Lecture 03: Change of Basis

Overview

Steven M. Wise builds a careful foundation in tensor algebra for continuum mechanics, defining transpose, symmetry, positive definiteness, invertibility, orthogonality, and rotations without relying on components. He proves the symmetric and skew-symmetric decomposition and the correspondence between skew-symmetric tensors and vectors in three dimensions, then derives orthogonal change-of-basis rules and reviews invariant trace, determinant, and tensor-exponential properties.

Key takeaways

The transpose of a tensor can be defined independently of any matrix representation by requiring (S U) · V = (Sᵀ V) · U for all vectors U and V.
Every second-order tensor decomposes uniquely as S = ½(S + Sᵀ) + ½(S - Sᵀ), separating its symmetric and skew-symmetric parts.
In three dimensions, skew-symmetric tensors and vectors are equivalent through W V = w × V; both representations have exactly three independent degrees of freedom.
A change of basis between right-handed orthonormal frames is represented by an orthogonal matrix A, and vector coordinates transform as v = A v′ and v′ = Aᵀv.
Trace and determinant are invariant under orthogonal changes of frame, and a tensor is invertible exactly when its determinant is nonzero.
Exponentiating a skew-symmetric tensor produces a rotation tensor, linking tensor exponentials directly to orientation-preserving rotations.

Chapters

0:00 Tensor Transpose, Symmetry, and Positive Definiteness

A second-order tensor transpose is defined without components by the identity (S U) · V = (Sᵀ V) · U for arbitrary vectors U and V.
A tensor is symmetric when Sᵀ = S and skew-symmetric when Sᵀ = -S.
Positive definiteness means V · (S V) > 0 for every nonzero V; symmetric positive definite tensors are abbreviated SPD.

5:15 Transpose Uniqueness, Inverses, and Singular Tensors

The transpose definition requires proving that a tensor satisfying the vector identity exists and is unique; frame vectors and orthonormality offer a route to the proof.
An invertible tensor S has a tensor S⁻¹ satisfying both S S⁻¹ = I and S⁻¹ S = I, and any such inverse is unique.
If S is invertible, then (Sᵀ)⁻¹ = (S⁻¹)ᵀ; tensors without inverses are called singular.

9:04 Orthogonal Tensors and Handedness-Preserving Rotations

An orthogonal tensor Q satisfies Qᵀ = Q⁻¹, equivalently QᵀQ = I; the discussion concerns real tensors rather than complex unitary matrices.
Orthogonality preserves dot products, while a rotation tensor also preserves the right-handed orientation of every right-handed orthonormal frame.
A rotation is characterized by orthogonality together with determinant +1; orthogonal tensors more generally have determinant magnitude 1.

12:12 Tensor Components and the Symmetric-Skew Decomposition Theorem

In any frame, the matrix representing Sᵀ is the transpose of the matrix representing S; for invertible S, the matrix of S⁻¹ is the inverse matrix.
Wise states the decomposition result: every second-order tensor S has unique tensors E and W such that S = E + W, with E symmetric and W skew-symmetric.
The proposed components are E = ½(S + Sᵀ) and W = ½(S - Sᵀ), whose sum recovers S.

16:05 Proving Existence and Uniqueness of the Tensor Decomposition

Existence follows by taking the transpose of E and W and using (S + T)ᵀ = Sᵀ + Tᵀ, (αS)ᵀ = αSᵀ, and (Sᵀ)ᵀ = S.
For uniqueness, suppose S = E₁ + W₁ = E₂ + W₂; then δ = E₁ - E₂ = W₂ - W₁ is both symmetric and skew-symmetric.
A tensor that is both symmetric and skew-symmetric must be zero, so E₁ = E₂ and W₁ = W₂.

22:50 Matrix Example of Symmetric and Skew-Symmetric Parts

Wise applies the formulas E = ½(S + Sᵀ) and W = ½(S - Sᵀ) to a displayed component matrix.
The symmetric part retains the original diagonal entries, while each pair of off-diagonal entries becomes equal in E and opposite in W.
The example checks that Eᵀ = E, Wᵀ = -W, and E + W reproduces the original matrix.

24:48 Cross Products and the Axial Vector Correspondence

The three-dimensional cross product is linear in both vectors and follows the orientation rules of a right-handed frame.
For each skew-symmetric tensor W, there is a unique vector w such that W V = w × V for every vector V.
Conversely, every vector w defines a unique skew-symmetric tensor whose action is w × V; component formulas use the permutation symbol.

