A′ = S⁻¹AS for an invertible S. Which of the following are necessarily the same for A and A′?

The components of each eigenvector

It is known that Lᵢ = Iᵢⱼωⱼ holds for every angular velocity ω, and that L and ω are vectors. What does this establish about Iᵢⱼ?

It is a rank-2 tensor, by the quotient law

It is a scalar multiple of δᵢⱼ

It is necessarily antisymmetric

Mathematical Physics I: Inner-Product Spaces, Completeness, Similarity Transformations and Tensors

The first of two chapters written for Section 2 of the GATE Physics (PH) paper, Mathematical Physics, and only for the parts of it no shared mathematics chapter teaches. The shared Linear Algebra chapter already teaches eigenvalues, eigenvectors and when a matrix can be diagonalised, and the Vector Spaces chapter teaches basis, independence and rank. What the physics syllabus adds, and what this chapter supplies, is the inner product: orthogonality, the Gram–Schmidt procedure, and completeness, written the way quantum mechanics uses it, Σ|eᵢ⟩⟨eᵢ| = 1. It then treats similarity transformations as a change of basis, with the Hermitian and unitary matrices that physics diagonalises every day, and ends with tensors: the transformation laws that define covariant and contravariant components, the Kronecker delta and Levi-Civita symbols, the metric, contraction and the quotient law. Every identity here is used again in the classical-mechanics, electromagnetism and quantum chapters of this paper.

1. Inner products, orthogonality and the Gram–Schmidt procedure

A linear vector space becomes geometry once it has an inner product ⟨u|v⟩: a complex number that is linear in the second slot, conjugate-linear in the first, satisfies ⟨u|v⟩ = ⟨v|u⟩*, and gives ⟨v|v⟩ > 0 for every non-zero v. The norm is ‖v‖ = √⟨v|v⟩, two vectors are orthogonal when ⟨u|v⟩ = 0, and the Cauchy–Schwarz inequality |⟨u|v⟩|² ≤ ⟨u|u⟩⟨v|v⟩ holds with equality only when one vector is a multiple of the other. In Cⁿ the inner product is ⟨u|v⟩ = Σ uᵢ* vᵢ; for functions on an interval it is ⟨f|g⟩ = ∫ f*(x) g(x) dx.

Gram–Schmidt turns any independent set v₁, v₂, … into an orthonormal one by subtracting, at each step, the components along the vectors already built: e₁ = v₁/‖v₁‖, then u₂ = v₂ − ⟨e₁|v₂⟩e₁ and e₂ = u₂/‖u₂‖, and so on. Worked: from v₁ = (1, 1, 0) and v₂ = (1, 0, 1), e₁ = (1, 1, 0)/√2, the component of v₂ along e₁ is 1/√2, so u₂ = (1, 0, 1) − ½(1, 1, 0) = (½, −½, 1) with ‖u₂‖ = √(3/2) ≈ 1.22, and e₂ = (1, −1, 2)/√6. Check: e₁·e₂ = (1 − 1 + 0)/√12 = 0.

⚠️ Subtract the projection on the NORMALISED vector

The projection of v₂ on v₁ is (⟨v₁|v₂⟩/⟨v₁|v₁⟩)v₁. Writing ⟨v₁|v₂⟩v₁ without dividing by ⟨v₁|v₁⟩ — here 2 — subtracts twice too much and leaves a vector that is not orthogonal to v₁. Either normalise first, as above, or keep the denominator.

2. Orthonormal bases and completeness

In an orthonormal basis {|eᵢ⟩} every vector expands as |v⟩ = Σ cᵢ|eᵢ⟩ with cᵢ = ⟨eᵢ|v⟩ — the coefficient is an inner product, not the solution of a linear system. Substituting back gives |v⟩ = Σ |eᵢ⟩⟨eᵢ|v⟩ for every v, which is the completeness (closure) relation Σ|eᵢ⟩⟨eᵢ| = 1. Taking the norm gives Parseval’s relation ‖v‖² = Σ|cᵢ|². Completeness says the basis leaves nothing out: the only vector orthogonal to every eᵢ is the zero vector.

Finite and infinite dimensions side by side
Idea	Cⁿ	Functions on (−π, π)
Inner product	Σ uᵢ* vᵢ	∫ f* g dx
Orthonormal set	unit vectors eᵢ	e^inx/√(2π), n = 0, ±1, ±2, …
Coefficient	cᵢ = ⟨eᵢ\|v⟩	cₙ = (1/√(2π)) ∫ e^−inx f dx
Completeness	Σ\|eᵢ⟩⟨eᵢ\| = 1	Σₙ φₙ(x) φₙ*(x′) = δ(x − x′)

In an infinite-dimensional space orthonormality alone does not guarantee completeness: the sines sin nx on (−π, π) are orthogonal, but they cannot represent an even function such as cos x, so they are not complete there — the cosines have to be added. Completeness of the full set e^inx is what lets a Fourier series converge to f in the mean, and it is what quantum mechanics assumes when it expands a state in the eigenfunctions of a Hermitian operator.

