Eigenvalues and Eigenvectors for CSIR NET and GATE Maths
Eigenvalues and eigenvectors are the scalars λ and non-zero vectors v satisfying Av = λv. For exams, the key check is whether a repeated eigenvalue still gives enough independent eigenvectors.
By Sunil Bansal, Founder · Updated 12 Nov 2026
Key takeaways
- An eigenvector is a non-zero vector that a square matrix only scales. Find the eigenvalues from det(A − λI) = 0, then the eigenvectors from (A − λI)v = 0 for each λ.
- Distinct eigenvalues always give a full set of independent eigenvectors. A repeated eigenvalue needs a rank check before you can call the matrix diagonalisable.
- The trace of a matrix equals the sum of its eigenvalues and its determinant equals their product, which gives a quick check on the characteristic polynomial before you find any eigenvectors.
- The GATE MA 2027 syllabus and the existing CSIR NET syllabus both name Jordan form beyond diagonalisation, so the repeated-eigenvalue case is in scope for both.
In this article
- 1What are eigenvalues and eigenvectors, and how do you find them for a 2×2 matrix?
- 2Which case is your matrix in? A decision table for eigenvectors and diagonalisation
- 3Why does a repeated eigenvalue stop a matrix being diagonalisable?
- 4Do the GATE MA and CSIR NET syllabi test eigenvalues and Jordan form?
- 5Can you place these four matrices in the right case?
- 6How do we help you practise eigenvalues for CSIR NET and GATE?
- 7Frequently asked questions
Two matrices can share a characteristic polynomial and still behave differently. The matrices [[2,1],[0,2]] and [[2,0],[0,2]] both give (λ − 2)², yet only the second can be diagonalised. Matrices on this page are written row by row, so [[2,1],[0,2]] has rows (2, 1) and (0, 2).
That gap is where a fast calculation turns into a wrong answer, because det(A − λI) = 0 is routine and the rank of A − λI settles the rest.
This page is for candidates preparing for CSIR NET (the Council of Scientific and Industrial Research National Eligibility Test) or GATE Mathematics (the MA paper of the Graduate Aptitude Test in Engineering). It gives a worked 2×2 example, a decision table that places any square matrix in the right case, and a four-matrix self-test. If you only need the definition and a 2×2 method, the first section gives both.
What are eigenvalues and eigenvectors, and how do you find them for a 2×2 matrix?
An eigenvector of a square matrix A is a non-zero vector v that A only scales, so Av = λv, and the scale factor λ is the eigenvalue. To find them, solve det(A − λI) = 0 for λ, then solve (A − λI)v = 0 for each λ you get.
Geometrically, v lies on a line through the origin that the matrix maps onto itself. An eigenvalue of 3 stretches v to three times its length. An eigenvalue of −2 doubles it and flips it. An eigenvalue of 0 collapses it to the zero vector. The zero vector is excluded because A·0 = λ·0 holds for every λ, so it tells you nothing about A.
Worked example with A = [[4,1],[2,3]]:
- A − λI has rows (4 − λ, 1) and (2, 3 − λ). Its determinant is (4 − λ)(3 − λ) − 2 = λ² − 7λ + 10 = (λ − 5)(λ − 2). The eigenvalues are 5 and 2.
- For λ = 5, A − 5I has rows (−1, 1) and (2, −2). Both rows say x = y, so v = (1, 1).
- For λ = 2, A − 2I has rows (2, 1) and (2, 1). Both rows say 2x + y = 0, so v = (1, −2).
- Check: A(1, 1) = (5, 5) and A(1, −2) = (2, −4) = 2(1, −2). The trace 4 + 3 = 7 equals 5 + 2, and the determinant 12 − 2 = 10 equals 5 × 2.
The trace and determinant check catches arithmetic slips in the characteristic polynomial before you spend time on eigenvectors. One more check is the Cayley-Hamilton theorem, which says a matrix satisfies its own characteristic polynomial: here A² − 7A + 10I is the zero matrix.
