GATE PH 2007 • Q1
MCQ • 1 Mark
Section 1: Mathematical Physics
Linear Algebra & Vector Calculus
Subtopic: Matrices, Eigenvalues & Eigenvectors
Question Statement:
The eigenvalues of a matrix are $i$, $-2i$ and $3i$. The matrix is
Options:
-
(A) unitary
-
(B) anti-unitary
-
(C) Hermitian
-
(D) anti-Hermitian
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2007 Q.1)**
**1. Theoretical Formulation & Physics Principles**:
- **Field & Concept**: *Section 1: Mathematical Physics* (`Matrices, Eigenvalues & Eigenvectors`)
- **Governing Principle**:
- A matrix $A$ is **Hermitian** if $A^\dagger = A$, and its eigenvalues are strictly real.
- A matrix $A$ is **skew-Hermitian (anti-Hermitian)** if $A^\dagger = -A$. Let $A ec{v} = \lambda ec{v}$. Taking the Hermitian conjugate:
$ ec{v}^\dagger A^\dagger = \lambda^* ec{v}^\dagger \implies - ec{v}^\dagger A ec{v} = \lambda^* ec{v}^\dagger ec{v} \implies -\lambda = \lambda^* \implies ext{Re}(\lambda) = 0$
Hence, the eigenvalues of an anti-Hermitian matrix are purely imaginary or zero.
- A matrix $A$ is **unitary** if $A^\dagger A = I$, which requires $|\lambda| = 1$ for all eigenvalues. Here $|-2i| = 2
eq 1$ and $|3i| = 3
eq 1$, so the matrix is not unitary.
**2. Detailed Step-by-Step Derivation & Evaluation**:
- Given eigenvalues: $\lambda_1 = i$, $\lambda_2 = -2i$, $\lambda_3 = 3i$.
- All three eigenvalues are purely imaginary numbers ($ ext{Re}(\lambda_i) = 0$).
- Therefore, the matrix is strictly **anti-Hermitian**.
**3. Official Verification & Conclusion**:
- **Official Master Answer**: **Option (D)** (anti-Hermitian).
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