Consider P=pmatrix1&2\\0&1pmatrix, Q=pmatrix1&0\\0&1pmatrix. Which statement(s) is/are true?

GATE 2024 PH – Question 43

Mathematical Physics · matrices: similarity transformations, diagonalization, eigenvalues and eigen vectors · 2 marks · Multiple select

Consider $$P=\begin{pmatrix}1&2\\0&1\end{pmatrix},\quad Q=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$ Which statement(s) is/are true?

P and Q have same set of eigenvalues
P and Q commute with each other
P and Q have different sets of linearly independent eigenvectors
P is diagonalizable

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Correct answer: (A) P and Q have same set of eigenvalues; (B) P and Q commute with each other; (C) P and Q have different sets of linearly independent eigenvectors

Explanation

$$P=\begin{pmatrix}1&2\\0&1\end{pmatrix},\qquad Q=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I .$$

**A. P and Q have the same set of eigenvalues.** True. P is triangular with diagonal entries 1, 1, so both eigenvalues are 1; Q = I also has 1, 1. ✓
**B. P and Q commute.** True. The identity commutes with every matrix. ✓
**C. P and Q have different sets of linearly independent eigenvectors.** True. For $P-I=\begin{pmatrix}0&2\\0&0\end{pmatrix}$ the only independent eigenvector is $(1,0)^T$; for $Q=I$ every vector is an eigenvector (two independent ones). ✓
**D. P is diagonalisable.** False. P is a Jordan block: it has a repeated eigenvalue but only one independent eigenvector, so it is defective.

Answer **A, B and C**.