Diagonalization of a Matrix: A Six-Step Check for CSIR NET and GATE

Diagonalization of a matrix means writing a square matrix A as A = PDP⁻¹, where D is a diagonal matrix and the columns of P are eigenvectors of A. It works exactly when A has n linearly independent eigenvectors. The diagonal entries of D are the matching eigenvalues, in the same order as the columns of P.

Five terms carry the whole topic:

GM is always between 1 and AM. A matrix is diagonalizable over a field when every eigenvalue lies in that field and GM equals AM for every eigenvalue, which is the same as the eigenspace dimensions adding up to n (Wikipedia, Diagonalizable matrix; Statlect, Matrix diagonalization).

Both exams test this. The GATE 2027 Mathematics syllabus from IIT Madras lists "diagonalization", "minimal polynomial", "diagonalization by a unitary matrix" and "Jordan canonical form" under Linear Algebra. CSIR's syllabus for Parts B and C lists "canonical forms, diagonal forms, triangular forms, Jordan forms". Check the current exam notice for any change. For how to match a whole linear algebra course to your exam, see our linear algebra course guide.

Run step 1 first: read whether the question is over ℝ or ℂ. Then check symmetry, triangular form and distinct eigenvalues, and stop at the first step that decides. Steps 1 to 5 need no eigenvectors. Only step 6 needs a rank calculation, and only for eigenvalues that repeat.

Step 5 is the shortcut that saves time in an exam. The minimal polynomial is the lowest-degree monic polynomial p with p(A) = 0, and A is diagonalizable exactly when it has no repeated roots. Our Cayley Hamilton theorem page covers how it relates to the characteristic polynomial.

No. A repeated eigenvalue only means the matrix needs step 6. It is not diagonalizable only if some eigenvalue has GM smaller than AM.

Take A = [[0, −9], [1, −6]]. Its characteristic polynomial is (x + 3)², so λ = −3 has AM 2. Then A + 3I = [[3, −9], [1, −3]] has rank 1, so the nullity is 1. GM is 1, smaller than 2, so A is not diagonalizable.

A common slip here is to take the rank of A itself instead of the rank of A − λI. A has determinant 9, so its rank is 2, which tells you nothing about the eigenspace.

Now take B = [[3, 0, 0], [0, 3, 1], [0, 0, 5]]. The eigenvalue 3 also repeats, with AM 2. But B − 3I = [[0, 0, 0], [0, 0, 1], [0, 0, 2]] has rank 1, so the nullity is 2. GM equals AM, so B is diagonalizable.

For a 2×2 matrix the answer is simpler: a repeated eigenvalue λ gives a diagonalizable matrix only if the matrix is λI. For 3×3 and larger, cases like B appear, and exam setters use them.

Among the 19,683 3×3 matrices with entries −1, 0 or 1, 31.6% are diagonalizable over ℝ and 79.4% over ℂ. Among the 14,641 2×2 matrices with entries −5 to 5, the figures are 68.4% and 97.4%. We worked this out ourselves in November 2026; the script below is the test we applied to each matrix.

Method. We took every matrix in three families: 2×2 with entries from −5 to 5 (14,641 matrices), 3×3 with entries 0 or 1 (512), and 3×3 with entries −1, 0 or 1 (19,683). We classified each one with exact integer arithmetic. A matrix is diagonalizable over ℂ when the squarefree part of its characteristic polynomial (the polynomial with repeated factors removed) evaluates to the zero matrix at A. It is diagonalizable over ℝ when, in addition, every root is real. We cross-checked the 512-matrix family with a floating-point eigenvalue and rank routine, and the counts matched (320 over ℂ, 268 over ℝ).

In each row the four middle counts add up to the number of matrices.

