Questions: Characteristic Polynomial and Diagonalization

Question 1

A 3×3 matrix has characteristic polynomial (λ − 2)(λ − 3)². Which additional fact would guarantee the matrix is diagonalizable?

• A. The eigenspace for λ = 2 is 1-dimensional
• B. The eigenspace for λ = 3 is 2-dimensional
• C. The trace of the matrix equals 8
• D. The determinant of the matrix equals 36

Answer: (B)

Diagonalizability requires that for every eigenvalue, the geometric multiplicity (dimension of the eigenspace) equals the algebraic multiplicity (multiplicity as a root of the characteristic polynomial). Here λ = 2 has algebraic multiplicity 1, so its eigenspace is automatically 1-dimensional — no problem there. The concern is λ = 3 with algebraic multiplicity 2: if its eigenspace is only 1-dimensional, diagonalization fails. Confirming a 2-dimensional eigenspace for λ = 3 ensures geometric = algebraic multiplicity for every eigenvalue. Trace and determinant tell you sums and products of eigenvalues, but nothing about eigenspace dimensions.

Question 2

What does the constant term of the characteristic polynomial of an n×n matrix always equal?

• A. The trace of the matrix (sum of diagonal entries)
• B. The sum of all eigenvalues
• C. The determinant of the matrix
• D. The product of the diagonal entries only

Answer: (C)

The characteristic polynomial is p(λ) = det(A − λI). At λ = 0, this becomes det(A − 0) = det(A). So the constant term of p(λ) is always the determinant of A. This gives a useful check: since the constant term also equals the product of all eigenvalues (evaluated at λ = 0), the product of the eigenvalues equals det(A). Similarly, the coefficient of λⁿ⁻¹ is always −tr(A), connecting eigenvalue sums to the trace. These relationships hold even without solving for eigenvalues explicitly.

Question 3

Two matrices with the same characteristic polynomial should be similar to each other.

• T. True
• F. False

Answer: False

Similarity implies the same characteristic polynomial — that direction holds. But the converse fails: two matrices can share a characteristic polynomial without being similar. The classic example involves the 2×2 identity matrix I and any diagonalizable matrix with both eigenvalues equal to 1; they have the same characteristic polynomial (λ−1)² but are similar only to I itself. More starkly, [[1,1],[0,1]] and [[1,0],[0,1]] have the same characteristic polynomial (λ−1)² but are not similar (one is diagonalizable, the other is not). The characteristic polynomial is a similarity invariant, but not a complete invariant.

Question 4

A square matrix with n distinct eigenvalues is always diagonalizable.

• T. True
• F. False

Answer: True

Eigenvectors corresponding to distinct eigenvalues are always linearly independent. If an n×n matrix has n distinct eigenvalues, it has n linearly independent eigenvectors, which form a basis of ℝⁿ. The columns of P (the eigenvector matrix) are then a complete basis, P is invertible, and A = PDP⁻¹ holds. Note that the converse is false: a matrix can be diagonalizable even with repeated eigenvalues, as long as the geometric multiplicity matches the algebraic multiplicity for each repeated eigenvalue.

Question 5

What is the difference between algebraic multiplicity and geometric multiplicity of an eigenvalue, and why does their relationship determine whether a matrix is diagonalizable?

Model answer: The algebraic multiplicity of eigenvalue λ is its multiplicity as a root of the characteristic polynomial det(A − λI). The geometric multiplicity is the dimension of the corresponding eigenspace — the number of linearly independent eigenvectors for that eigenvalue. Geometric multiplicity ≤ algebraic multiplicity always. Diagonalization requires a complete basis of eigenvectors, so you need n total linearly independent eigenvectors. If geometric multiplicity is strictly less than algebraic multiplicity for any eigenvalue, there are not enough eigenvectors to span ℝⁿ, and diagonalization fails.

The matrix [[2,1],[0,2]] illustrates the failure: characteristic polynomial (λ−2)² gives algebraic multiplicity 2, but solving (A−2I)v = 0 yields only a 1-dimensional eigenspace (geometric multiplicity 1). One eigenvector short of a basis means no diagonalization is possible. The matrix is instead similar to a Jordan block — the next topic in the theory of canonical forms.