Linear Algebra — Rank, Consistency, Eigenvalues and Eigenvectors

Linear algebra is the cheapest block of Engineering Mathematics, and eigenvalues are the cheapest part of it, because most eigenvalue questions can be answered without solving anything. Two facts do the work: the sum of the eigenvalues is the trace and their product is the determinant. For a 2×2 matrix that is already the whole characteristic equation, and for a 3×3 one it settles most multiple-choice options before you write a determinant. Learn those two before any method. Beneath them sits one idea that unifies the rest of the chapter — rank, the number of genuinely independent rows. A system of equations is consistent when the coefficient matrix and the augmented matrix have the same rank; it has a unique solution when that common rank also equals the number of unknowns, and infinitely many when it falls short. A square matrix is invertible when its rank is full, which is the same as saying its determinant is non-zero, which is the same as saying zero is not one of its eigenvalues. Those are not four facts to memorise. They are one fact in four costumes, and recognising the costume is the examinable skill. The examiner's favourite trap in this chapter is the repeated eigenvalue: a matrix with a repeated eigenvalue may or may not be diagonalisable, and the question is always whether it has enough independent eigenvectors to go round — so count eigenvectors, never eigenvalues, when diagonalisability is asked.

Rank Decides Everything About a Linear System

Write the system as Ax = b with n unknowns. Row-reduce the coefficient matrix A and the augmented matrix [A | b] and compare the two ranks. There are exactly three outcomes, and no fourth is possible.

The consistency test in full
RanksWhat it meansNumber of solutions
rank(A) < rank[A | b]A row reduced to 0 = non-zero. The equations contradict each other.Inconsistent — none at all
rank(A) = rank[A | b] = nAs many independent equations as unknowns.Exactly one
rank(A) = rank[A | b] = r < nConsistent but short of information: n − r unknowns stay free.Infinitely many, with n − r free parameters
⚠️ A zero determinant does not mean "no solution"
When |A| = 0 the system has either no solution or infinitely many, and the determinant alone cannot tell you which — you have to look at the augmented matrix. This is the single commonest error in the topic, and examiners set it deliberately: they give a system with a parameter, ask for the value that makes it singular, and then offer both "no solution" and "infinitely many solutions" as options. Compute the rank of [A | b], not just |A|.

For a homogeneous system Ax = 0 the augmented column is zero, so the ranks can never differ and the system is always consistent — x = 0 is a solution whatever A is. The only question worth asking is whether there is another one, and there is exactly when rank(A) < n, which for a square A is exactly when |A| = 0. That is why "the system has a non-trivial solution" and "the determinant vanishes" are interchangeable phrasings in a question stem.

Eigenvalues Without Solving a Characteristic Equation

  • Sum = trace, product = determinant. For A = [[4, 1], [2, 3]] the trace is 7 and the determinant is 10, so the eigenvalues are the two numbers adding to 7 and multiplying to 10: 5 and 2. No characteristic polynomial was written.
  • Triangular means read the diagonal. If A is upper or lower triangular — and a diagonal matrix is both — its eigenvalues are its diagonal entries, because |A − λI| is then a product of the diagonal terms.
  • Functions of A shift the eigenvalues, not the eigenvectors. If Ax = λx then Ak x = λk x, (A + cI)x = (λ + c)x, and A⁻¹x = (1/λ)x when λ ≠ 0. So a matrix with eigenvalues 1, 2, 3 has A² with eigenvalues 1, 4, 9 and hence trace 14 — computed without ever forming A².
  • Symmetry constrains the answer. A real symmetric matrix has real eigenvalues and orthogonal eigenvectors; a real skew-symmetric matrix has eigenvalues that are purely imaginary or zero. [[0, 2], [−2, 0]] has eigenvalues ±2i, and an option list containing real numbers can be discarded on sight.
  • Cayley–Hamilton turns high powers into low ones. Every matrix satisfies its own characteristic equation. For B = [[1, 2], [3, 4]], whose characteristic polynomial is λ² − 5λ − 2, this gives B² = 5B + 2I — and both sides really are [[7, 10], [15, 22]]. Any question asking for B⁴ or B⁻¹ in terms of B is this identity applied twice.
🎯 Why a repeated eigenvalue is the question, not the answer
An eigenvalue repeated k times in the characteristic polynomial has algebraic multiplicity k; the number of independent eigenvectors it actually supplies is its geometric multiplicity, and that can be smaller. [[3, 1], [0, 3]] has the single eigenvalue 3 with algebraic multiplicity 2, but solving (A − 3I)x = 0 gives a one-dimensional space — geometric multiplicity 1. With only one eigenvector for a 2×2 matrix there is no basis of eigenvectors, so the matrix is not diagonalisable. A real symmetric matrix never has this problem, which is why symmetry is always worth noticing first.