3. Similarity transformations, Hermitian and unitary matrices

A similarity transformation A′ = S⁻¹AS is the same linear map written in a new basis whose vectors are the columns of S. Anything that does not depend on the basis survives it: the characteristic polynomial, hence the eigenvalues, the trace, the determinant and the rank. The eigenvectors’ components do change, because they are written in the new basis. Diagonalisation is the special choice of S whose columns are n independent eigenvectors: then S⁻¹AS = D = diag(λ₁, …, λₙ).

Hermitian (H = H†): real eigenvalues, eigenvectors of distinct eigenvalues orthogonal, and always diagonalisable by a unitary U: U†HU = D. Physical observables are Hermitian for exactly this reason.
Unitary (U†U = 1): preserves inner products and norms, and every eigenvalue has |λ| = 1, so λ = e^iθ. A real unitary matrix is orthogonal (rotations and reflections, det = ±1).
Commuting Hermitian matrices can be diagonalised by one and the same unitary matrix — they share a complete set of eigenvectors. This is the matrix form of "compatible observables".
Functions of a matrix follow the eigenvalues: f(A) = S f(D) S⁻¹. Hence det(e^A) = e^{tr A}, and for the Pauli matrix σ_x (σ_x² = 1) the series gives e^iθσ_x = cos θ · 1 + i sin θ · σ_x.

🧠 Eigenvalues first, matrices never

H = [[2, i], [−i, 2]] is Hermitian with trace 4 and determinant 4 − 1 = 3, so its eigenvalues are 1 and 3. Then det(e^H) = e¹·e³ = e⁴ ≈ 54.6 and tr(H²) = 1 + 9 = 10, without forming e^H or H². Similarly [[1, 2], [2, 1]] has eigenvalues 3 and −1, so tr(A⁴) = 81 + 1 = 82.

4. Tensors: transformation laws, covariant and contravariant components

A tensor is defined by how its components change when the coordinates change from xⁱ to x′ⁱ. A contravariant vector transforms like a displacement dxⁱ: A′ⁱ = (∂x′ⁱ/∂xʲ) Aʲ. A covariant vector transforms like a gradient ∂φ/∂xⁱ: A′ᵢ = (∂xʲ/∂x′ⁱ) Aⱼ. Upper indices are contravariant, lower indices covariant, and a repeated upper–lower pair is summed (Einstein convention). A rank-2 tensor carries one factor per index: T′ⁱʲ = (∂x′ⁱ/∂xᵏ)(∂x′ʲ/∂xˡ) Tᵏˡ, and a mixed tensor T′ⁱⱼ takes one factor of each kind.

Worked in one dimension: under the stretch x′ = 2x, ∂x′/∂x = 2. A contravariant component doubles (A′ = 2A — a displacement measured in half-size units is twice the number), while a covariant component halves (A′ = A/2 — a slope per new unit is half as steep). A covariant component of 6 therefore becomes 3, and a contravariant one of 6 becomes 12. Under rotations of Cartesian axes the transformation matrix is orthogonal, its inverse is its transpose, and the two kinds coincide — which is why elementary physics never needed the distinction.

Tensors a physics paper uses
Tensor	Rank and symmetry	Where it appears
Inertia tensor Iᵢⱼ	2, symmetric	L = Iω for a rigid body
Stress tensor σᵢⱼ	2, symmetric	force per area on a surface
Metric gᵢⱼ	2, symmetric	ds² = gᵢⱼ dxⁱ dxʲ; raises and lowers indices
Field tensor F^μν	2, antisymmetric, 6 components	E and B together in relativity

5. Tensor algebra: δ, ε, contraction, symmetry and the quotient law

Kronecker delta δⁱⱼ is a mixed rank-2 tensor with the same components (1 or 0) in every coordinate system. In n dimensions δᵢᵢ = n, and δᵢⱼAⱼ = Aᵢ.
Levi-Civita symbol εᵢⱼₖ is +1 for even permutations of 123, −1 for odd ones and 0 when an index repeats. It is a pseudo-tensor (it changes sign under reflection). (A × B)ᵢ = εᵢⱼₖAⱼBₖ, and εᵢⱼₖεᵢₘₙ = δⱼₘδₖₙ − δⱼₙδₖₘ, whose full contraction gives εᵢⱼₖεᵢⱼₖ = 6.
Contraction — setting one upper index equal to one lower index and summing — lowers the rank by two: Tⁱᵢ is a scalar (the trace), and AⁱBᵢ is the invariant scalar product.
Symmetry counts. A rank-2 tensor in n dimensions splits uniquely into symmetric and antisymmetric parts, with n(n + 1)/2 and n(n − 1)/2 independent components. In four dimensions that is 10 and 6 — the metric and the field tensor.
Metric. gᵢⱼ lowers an index, Aᵢ = gᵢⱼAʲ, and its inverse gⁱʲ raises one. In Minkowski space with signature (+, −, −, −), x_μ = (ct, −x, −y, −z) while x^μ = (ct, x, y, z).

