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

Another way to see this is by directly bounding the degree of the characteristic equation and then testing a boundary case.

Step 1: Degree bound. For an \(n \times n\) matrix \(A\), the characteristic polynomial \(p(x) = \det(xI - A)\) is a monic polynomial of degree exactly \(n\) in \(x\). By the fundamental theorem of algebra, \(p(x)\) factors (over \(\mathbb{C}\)) into exactly \(n\) linear factors, counted with multiplicity. Hence no root (eigenvalue) can repeat more than \(n\) times - this rules out any answer larger than \(n\), eliminating the option "\(n+1\)".

Step 2: Rule out that the bound is never attained. If the true maximum were less than \(n\) (say \(n-1\) or \(1\)), there would need to be some structural reason every real matrix avoids a fully repeated eigenvalue. But diagonal (or scalar) real matrices exist freely, so no such restriction applies.

Step 3: Exhibit a tight example. Take \(A = 5I_n\) (or even the zero matrix, \(c = 0\)). Its only eigenvalue is \(5\) (or \(0\)), with multiplicity \(n\), since \(\det(xI - A) = (x-5)^n\).

Step 4: Conclude. The upper bound \(n\) from Step 1 is achieved in Step 3, so the maximum possible multiplicity equals \(n\) exactly, for every \(n > 1\).

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