Let V be a finite-dimensional vector space and T: V → V linear. Which statements are always true?

T is injective if and only if T is surjective

if T² = 0 then rank T ≤ (dim V)/2

T is injective if and only if T is surjective

if T² = 0 then rank T ≤ (dim V)/2

The system x + y + z = 1, x + 2y + 3z = 2, x + 3y + 5z = k in real unknowns

is consistent only for k = 3, and then has infinitely many solutions

is consistent for every k and has a unique solution

is consistent only for k = 3, and then has a unique solution

For A = [[1, 1], [0, 1]], which statements are true?

A is not diagonalisable over ℂ

the minimal polynomial of A is (x − 1)²

A is not diagonalisable over ℂ

the minimal polynomial of A is (x − 1)²

the geometric multiplicity of the eigenvalue 1 is 2

Let A be an n × n complex matrix. Which statements are true?

A is diagonalisable if and only if its minimal polynomial has no repeated root

the minimal and characteristic polynomials of A have the same roots

if A³ = A then A is diagonalisable

A is diagonalisable if and only if its minimal polynomial has no repeated root

the minimal and characteristic polynomials of A have the same roots

if A³ = A then A is diagonalisable

if A² = 0 and A ≠ 0 then A is diagonalisable

Vector Spaces, Linear Maps, Eigenvalues and the Minimal Polynomial

Linear Algebra is Section 2 of the GATE Mathematics paper and one of its most reliable sources of marks, because nearly every question reduces to a handful of invariants: dimension, rank and nullity, trace and determinant, the characteristic and minimal polynomials. This chapter covers the first half of the section — finite-dimensional vector spaces over ℝ or ℂ, linear transformations and their matrices, the rank–nullity theorem, systems of linear equations, eigenvalues and eigenvectors, diagonalisation, the minimal polynomial and the Cayley–Hamilton theorem. The second chapter takes inner products, the special classes of matrices, unitary diagonalisation, the Jordan form and quadratic forms. Throughout, the habit to build is to answer with an invariant rather than a computation: the rank of an idempotent matrix is its trace, a matrix is diagonalisable exactly when its minimal polynomial has no repeated root, and a high power of a matrix is a linear polynomial in it once Cayley–Hamilton is applied.

1. Subspaces, bases and dimension

A basis of a vector space V is a linearly independent spanning set, and every basis of a finite-dimensional V has the same number of elements, its dimension. Any independent set extends to a basis and any spanning set contains one. For subspaces U, W of V: U ∩ W and U + W are subspaces, dim(U + W) = dim U + dim W − dim(U ∩ W), and U ∪ W is a subspace only if one of them contains the other.

Dimensions the paper keeps asking
Space	Dimension
n × n real matrices	n²
symmetric n × n matrices	n(n + 1)/2
skew-symmetric n × n matrices	n(n − 1)/2
n × n matrices of trace 0	n² − 1
n × n upper triangular matrices	n(n + 1)/2
real polynomials of degree ≤ n	n + 1
ℂⁿ as a vector space over ℝ	2n
n × n complex Hermitian matrices, over ℝ	n²

⚠️ The field changes the dimension

ℂ is one-dimensional over ℂ and two-dimensional over ℝ, with basis {1, i}. The Hermitian matrices are not a complex vector space at all (iH is skew-Hermitian), but over ℝ they form a space of dimension n²: n real diagonal entries plus 2 real parameters for each of the n(n − 1)/2 entries above the diagonal. Always ask over which field the question is set.

2. Linear transformations, matrices and rank–nullity

A linear T: V → W between finite-dimensional spaces, once bases are fixed, is a matrix, and a change of bases in V = W replaces the matrix A by P⁻¹AP, a similar matrix with the same rank, trace, determinant, characteristic and minimal polynomials. The rank–nullity theorem says dim V = rank T + nullity T, where rank T = dim T(V) and nullity T = dim ker T. Row rank equals column rank, and rank(AB) ≤ min(rank A, rank B).