Gauss Elimination and LU — What GATE Actually Asks

Elimination questions are almost never "solve this system". They ask for one intermediate quantity: the value of a pivot after two steps, the number of row interchanges partial pivoting would need, an entry of L or U, or the rank that emerges. So carry out the reduction in the order the method prescribes and stop the moment the asked-for number appears, rather than solving through to x, y, z and then hunting for the answer.

  • Elementary row operations do not change the rank, and they do not change the solution set either. They do change the determinant in known ways: swapping two rows flips its sign, scaling a row by c multiplies it by c, and adding a multiple of one row to another leaves it alone.
  • Partial pivoting exists for arithmetic, not for algebra. Swapping in the row with the largest pivot does not change the exact answer at all; it stops a small pivot from amplifying rounding error. A question that asks why pivoting is used is asking about numerical stability, and the word to look for in the options is not "accuracy of the method" but "growth of round-off".
  • A rank-one matrix is spotted, not computed. In [[1, 2, 3], [2, 4, 6], [3, 6, 9]] every row is a multiple of the first, so the rank is 1 — and therefore two eigenvalues are 0 and the third is the trace, 14. Recognising proportional rows saves the whole reduction.

Key takeaways

  • Sum of eigenvalues = trace, product = determinant. For a 2×2 matrix that is the whole characteristic equation, and for a 3×3 it eliminates most options before any work.
  • Consistency is a rank comparison: unequal ranks means no solution, equal ranks equal to n means one, equal ranks below n means infinitely many with n − r free parameters.
  • A vanishing determinant does not decide between "no solution" and "infinitely many" — only the augmented matrix does. Examiners offer both options on purpose.
  • Powers, inverses and shifts of A keep the eigenvectors and transform the eigenvalues: λ becomes λk, 1/λ or λ + c. Trace(A²) follows from the eigenvalues alone.
  • Diagonalisability is about eigenvectors, not eigenvalues. [[3, 1], [0, 3]] repeats the eigenvalue 3 but supplies only one eigenvector, so it is not diagonalisable; a real symmetric matrix always is.
  • Cayley–Hamilton replaces a high power of a matrix with a linear combination of lower ones: for [[1, 2], [3, 4]], A² = 5A + 2I.
  • Proportional rows mean rank 1, which forces all but one eigenvalue to be zero and makes the survivor equal to the trace.

Practice questions (10)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. The eigenvalues of the matrix [[4, 1], [2, 3]] are:

    1. 5 and 2
    2. 4 and 3
    3. 7 and 10
    4. 6 and 1
    Show answer

    Answer: A — 5 and 2

    Do not build the characteristic polynomial. The trace is 4 + 3 = 7 and the determinant is 4·3 − 1·2 = 10, so the eigenvalues are the pair that adds to 7 and multiplies to 10: 5 and 2. Option B is the diagonal, which would be the answer only if the matrix were triangular — it is not, because the entry 1 above the diagonal is non-zero. Option C mistakes the trace and determinant themselves for the eigenvalues, which is the commonest slip here: they are the sum and product of the eigenvalues, not the eigenvalues.

