 \]

Step 2: Solve the cubic for \(\lambda\).
Test \(\lambda=1\):
\[ 1-4+5-2=0 \] so \(\lambda=1\) is a root. Divide the cubic by \((\lambda-1)\):
\[ \lambda^3-4\lambda^2+5\lambda-2=(\lambda-1)(\lambda^2-3\lambda+2) \] Factor the quadratic:
\[ \lambda^2-3\lambda+2=(\lambda-1)(\lambda-2) \] So the full factorisation is
\[ (\lambda-1)^2(\lambda-2)=0 \] giving roots \(\lambda=1\) (repeated) and \(\lambda=2\).

Step 3: Read off the possible eigenvalues.
Since \(P\) is a \(3\times3\) matrix, it has \(3\) eigenvalues (counted with multiplicity), and each of them must be a root of \(\lambda^3-4\lambda^2+5\lambda-2=0\), so each eigenvalue is either \(1\) or \(2\). The three eigenvalues could be any mix, such as \((1,1,1)\), \((1,1,2)\), \((1,2,2)\) or \((2,2,2)\); all are consistent with the given equation, since the question does not pin down the exact multiplicities. So the set of values an eigenvalue can take is \(\{1,2\}\).

Step 4: Rule out the other options.

(A) \(\{1,2,3\}\): The value \(3\) is not a root of the cubic, so \(3\) can never be an eigenvalue of \(P\). Incorrect.

(B) \(\{1\}\): This misses the valid root \(\lambda=2\), which the equation clearly allows. Incorrect.

(D) \(\{2\}\): This misses the valid repeated root \(\lambda=1\). Incorrect.

Final Answer:
The set of all possible eigenvalues of \(P\) is
\[ \boxed{\{1,2\}} \]