Question

For 𝑛 > 1, the maximum multiplicity of any eigenvalue of an 𝑛× 𝑛 matrix with
elements from ℝ is

• A. 𝑛
• B. 𝑛−1
• C. 1
• D. 𝑛+ 1

Hint:

The characteristic polynomial of an nxn matrix has degree n, so no eigenvalue can repeat more than n times; a scalar matrix cI achieves multiplicity n exactly.

Answer 1

The Correct Option is A

We are asked for the largest possible algebraic multiplicity that any eigenvalue can have in an \( n \times n \) matrix with real entries, for \( n > 1 \).

Step 1: Recall what "multiplicity" means here.

The algebraic multiplicity of an eigenvalue \( \lambda \) is the number of times \( (x - \lambda) \) appears as a factor of the characteristic polynomial \( \det(A - xI) \). Since the characteristic polynomial of an \( n \times n \) matrix has degree exactly \( n \), any single eigenvalue's multiplicity can be at most \( n \) (it cannot exceed the total degree).

Step 2: Check whether multiplicity \( n \) is actually achievable.

Consider the scalar matrix \( A = cI_n \), where \( I_n \) is the \( n \times n \) identity matrix and \( c \in \mathbb{R} \) is any real constant. Its characteristic polynomial is

\[ \det(A - xI) = (c - x)^n \]

so \( \lambda = c \) is an eigenvalue with algebraic multiplicity exactly \( n \). This matrix has real entries, satisfying the problem's condition.

Step 3: Conclude.

Since the multiplicity can never exceed \( n \) (Step 1) and we exhibited a real matrix achieving multiplicity exactly \( n \) (Step 2), the maximum possible multiplicity of any eigenvalue is \( n \).

Final Answer: \( n \), option (A).