Question

For \(n > 1\), the maximum multiplicity of any eigenvalue of an \(n \times n\) matrix with elements from \(\mathbb{R}\) is

• A. \(n\)
• B. \(n - 1\)
• C. \(1\)
• D. \(n + 1\)

Hint:

The characteristic polynomial of an \(n \times n\) matrix has degree \(n\), and a scalar matrix like \(cI_n\) gives one eigenvalue with multiplicity exactly \(n\).

Answer 1

The Correct Option is A

Step 1: A useful way to see this without directly writing the characteristic polynomial is to think in terms of the Jordan canonical form of $A$ (over $\mathbb{C}$, which every real matrix also possesses after extending scalars).

Step 2: Every $n \times n$ matrix $A$ has a Jordan form made of Jordan blocks whose sizes add up to exactly $n$. If a matrix has a single eigenvalue $c$, the simplest case is one Jordan block of size $n$ for $c$, attained for instance when $A$ is diagonalizable with every eigenvalue equal to $c$, i.e. $A = cI_n$, which gives that eigenvalue an algebraic multiplicity of exactly $n$, using up the entire dimension of the matrix.

Step 3: Because the sum of all block sizes across all eigenvalues must equal $n$ exactly, this is just the dimension of the space the matrix acts on, a single eigenvalue can never claim more than $n$ of these dimensions, so $n$ is a hard ceiling on multiplicity.

Step 4: Since $A = cI_n$ is a genuine real matrix that reaches this ceiling, with $c$ appearing $n$ times, the ceiling of $n$ is not just a bound, it is actually reached. So the maximum multiplicity possible is $n$, ruling out $n-1$ (too small, since $n$ is achievable), $1$ (the multiplicity in the opposite extreme case of all-distinct eigenvalues, not the maximum), and $n+1$ (exceeds the degree of the characteristic polynomial and can never occur).

\[ \boxed{n} \]