  2. A 3×3 matrix A has eigenvalues 1, 2 and 3. The trace of A² is:

    1. 36
    2. 14
    3. 6
    4. 12
    Show answer

    Answer: B — 14

    Squaring a matrix squares its eigenvalues and leaves the eigenvectors alone, so A² has eigenvalues 1, 4 and 9. The trace is the sum of the eigenvalues, so trace(A²) = 1 + 4 + 9 = 14. A itself was never needed. Option A squares the trace of A (6² = 36), which is a different quantity — the trace of a square is not the square of the trace. Option C is trace(A), and option D is the determinant of A² (1·4·9 = 36) misremembered; note that det(A²) = 36 while trace(A²) = 14, and a question offering both is testing whether you read which was asked.

  3. For the system x + y + z = 1, x + 2y + 3z = 4, x + 3y + λz = 6, the value λ = 5 makes the system:

    1. Have infinitely many solutions
    2. Have a unique solution
    3. Inconsistent, with no solution
    4. Have exactly two solutions
    Show answer

    Answer: C — Inconsistent, with no solution

    The determinant of the coefficient matrix works out to λ − 5, so λ = 5 makes it singular and the unique-solution case is gone. That is only half the work. Row-reduce the augmented matrix at λ = 5: subtracting the first row from the second gives (0, 1, 2 | 3) and from the third gives (0, 2, 4 | 5), and twice the first of those is (0, 2, 4 | 6) — which contradicts (0, 2, 4 | 5). So rank(A) = 2 while rank[A | b] = 3, and the system is inconsistent. Option A is the trap: a zero determinant is consistent with either A or C, and only the augmented matrix decides between them. Option D can be dismissed on principle — a linear system never has finitely many solutions greater than one.

  4. The system x + y + z = 6, 2x + 2y + 2z = 12, x + 2y + 3z = 14 has:

    1. A unique solution
    2. Infinitely many solutions, with one free parameter
    3. No solution
    4. A unique solution only if a fourth equation is added
    Show answer

    Answer: B — Infinitely many solutions, with one free parameter

    The second equation is exactly twice the first, so it carries no information the first did not. Two independent equations survive, giving rank(A) = 2, and the augmented matrix reduces to rank 2 as well — the duplicate row is consistent, not contradictory. With three unknowns and rank 2, there are n − r = 3 − 2 = 1 free parameter and therefore infinitely many solutions. Option C would need the ranks to differ, which happens when a duplicated row has a different right-hand side: change 12 to 13 and the same system becomes inconsistent. That single digit is the whole difference between the two answers, and it is where examiners set this question.

  5. The matrix [[3, 1], [0, 3]] is:

    1. Diagonalisable, with eigenvalues 3 and 1
    2. Diagonalisable, with the repeated eigenvalue 3
    3. Not diagonalisable, because it has only one independent eigenvector
    4. Not diagonalisable, because its determinant is zero
    Show answer

    Answer: C — Not diagonalisable, because it has only one independent eigenvector

    The matrix is triangular, so the eigenvalues are the diagonal entries: 3 twice — algebraic multiplicity 2. Now solve (A − 3I)x = 0, that is [[0, 1], [0, 0]]x = 0, which forces the second component to be zero and leaves the first free: a one-dimensional eigenspace, geometric multiplicity 1. A 2×2 matrix needs two independent eigenvectors to be diagonalisable, so this one is not. Option A misreads the off-diagonal 1 as an eigenvalue. Option D is a false reason attached to a true conclusion: the determinant is 9, not 0, and in any case a zero determinant does not obstruct diagonalisation — the zero matrix is diagonal already.