🎯 The quotient law

If the product of a set of numbers Tᵢⱼ with an arbitrary tensor gives a tensor — say TᵢⱼBʲ is a covariant vector for every contravariant Bʲ — then Tᵢⱼ is itself a tensor of the rank the indices show. The inertia tensor is identified this way: L = Iω holds for every ω, and both L and ω are vectors, so I must be a rank-2 tensor.

Key takeaways

In an orthonormal basis the expansion coefficient is an inner product, cᵢ = ⟨eᵢ|v⟩, and Σ|eᵢ⟩⟨eᵢ| = 1 is completeness; ‖v‖² = Σ|cᵢ|² is Parseval.
Gram–Schmidt subtracts the projection on the normalised vectors already built; forgetting the ⟨v₁|v₁⟩ denominator is the standard error.
A similarity transformation keeps the characteristic polynomial, eigenvalues, trace, determinant and rank; it changes the components of the eigenvectors.
Hermitian: real eigenvalues, unitary diagonalisation. Unitary: |λ| = 1. Functions of matrices act on eigenvalues, so det(e^A) = e^{tr A}.
Contravariant components transform with ∂x′/∂x, covariant with ∂x/∂x′; contraction drops the rank by two; a rank-2 tensor in n dimensions has n(n + 1)/2 symmetric and n(n − 1)/2 antisymmetric components.

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.