Which case is your matrix in? A decision table for eigenvectors and diagonalisation
Count the independent eigenvectors for each eigenvalue and compare that with how often the eigenvalue repeats. Distinct eigenvalues always pass. A repeated eigenvalue passes only if n − rank(A − λI) equals its repeat count. Real symmetric matrices always pass. A real matrix with complex eigenvalues has no real eigenvector for them.
Three terms make this precise. The algebraic multiplicity of λ is how many times λ is a root of the characteristic polynomial. The geometric multiplicity of λ is the number of independent eigenvectors for λ, which equals n − rank(A − λI) for an n×n matrix, and it never exceeds the algebraic multiplicity. A matrix is diagonalisable when it has n independent eigenvectors, which happens exactly when the two multiplicities agree for every eigenvalue. Otherwise it is defective.
| Case | Example | Eigenvalues | Test | Verdict |
|---|---|---|---|---|
| Distinct real eigenvalues | [[4,1],[2,3]] | 5, 2 | None needed | Diagonalisable |
| Repeated, one eigenvector direction | [[2,1],[0,2]] | 2, 2 | rank(A − 2I) = 1, so 2 − 1 = 1 eigenvector | Defective |
| Repeated, full set | [[2,0],[0,2]] | 2, 2 | rank(A − 2I) = 0, so 2 eigenvectors | Diagonalisable |
| Complex pair | [[0,−1],[1,0]] (90° rotation) | i, −i | No real eigenvector | Diagonalisable over the complex numbers, not the reals |
| Real symmetric | [[2,−1,0],[−1,2,−1],[0,−1,2]] | 2 + √2, 2, 2 − √2 | Symmetric | Diagonalisable |
| Repeated zero eigenvalue | 3×3 matrix with every entry 2 | 6, 0, 0 | rank(A − 0I) = 1, so 3 − 1 = 2 eigenvectors for 0 | Diagonalisable |
The routine, with a pass or fail at each step:
- Write det(A − λI). Pass: the sum of the roots equals the trace and their product equals the determinant. Fail: recompute the polynomial.
- Factor it and list each eigenvalue with its algebraic multiplicity.
- For each repeated λ, row-reduce A − λI to find its rank r. The geometric multiplicity is n − r.
- Pass: geometric equals algebraic for every λ, so A = PDP⁻¹ with D diagonal. Fail: the matrix is defective, and you need a Jordan form with generalised eigenvectors.
Why does a repeated eigenvalue stop a matrix being diagonalisable?
Diagonalising A as PDP⁻¹ needs n independent eigenvectors to fill the columns of P, and a repeated eigenvalue can supply fewer. For B = [[2,1],[0,2]], B − 2I has rows (0, 1) and (0, 0). Its rank is 1, so only one independent eigenvector exists, although 2 is a double root.
Solving (B − 2I)v = 0 gives y = 0, so every eigenvector is a multiple of (1, 0). Compare 2I = [[2,0],[0,2]], where 2I − 2I is the zero matrix. Every non-zero vector is an eigenvector, so there are two independent ones.
B is itself a Jordan block, the standard form for defective matrices. When an eigenvalue is repeated and you are asked whether a matrix is diagonalisable, our recommendation is to test the rank before anything else.
Two other cases need no rank check. A real symmetric matrix is always diagonalisable, and its eigenvectors for different eigenvalues are orthogonal. A matrix with all-distinct eigenvalues, real or complex, is always diagonalisable over the complex numbers.
Do the GATE MA and CSIR NET syllabi test eigenvalues and Jordan form?