  6. Using the Cayley–Hamilton theorem for A = [[1, 2], [3, 4]], A² equals:

    1. 5A + 2I
    2. 5A − 2I
    3. 2A + 5I
    4. −5A + 2I
    Show answer

    Answer: A — 5A + 2I

    The trace is 5 and the determinant is 1·4 − 2·3 = −2, so the characteristic equation is λ² − (trace)λ + (det) = λ² − 5λ − 2 = 0. Cayley–Hamilton says the matrix satisfies it: A² − 5A − 2I = 0, hence A² = 5A + 2I. Check both sides: A² = [[7, 10], [15, 22]] and 5A + 2I = [[5 + 2, 10], [15, 20 + 2]] = [[7, 10], [15, 22]]. Option B is what you get by dropping the sign of the negative determinant, which is the only real difficulty in the question — the constant term of the characteristic polynomial is +det, and det is itself negative here.

  7. The eigenvalues of the real skew-symmetric matrix [[0, 2], [−2, 0]] are:

    1. 2 and −2
    2. 2i and −2i
    3. 0 and 0
    4. 4 and −4
    Show answer

    Answer: B — 2i and −2i

    The trace is 0 and the determinant is 0·0 − 2·(−2) = 4, so the eigenvalues sum to 0 and multiply to 4 — they must be ±2i, since no pair of real numbers adds to zero and multiplies to a positive number. That is the general rule worth carrying: a real skew-symmetric matrix has eigenvalues that are purely imaginary or zero, exactly as a real symmetric matrix has real ones. Option A is the pair that sums to 0 but multiplies to −4, and option C is what a zero trace suggests if you forget that the product must also be matched.

  8. The rank of the matrix [[1, 2, 3], [2, 4, 6], [3, 6, 9]] is _____ (answer as a whole number).

    Numerical answer — type the value.

    Show answer

    Answer: 1

    The second row is twice the first and the third is three times it, so there is only one independent row and the rank is 1 — no reduction needed once you see the proportionality. Two consequences follow immediately and are often what the next question asks: since the rank is 1 the nullity is 3 − 1 = 2, so two eigenvalues are 0; and since the eigenvalues must sum to the trace, 1 + 4 + 9 = 14, the remaining eigenvalue is 14. Note also that this is a numerical-answer question, so a wrong entry costs nothing — never leave one blank.

  9. The sum of the eigenvalues of the matrix [[2, 1, 0], [1, 3, 1], [0, 1, 2]] is _____.

    Numerical answer — type the value.

    Show answer

    Answer: 7

    The sum of the eigenvalues is the trace, so it is 2 + 3 + 2 = 7, and the characteristic polynomial never needs to be formed. For the record the eigenvalues are 1, 2 and 4, which do sum to 7 and multiply to 8 — the determinant. The matrix is symmetric, which guarantees in advance that all three are real, so an option list or a computed answer containing a complex number would be wrong before any arithmetic. Reading the trace off the diagonal is the two-second route, and it is available whatever the size of the matrix.

  10. A is a real 3×3 matrix with |A| = 0. Which of the following statements must be true? (More than one option may be correct.)

    1. At least one eigenvalue of A is zero
    2. The system Ax = b has no solution for every b
    3. The system Ax = 0 has a non-trivial solution
    4. A cannot be diagonalised
    Show answer

    Answer: A — At least one eigenvalue of A is zero; C — The system A**x** = **0** has a non-trivial solution

    The determinant is the product of the eigenvalues, so |A| = 0 forces at least one of them to be zero — option A is true. And a zero determinant means the rank is below 3, so the homogeneous system has a solution beyond x = 0 — option C is true. Option B over-reads it: with |A| = 0 the non-homogeneous system has either no solution or infinitely many, and which one depends on b, so "for every b" is false. Option D over-reads it in the other direction: the zero matrix has determinant 0 and is already diagonal. Since GATE gives no partial credit on an MSQ, adding either over-reading to the two correct options scores nothing at all — and since a wrong MSQ costs nothing either, the right tactic is to test all four statements rather than stopping at the first two that look right.