With e₁ = (1, 1, 0)/√2, the component ⟨e₁|v⟩ of v = (1, 0, 1) along e₁ is:
1. 1/√2
2. 1
3. √2
4. 1/2
Show answer
Answer: A — 1/√2
⟨e₁|v⟩ = (1·1 + 1·0 + 0·1)/√2 = 1/√2. The value 1 is ⟨v₁|v⟩ with the unnormalised v₁ = (1, 1, 0), and 1/2 is the projection coefficient ⟨v₁|v⟩/⟨v₁|v₁⟩ — both belong to the unnormalised vector.
Gram–Schmidt is applied to v₁ = (1, 1, 0) and v₂ = (1, 0, 1). The norm of u₂ = v₂ − ⟨e₁|v₂⟩e₁, before it is normalised, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.22
u₂ = (1, 0, 1) − (1/√2)(1, 1, 0)/√2 = (½, −½, 1), so ‖u₂‖ = √(¼ + ¼ + 1) = √1.5 = 1.22. Subtracting ⟨v₁|v₂⟩v₁ = (1, 1, 0) without the denominator gives (0, −1, 1) of norm 1.41, which is not orthogonal to v₁.
For a complete orthonormal set {|eᵢ⟩}, which statement is the completeness relation?
1. Σᵢ |eᵢ⟩⟨eᵢ| = 1
2. ⟨eᵢ|eⱼ⟩ = δᵢⱼ
3. Σᵢ ⟨eᵢ|eᵢ⟩ = 1
4. |eᵢ⟩⟨eⱼ| = δᵢⱼ
Show answer
Answer: A — Σᵢ |eᵢ⟩⟨eᵢ| = 1
Completeness is the resolution of the identity, Σ|eᵢ⟩⟨eᵢ| = 1, which lets every vector be written as Σ|eᵢ⟩⟨eᵢ|v⟩. ⟨eᵢ|eⱼ⟩ = δᵢⱼ is orthonormality, a different property that an incomplete set can also have; Σ⟨eᵢ|eᵢ⟩ equals the dimension, not 1.
A′ = S⁻¹AS for an invertible S. Which of the following are necessarily the same for A and A′?
1. The trace
2. The determinant
3. The components of each eigenvector
4. The characteristic polynomial
Show answer
Answer: A — The trace; B — The determinant; D — The characteristic polynomial
det(S⁻¹AS − λ1) = det(S⁻¹(A − λ1)S) = det(A − λ1), so the characteristic polynomial — and with it the trace, determinant and eigenvalues — is unchanged. If Ax = λx then A′(S⁻¹x) = λ(S⁻¹x): the eigenvector becomes S⁻¹x, so its components change.
For A = [[1, 2], [2, 1]], the trace of A⁴ is ____.
Numerical answer — type the value.
Show answer
Answer: 82
Trace 2 and determinant −3 give eigenvalues 3 and −1, so tr(A⁴) = 3⁴ + (−1)⁴ = 81 + 1 = 82. Direct check: A² = [[5, 4], [4, 5]] and A⁴ = [[41, 40], [40, 41]], trace 82. Squaring the trace, (tr A)⁴ = 16, is the tempting wrong route.
H = [[2, i], [−i, 2]]. The value of det(e^H), to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 54.6
H is Hermitian with eigenvalues 1 and 3 (trace 4, determinant 3). e^H has eigenvalues e¹ and e³, so det(e^H) = e⁴ = 54.6 — the identity det(e^A) = e^{tr A}. Exponentiating each entry of H is not the matrix exponential and gives a meaningless number.
Every eigenvalue λ of a unitary matrix satisfies:
1. |λ| = 1
2. λ is real
3. λ = ±1 only
4. λ is purely imaginary
Show answer
Answer: A — |λ| = 1
If Ux = λx then ‖Ux‖ = ‖x‖ because U preserves norms, so |λ|‖x‖ = ‖x‖ and |λ| = 1: λ = e^iθ anywhere on the unit circle. Real eigenvalues belong to Hermitian matrices; ±1 is only the special case of a real symmetric orthogonal matrix.
For the Pauli matrix σ_x = [[0, 1], [1, 0]], e^iθσ_x equals:
1. cos θ · 1 + i sin θ · σ_x
2. [[e^iθ, 0], [0, e^iθ]]
3. cosh θ · 1 + sinh θ · σ_x
4. [[1, e^iθ], [e^iθ, 1]]
Show answer
Answer: A — cos θ · 1 + i sin θ · σ_x
σ_x² = 1, so the exponential series splits into even powers, Σ(iθ)^2k/(2k)! = cos θ, times 1, and odd powers, i sin θ, times σ_x. The hyperbolic form belongs to e^θσ_x without the i, and exponentiating entry by entry is not a matrix function.
The number of independent components of a symmetric rank-2 tensor in four dimensions is ____.
Numerical answer — type the value.
Show answer
Answer: 10
A symmetric tensor is fixed by its diagonal (n entries) and one triangle (n(n − 1)/2), so n(n + 1)/2 = 4·5/2 = 10 — the count for the metric of space-time. The antisymmetric part has n(n − 1)/2 = 6, the count for the electromagnetic field tensor; 16 is the unrestricted count.
In one dimension the coordinate is changed from x to x′ = 2x. A covariant vector has component A₁ = 6 in the old coordinate. Its component in the new coordinate is ____.
Numerical answer — type the value.
Show answer
Answer: 3
Covariant: A′ = (∂x/∂x′)A = (1/2)(6) = 3, as a gradient per unit length halves when the unit halves. A contravariant component would use ∂x′/∂x = 2 and become 12 — reversing the two laws is the trap.
In three dimensions, with summation over repeated indices, which of the following are correct?
1. δᵢᵢ = 3
2. εᵢⱼₖδⱼₖ = 0
3. εᵢⱼₖεᵢⱼₖ = 6
4. AᵢBᵢ is a vector
Show answer
Answer: A — δᵢᵢ = 3; B — εᵢⱼₖδⱼₖ = 0; C — εᵢⱼₖεᵢⱼₖ = 6
δᵢᵢ = 1 + 1 + 1 = 3. εᵢⱼₖδⱼₖ = εᵢⱼⱼ = 0 because ε vanishes when two indices repeat. εᵢⱼₖεᵢⱼₖ counts the 6 non-zero entries, each squared to 1. AᵢBᵢ has no free index, so it is the scalar A·B, not a vector.
Contracting the mixed tensor Tⁱⱼₖ on its indices i and j gives:
1. A covariant vector
2. A scalar
3. A contravariant vector
4. A rank-2 tensor
Show answer
Answer: A — A covariant vector
Contraction removes one upper and one lower index and lowers the rank by two: rank 3 becomes rank 1, and the surviving index k is a lower one, so Tⁱᵢₖ is a covariant vector. A scalar needs the rank to fall to zero, which takes a rank-2 tensor.
It is known that Lᵢ = Iᵢⱼωⱼ holds for every angular velocity ω, and that L and ω are vectors. What does this establish about Iᵢⱼ?
1. It is a rank-2 tensor, by the quotient law
2. It is a scalar multiple of δᵢⱼ
3. It is necessarily antisymmetric
4. Nothing, unless ω is fixed
Show answer
Answer: A — It is a rank-2 tensor, by the quotient law
The quotient law: if contracting Iᵢⱼ with an arbitrary vector always yields a vector, Iᵢⱼ transforms as a rank-2 tensor. It is symmetric, not antisymmetric, and it is a multiple of δᵢⱼ only for bodies such as a sphere or a cube about its centre. The "for every ω" is what makes the law apply.