For T: V → V with V finite-dimensional, injective ⇔ surjective ⇔ invertible, straight from rank–nullity. On an infinite-dimensional space this fails: the shift (x₁, x₂, …) ↦ (0, x₁, x₂, …) is injective and not onto.
Differentiation D on the polynomials of degree ≤ n has kernel the constants, so nullity 1 and rank n; D² has nullity 2 and rank n − 1. D is nilpotent: Dⁿ⁺¹ = 0.
If T² = 0 then T(V) ⊆ ker T, so rank T ≤ nullity T and rank T ≤ (dim V)/2.

Linear systems. Ax = b with A of size m × n is consistent exactly when rank A = rank [A | b]; it then has a unique solution if this common rank is n, and otherwise a family of solutions of dimension n − rank A, namely one particular solution plus the null space of A.

3. Eigenvalues, eigenvectors and diagonalisation

λ is an eigenvalue of A when A − λI is singular, i.e. a root of the characteristic polynomial p_A(x) = det(xI − A). The sum of the eigenvalues (with algebraic multiplicity) is the trace and their product is the determinant. The algebraic multiplicity of λ is its multiplicity as a root of p_A; its geometric multiplicity is dim ker(A − λI), and 1 ≤ geometric ≤ algebraic.

A is diagonalisable (A = PDP⁻¹) ⇔ it has n independent eigenvectors ⇔ geometric = algebraic multiplicity for every eigenvalue. n distinct eigenvalues are sufficient, not necessary (I is diagonal with one eigenvalue).
If Av = λv then p(A)v = p(λ)v for any polynomial p, and A⁻¹v = λ⁻¹v when A is invertible; so the eigenvalues of A² + I are λ² + 1.
A rank-one matrix uvᵀ has eigenvalues vᵀu and 0 (with multiplicity n − 1); the all-ones n × n matrix has eigenvalues n and 0. So aI + bJ has eigenvalues a + nb and a (n − 1 times).

⚠️ The shear [[1, 1], [0, 1]]

Its only eigenvalue is 1, with algebraic multiplicity 2, but A − I = [[0, 1], [0, 0]] has rank 1, so the geometric multiplicity is 1: one independent eigenvector, (1, 0), and A is not diagonalisable over ℝ or ℂ, although it is invertible. Invertibility and diagonalisability are independent properties; [[0, 0], [0, 1]] is diagonal and singular.

4. The minimal polynomial and the Cayley–Hamilton theorem

Cayley–Hamilton: every square matrix satisfies its characteristic polynomial, p_A(A) = 0. The minimal polynomial m_A is the monic polynomial of least degree with m_A(A) = 0; it divides every polynomial that kills A, in particular p_A, and it has exactly the same roots as p_A (each eigenvalue appears, possibly with smaller multiplicity).

What the minimal polynomial tells you
Fact	Consequence
m_A has distinct roots (splits into distinct linear factors)	A is diagonalisable, and conversely
A² = A (idempotent)	m_A divides x(x − 1): diagonalisable with eigenvalues 0, 1, and rank A = trace A
A³ = A	m_A divides x(x − 1)(x + 1): diagonalisable, eigenvalues in {0, 1, −1}
A^k = 0 for some k (nilpotent)	every eigenvalue is 0 and Aⁿ = 0; a non-zero nilpotent is never diagonalisable
the largest power of (x − λ) in m_A	the size of the largest Jordan block for λ

🧠 Powers and inverses from Cayley–Hamilton

For a 2 × 2 matrix, p_A(x) = x² − (tr A)x + det A. With A = [[1, 2], [3, 4]], tr A = 5 and det A = −2, so A² = 5A + 2I. Then A³ = 5A² + 2A = 27A + 10I and A⁴ = 27A² + 10A = 145A + 54I, with no matrix multiplication at all. Dividing A² − 5A − 2I = 0 by A gives A⁻¹ = (A − 5I)/2.