Yes. Both syllabi name eigenvalues and eigenvectors, and both go on to Jordan form. The GATE 2027 syllabus is published by IIT Madras, the organising institute for GATE 2027. Its Mathematics paper, Section 2 (Linear Algebra), lists "characteristic polynomial, eigenvalues and eigen-vectors, diagonalization, minimal polynomial, Cayley-Hamilton Theorem". It also lists "diagonalization by a unitary matrix" and "Jordan canonical form". Eigenvalues appear in Section 5 as well, under "Sturm-Liouville eigenvalue problems" and "Stability of stationary points for linear systems with constant coefficients" in ordinary differential equations (ODEs). The full text is in the GATE 2027 Mathematics syllabus.
The CSIR NET Mathematical Sciences syllabus, Unit 1, lists "Eigenvalues and eigenvectors, Cayley-Hamilton theorem". It then lists "Matrix representation of linear transformations. Change of basis, canonical forms, diagonal forms, triangular forms, Jordan forms." The wording is in the CSIR Human Resource Development Group (CSIR-HRDG) Mathematical Sciences syllabus. As of November 2026, CSIR-HRDG has published a proposed revised Mathematical Sciences syllabus (feedback closed 14 May 2026). Eigenvalues, Cayley-Hamilton and Jordan forms appear in the existing syllabus; confirm the syllabus named in the December 2026 information bulletin.
Our comparison of the GATE and IIT JAM (Joint Admission Test for Masters) mathematics syllabi shows where the two exams differ. If you are also deciding how many questions to attempt, see our CSIR NET exam pattern guide.
Can you place these four matrices in the right case?
Yes, if you can run the repeated-eigenvalue rank test from the decision table. Classify each matrix below, then check your multiplicities against the answers.
- (a) [[1,2],[2,1]]
- (b) [[3,1],[0,3]]
- (c) [[1,1],[−1,1]]
- (d) [[2,0,0],[0,2,0],[0,0,5]]
Answers:
- (a) Eigenvalues are 3 with v = (1, 1) and −1 with v = (1, −1). The matrix is symmetric and diagonalisable.
- (b) The eigenvalue is 3, repeated twice. A − 3I has rank 1, so there is one eigenvector and the matrix is defective.
- (c) The characteristic polynomial is λ² − 2λ + 2, so the eigenvalues are 1 + i and 1 − i. There is no real eigenvector. The matrix is diagonalisable over the complex numbers.
- (d) The eigenvalues are 2, 2 and 5. A − 2I has rank 1, so the eigenvalue 2 has 3 − 1 = 2 eigenvectors. That equals its algebraic multiplicity, so the matrix is diagonalisable.
For one main textbook per section, see our best GATE mathematics books reading list.
How do we help you practise eigenvalues for CSIR NET and GATE?
We teach mathematics for CSIR NET, GATE, IIT JAM and CUET PG (Common University Entrance Test, Postgraduate) online, through live classes and recorded courses in Hindi and English. Start with the demo classes on our free content page, where you can try our teaching before paying. It also has a PDF Drive of free resources, and we publish notes and previous-year questions. Before choosing any course, our linear algebra course online guide lists what to check against your exam.
Our CSIR-NET and GATE Mathematics 2027 course includes PDF notes, assignments, discussion of previous-year questions, mock tests and analytics. It also includes continuous doubt support and weekly doubt sessions or personal mentoring, which is the right place to bring a matrix you cannot classify. Our Core Concept Diagnostic test series is designed by SB Sir, who secured All India Rank (AIR) 11 in CSIR NET in December 2017 and is GATE MA qualified.
As of November 2026, the 2027 course costs ₹15,999 one time with access through 30 June 2027, or four instalments of ₹3,999.75 (₹15,999 in total). A ₹999 Demo Plan gives 10 days of access. Our digital products carry a no-refund policy, so try a demo class or the Demo Plan first.
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Frequently asked questions
Use the same method. A covariance matrix is symmetric and positive semidefinite, so its eigenvalues are real and non-negative and its eigenvectors are orthogonal. In principal component analysis (PCA), the eigenvector for the largest eigenvalue is the direction of greatest variance.
