Discussion
Let \(T : \mathbb{R}^{4} \to \mathbb{R}^{4}\) be a linear transformation and the null space of T be the subspace of \(\mathbb{R}^{4}\) given by
\[ \begin{aligned} \{(x_{1},x_{2},x_{3},x_{4}) &\in \mathbb{R}^{4} : \\ 4x_{1}+3x_{2}+2x_{3}+x_{4} &= 0\}. \end{aligned} \]
If \(Rank(T - 3I) = 3\), where I is the identity map on \(\mathbb{R}^{4}\), then the minimal polynomial of T is
Answer: A
The minimal polynomial is x(x - 3).
The null space of T is a hyperplane in \(\mathbb{R}^{4}\), of dimension 3, so 0 is an eigenvalue with geometric multiplicity 3. Rank(T - 3I) = 3 means T - 3I has nullity 1, so 3 is an eigenvalue with geometric multiplicity 1.
The algebraic multiplicities are at least the geometric ones and add up to 4, so they are exactly 3 and 1. Each eigenvalue then has as many independent eigenvectors as its multiplicity, so T is diagonalizable with eigenvalues 0, 0, 0, 3, and its minimal polynomial has only simple factors: x(x - 3).
The options with higher powers would describe non-diagonalizable maps, which would need fewer eigenvectors than the null space supplies.
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