Key takeaways

dim(U + W) = dim U + dim W − dim(U ∩ W); the union of two subspaces is a subspace only when one contains the other.
rank T + nullity T = dim V; on a finite-dimensional space injective, surjective and invertible coincide.
Ax = b is consistent iff rank A = rank [A | b], and the solution set has dimension n − rank A.
Trace = sum and determinant = product of eigenvalues; diagonalisable iff geometric equals algebraic multiplicity for each eigenvalue.
The minimal polynomial divides the characteristic polynomial, shares its roots, and has distinct roots exactly when A is diagonalisable.

Practice questions (13)

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.

The dimension of the real vector space of all 4 × 4 real skew-symmetric matrices is ____.
Numerical answer — type the value.
Show answer
Answer: 6
A skew-symmetric matrix has zero diagonal and is fixed by its entries above the diagonal, of which there are n(n − 1)/2 = 4 × 3/2 = 6. The symmetric matrices would give n(n + 1)/2 = 10, and 6 + 10 = 16 = n², as it must.
Let V be the space of real polynomials of degree at most 4 and T: V → V the map T(p) = p″ (the second derivative). The rank of T is ____.
Numerical answer — type the value.
Show answer
Answer: 3
dim V = 5 (basis 1, x, x², x³, x⁴). ker T is the polynomials of degree ≤ 1, of dimension 2, so by rank–nullity rank T = 5 − 2 = 3; indeed the image is spanned by 2, 6x, 12x². Answering 4 treats T as a single derivative.
U and W are subspaces of ℝ⁷ with dim U = 4 and dim W = 5. The smallest possible value of dim(U ∩ W) is ____.
Numerical answer — type the value.
Show answer
Answer: 2
dim(U ∩ W) = dim U + dim W − dim(U + W) ≥ 4 + 5 − 7 = 2, because U + W sits inside ℝ⁷. The bound is attained by U = span(e₁, …, e₄), W = span(e₃, …, e₇), whose intersection is span(e₃, e₄).
Let V be a finite-dimensional vector space and T: V → V linear. Which statements are always true?
1. T is injective if and only if T is surjective
2. if T² = 0 then rank T ≤ (dim V)/2
3. rank T² = rank T
4. rank T + nullity T = dim V
Show answer
Answer: A — T is injective if and only if T is surjective; B — if T² = 0 then rank T ≤ (dim V)/2; D — rank T + nullity T = dim V
(1) and (4) are rank–nullity: nullity 0 ⇔ rank = dim V. (2) T² = 0 gives T(V) ⊆ ker T, so rank ≤ nullity = dim V − rank. (3) is false: T = [[0, 1], [0, 0]] has rank 1 and T² = 0 has rank 0.
The system x + y + z = 1, x + 2y + 3z = 2, x + 3y + 5z = k in real unknowns
1. is consistent only for k = 3, and then has infinitely many solutions
2. is consistent for every k and has a unique solution
3. is consistent only for k = 3, and then has a unique solution
4. is inconsistent for every k
Show answer
Answer: A — is consistent only for k = 3, and then has infinitely many solutions
R₂ − R₁ = (0, 1, 2 | 1) and R₃ − R₂ = (0, 1, 2 | k − 2). These agree only if k − 2 = 1, i.e. k = 3; otherwise rank A = 2 < rank [A | b] = 3. For k = 3, rank A = 2 < 3 unknowns, so the solutions form a one-parameter family. The determinant of A is 0, which rules out uniqueness for every k.
For A = [[2, 1, 1], [1, 2, 1], [1, 1, 2]], the sum of the eigenvalues of A⁻¹ is ____.
Numerical answer — type the value.
Show answer
Answer: 2.25
