Question

Let P and Q be two square matrices such that PQ = I, where I is an identity matrix. Then zero is an eigen value of

• A. P but not Q
• B. Q but not P
• C. Both P and Q
• D. Neither P nor Q

Hint:

Remember the fundamental connection: A matrix \(M\) is singular \(\iff\) \(\det(M) = 0\) \(\iff\) \(M\) is not invertible \(\iff\) \(\lambda = 0\) is an eigenvalue of \(M\). If two matrices multiply to the identity matrix, they are both invertible, so their determinants are non-zero.

Answer 1

The Correct Option is D

Step 1: Concept Overview:
This problem explores the relationship between eigenvalues, matrix invertibility, and determinants. A square matrix has a zero eigenvalue if and only if it's singular, meaning its determinant is zero.

Step 2: Core Principles:
1. \(PQ = I\) signifies that \(P\) and \(Q\) are inverses of each other; hence, both are invertible.
2. A matrix has a zero eigenvalue (\(\lambda = 0\)) if and only if its determinant is zero.
3. Apply the determinant property: \(\det(AB) = \det(A)\det(B)\).

Step 3: Detailed Solution:
Given the matrix relationship: \[ PQ = I \] Taking the determinant of both sides: \[ \det(PQ) = \det(I) \] Using the property \(\det(PQ) = \det(P)\det(Q)\) and knowing that \(\det(I) = 1\): \[ \det(P) . \det(Q) = 1 \] This implies that neither \(\det(P)\) nor \(\det(Q)\) can be zero, because their product must equal 1. \[ \det(P) eq 0 \quad \text{and} \quad \det(Q) eq 0 \] Since a matrix has a zero eigenvalue if and only if its determinant is zero, and since \(\det(P)\) and \(\det(Q)\) are non-zero, neither \(P\) nor \(Q\) can have a zero eigenvalue.

Step 4: Conclusion:
Therefore, zero is not an eigenvalue of either \(P\) or \(Q\).