Three things stand out:

1. Repeated eigenvalues are not a death sentence. Among matrices with a repeated eigenvalue, 11 of 387 (2.8%) were diagonalizable in the 2×2 family, 64 of 256 (25.0%) in the 3×3 {0, 1} family, and 573 of 4,629 (12.4%) in the 3×3 {−1, 0, 1} family. The 11 in the 2×2 family are exactly the scalar matrices λI.
2. The field changes the answer a lot. In the 3×3 {−1, 0, 1} family only 31.6% were diagonalizable over ℝ, against 79.4% over ℂ. The rotation matrix [[0, −1], [1, 0]] is the standard case: its eigenvalues are ±i, so it diagonalizes over ℂ and not over ℝ.
3. The counts describe these small integer families only. Exam questions are chosen, not drawn at random, so read the table as a picture of how the cases are distributed, not as exam odds.

Once A = PDP⁻¹, the power is Aᵏ = PDᵏP⁻¹, and Dᵏ just raises each diagonal entry to the kth power (Statlect).

Let A = [[4, 1], [2, 3]]. The trace is 7 and the determinant is 10, so the eigenvalues are 2 and 5, which are distinct, so step 4 already says A is diagonalizable. For λ = 2, the vector (−1, 2) works, because A(−1, 2) = (−2, 4). For λ = 5, the vector (1, 1) works, because A(1, 1) = (5, 5).

So P = [[−1, 1], [2, 1]], D = diag(2, 5) and P⁻¹ = [[−1/3, 1/3], [2/3, 1/3]]. Then D¹⁰ = diag(1024, 9765625), and

PD¹⁰ = [[−1024, 9765625], [2048, 9765625]].

Multiplying by P⁻¹ gives A¹⁰ = [[6510758, 3254867], [6509734, 3255891]].

Try each with the six steps before reading the answers. Exam papers ask this kind of question in the form "which of the following is diagonalizable?" (GATE Maths previous papers with solutions is where to practise the real format.)

1. M1 = [[3, 1, 0], [0, 3, 0], [0, 0, 5]]
2. M2 = [[2, 0, 0], [0, 2, 0], [1, 0, 3]]
3. M3 = [[0, −1], [1, 0]], over ℝ and over ℂ
4. M4 = [[0, 1, 0], [0, 0, 1], [0, 0, 0]]

Answers.

1. M1 is not diagonalizable. It is triangular with eigenvalues 3, 3, 5. M1 − 3I has rank 2, so nullity 1, which is less than AM 2.
2. M2 is diagonalizable. It is triangular with eigenvalues 2, 2, 3. The eigenvalue 2 repeats, but M2 − 2I has rank 1, so nullity 2 equals AM.
3. M3 is diagonalizable over ℂ and not over ℝ. Its characteristic polynomial is x² + 1, with roots ±i.
4. M4 is not diagonalizable. It is a non-zero nilpotent matrix: its characteristic polynomial is x³, but M4² is not zero, so the minimal polynomial is x³, which has a repeated root.

Yes, with exact arithmetic. Floating-point routines such as numpy.linalg.eig return approximate eigenvectors, and for a repeated eigenvalue the rounding can make nearly dependent eigenvectors look independent. This script applies step 5's logic exactly: it tests whether the squarefree part of the characteristic polynomial sends A to zero.

from sympy import Matrix, Poly, symbols, gcd, quo, eye, zeros
x = symbols('x')

def diagonalizable(A, over_reals=False):
    cp = Poly(A.charpoly(x).as_expr(), x)
    sqfree = quo(cp, gcd(cp, cp.diff(x)))   # remove repeated factors
    n = A.shape[0]
    value, power = zeros(n, n), eye(n)
    for c in reversed(sqfree.all_coeffs()):  # evaluate sqfree at A
        value += c * power
        power = power * A
    ok = value == zeros(n, n)
    if over_reals:                            # also need every root real
        ok = ok and sqfree.count_roots() == sqfree.degree()
    return ok

print(diagonalizable(Matrix([[0, -1], [1, 0]])))                   # True  (over C)
print(diagonalizable(Matrix([[0, -1], [1, 0]]), over_reals=True))  # False (over R)

It gives the answers listed above for M1 to M4 and for the matrix [[0, −9], [1, −6]].

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