27:32 Proving the Skew Tensor and Vector Mapping

Starting from a skew-symmetric W, the proof constructs w from W's components and uses the permutation-symbol identity to recover W.
Skew symmetry supplies the sign change needed in the component calculation, confirming that the constructed vector reproduces the tensor's action.
Uniqueness follows by assuming two vectors produce the same action, subtracting their component identities, and showing all three component differences vanish.
In the reverse direction, the component formula builds a skew-symmetric tensor from any w and verifies its action matches w × V.

38:15 Axial Vector Example and Three Degrees of Freedom

A displayed skew-symmetric tensor acts on a sample vector to give the same components as the cross product of its axial vector with that vector.
A three-dimensional skew-symmetric tensor has three independent degrees of freedom: zero diagonal entries and paired off-diagonal entries related by sign.
Wise names the symmetric-part and skew-part maps, and defines vectorization and tensorization as inverse maps between axial vectors and skew-symmetric tensors.

45:59 Change-of-Basis Tensor and Frame Vector Relations

Wise introduces two right-handed orthonormal frames, F and F′, and defines a change-of-basis tensor using their basis vectors and direction cosines.
The primed basis vector can be expanded in the unprimed frame as e′ⱼ = aᵢⱼ eᵢ, with coefficients found by dotting against the unprimed frame vectors.
The reverse frame expansion expresses each unprimed basis vector using the primed basis and the corresponding direction-cosine coefficients.

55:48 Orthogonality of the Change-of-Basis Matrix

Substituting the two frame expansions into one another yields sums of direction-cosine products equal to Kronecker deltas.
In matrix notation, the change-of-basis matrix A satisfies AᵀA = I and AAᵀ = I, so it is orthogonal.
The component derivation relies on uniqueness of basis representations and careful distinction between free and summed indices.

1:01:11 Vector Coordinate Changes Between Frames

For a vector with components v in F and v′ in F′, the change-of-basis matrix gives vᵢ = aᵢⱼv′ⱼ.
The reverse conversion is v′ₖ = aᵢₖvᵢ, which corresponds to multiplication by Aᵀ.
The lecture simplifies cumbersome frame-subscript notation by using primes to indicate components in F′.

1:05:36 Second-Order Tensor Change of Basis by Similarity

A tensor's matrix representations in F and F′ are related by an orthogonal similarity transformation.
To convert the representation from primed to unprimed coordinates, the formula sandwiches the primed matrix between A and Aᵀ.
The reverse conversion uses the corresponding Aᵀ S A product.

1:06:59 Frame-Invariant Trace and Determinant Properties

The trace is the sum of diagonal entries in a frame, but the orthogonal similarity transformation shows that this sum is unchanged between F and F′.
The determinant of a tensor equals the determinant of its matrix representation in any frame and is invariant under the change of basis.
A tensor is invertible exactly when its determinant is nonzero; determinants multiply across tensor products, and det(ST) = det(TS).
An orthogonal tensor has determinant ±1, while a rotation tensor has determinant +1.

1:12:55 Tensor Exponential of a Skew-Symmetric Tensor

The tensor exponential is defined by a power series that converges for every second-order tensor.
For skew-symmetric W, exp(W) is a rotation tensor, so it is orthogonal and has determinant +1.
The transpose commutes with tensor exponentiation in the sense that (exp W)ᵀ = exp(Wᵀ).

Transcript excerpt

0:04 Well, welcome back to Continuum Mechanics. We're still in chapter 1. This is our third lecture. And I at first I guess that we would do three lectures, maybe even two. But uh I'm realizing at the pace I want to go, I'm going to it's going to take me five lectures to cover what's in chapter one. The reason is is because I want to lay a nice clear careful foundation so that everything we do after that isn't just us traveling at lightning speed and and um you not being able to hold on or understand what we're doing. So let's form a good foundation in tensor algebra and then uh maybe we'll go a little faster as we go along. But if if you're

0:52 comfortable at this pace, then uh we'll we'll stick with it. I just want to make sure that we c that we cover the big equations by the end of the of the end of the the the semester. That is Navier Stokes equations and the um Koshi Navier equations for elasticity. Anyway, part three of five. Today we're talking about special classes of tensors. And we've already you you know you know pretty much everything you need to know in this course about matrices because you come with good prerequisite knowledge in in matrices. What does it mean to uh take the transpose of a tensor that has no components for

1:40 example? No. Remember when we think about tensors, we don't have to think about their components. They're just objects. Whenever we introduce a frame, then we can talk about the components of that tensor in the frame. But a tensor is just an object that does something. Okay, it acts on vectors to produce vectors. So what does it mean then if I

This is the start of the transcript. Save this video free for the whole transcript, searchable and time-coded.

In this course

Related notes

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Steven M. Wise.