A = I + J with J the all-ones matrix, whose eigenvalues are 3, 0, 0; so A has eigenvalues 4, 1, 1 (check: trace 6, determinant 4). A⁻¹ has eigenvalues 1/4, 1, 1, summing to 2.25. The reciprocal of the trace, 1/6, is the tempting wrong answer.
For A = [[1, 1], [0, 1]], which statements are true?
1. A is not diagonalisable over ℂ
2. the minimal polynomial of A is (x − 1)²
3. the geometric multiplicity of the eigenvalue 1 is 2
4. A is invertible
Show answer
Answer: A — A is not diagonalisable over ℂ; B — the minimal polynomial of A is (x − 1)²; D — A is invertible
A − I = [[0, 1], [0, 0]] ≠ 0 but (A − I)² = 0, so m_A = (x − 1)², which has a repeated root: (1) and (2) true. ker(A − I) = span{(1, 0)} has dimension 1, so (3) is false. det A = 1 ≠ 0, so (4) is true.
A real 3 × 3 matrix A has eigenvalues 1, 2 and 3. The eigenvalues of A² + I are
1. 2, 5, 10
2. 2, 3, 4
3. 1, 4, 9
4. 3, 6, 11
Show answer
Answer: A — 2, 5, 10
If Av = λv then (A² + I)v = (λ² + 1)v, so the eigenvalues are 1 + 1, 4 + 1, 9 + 1 = 2, 5, 10. Since A has three distinct eigenvalues these are all of them. Adding 1 to λ instead of λ² gives 2, 3, 4.
Let A = [[1, 2], [3, 4]]. By the Cayley–Hamilton theorem A⁴ = αA + βI for real numbers α, β. The value of α is ____.
Numerical answer — type the value.
Show answer
Answer: 145
tr A = 5, det A = −2, so A² = 5A + 2I. Then A³ = 5A² + 2A = 5(5A + 2I) + 2A = 27A + 10I and A⁴ = 27A² + 10A = 27(5A + 2I) + 10A = 145A + 54I. Check the (1, 1) entry: A⁴ has 199 = 145 + 54.
Let A be an n × n complex matrix. Which statements are true?
1. A is diagonalisable if and only if its minimal polynomial has no repeated root
2. the minimal and characteristic polynomials of A have the same roots
3. if A³ = A then A is diagonalisable
4. if A² = 0 and A ≠ 0 then A is diagonalisable
Show answer
Answer: A — A is diagonalisable if and only if its minimal polynomial has no repeated root; B — the minimal and characteristic polynomials of A have the same roots; C — if A³ = A then A is diagonalisable
(1) is the standard criterion over ℂ. (2) m_A divides p_A, and every eigenvalue λ is a root of m_A because m_A(A)v = m_A(λ)v = 0 for an eigenvector v. (3) m_A divides x³ − x = x(x − 1)(x + 1), which has distinct roots. (4) m_A = x², a repeated root; e.g. [[0, 1], [0, 0]].
Let u = (1, 2, 3)ᵀ, v = (1, 1, 1)ᵀ and A = uvᵀ. The trace of A³ is ____.
Numerical answer — type the value.
Show answer
Answer: 216
A = uvᵀ has rank 1 and A² = u(vᵀu)vᵀ = (vᵀu)A with vᵀu = 6. So A³ = 36A, and trace A³ = 36 × trace A = 36 × 6 = 216; equivalently the eigenvalues of A³ are 6³, 0, 0.
A 5 × 5 real matrix P satisfies P² = P and trace P = 3. The rank of P is
1. 3
2. 2
3. 5
4. not determined by the data
Show answer
Answer: A — 3
m_P divides x(x − 1), so P is diagonalisable with eigenvalues 0 and 1. Its rank is the number of eigenvalues equal to 1, which is the trace: 3. An idempotent is the projection onto its image along its kernel.
The number of linearly independent eigenvectors of [[2, 1, 0], [0, 2, 0], [0, 0, 3]] is ____.
Numerical answer — type the value.
Show answer
Answer: 2
Eigenvalue 2 has algebraic multiplicity 2, but A − 2I = [[0, 1, 0], [0, 0, 0], [0, 0, 1]] has rank 2, so its kernel is one-dimensional. Eigenvalue 3 contributes one more. Total 2, so the matrix is not diagonalisable.