Lecture 04: Eigenvalues and Special Decompositions
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Overview
Steven M. Wise develops eigenvalue theory for second-order tensors, showing how symmetric tensors admit orthogonal eigenvector frames and how trace, determinant, and a third principal invariant encode their characteristic polynomial. He then derives positive-definite square roots and both forms of polar decomposition, defines the tensor inner product, and introduces fourth-order tensors as linear maps on second-order tensors, including their 81-component coordinate representation.
Key takeaways
- For a symmetric tensor, eigenvectors belonging to distinct eigenvalues are orthogonal, and the spectral theorem supplies a right-handed orthonormal eigenvector frame in which the tensor is diagonal.
- The three principal invariants encode the characteristic polynomial: I₁ = tr(S), I₂ = ½[(tr S)² − tr(S²)], and I₃ = det(S).
- Every SPD tensor has a unique SPD square root, obtained by preserving its eigenvectors and taking the positive square root of each eigenvalue.
- For invertible F with positive determinant, F = RU = VR, where U = √(FᵀF), V = √(FFᵀ), and both decompositions use the same proper rotation R.
- The tensor inner product tr(SᵀD) becomes a sum of corresponding components in an orthonormal frame; symmetric and skew-symmetric tensors are orthogonal under it.
- A fourth-order tensor in three dimensions has 81 components and acts linearly on second-order tensors through Uᵢⱼ = CᵢⱼₖₗTₖₗ.
Chapters
- An eigenpair consists of a scalar λ and unit vector e satisfying Se = λe.
- Applying S to e only stretches or reverses that direction, so its one-dimensional subspace remains invariant.
- Eigenvectors can be normalized by dividing by their norm, and changing their sign gives another unit eigenvector.
- For a second-order tensor S, the characteristic polynomial is P(α) = det(S − αI), a cubic in three dimensions.
- An eigenvalue λ makes S − λI singular, so the homogeneous system (S − λI)e = 0 has nontrivial solutions.
- A real tensor may have complex eigenvalues, which occur in conjugate pairs; the lecture sets them aside for continuum mechanics.
- For an SPD tensor, e·Se = λ for a unit eigenvector e, and positive definiteness therefore implies λ > 0.
- For symmetric S, eigenvectors u and w associated with distinct eigenvalues λ and μ satisfy (λ − μ)(u·w) = 0.
- Since λ ≠ μ, the eigenvectors are orthogonal; these results apply to real eigenpairs.
- The spectral decomposition theorem gives every symmetric second-order tensor a right-handed orthonormal frame of eigenvectors.
- The three eigenvalues are fixed up to ordering and are counted with multiplicity, including repeated values such as three eigenvalues equal to 2.
- In the eigenvector frame, S is diagonal, or equivalently S = Σᵢ λᵢ eᵢ ⊗ eᵢ.
- The first principal invariant is I₁(S) = tr(S), and the third is I₃(S) = det(S).
- The second invariant is I₂(S) = ½[(tr S)² − tr(S²)].
- Trace, determinant, and trace of S² are basis-invariant quantities, so all three principal invariants are independent of the chosen frame.
- The characteristic polynomial can be written P(α) = −α³ + I₁α² − I₂α + I₃.
- For a symmetric tensor with eigenvalues λ₁, λ₂, λ₃, I₁ is their sum and I₃ is their product.
- The second invariant is the pairwise sum λ₁λ₂ + λ₂λ₃ + λ₃λ₁, connecting the polynomial coefficients directly to eigenvalues.
- Every symmetric positive-definite tensor C has a unique symmetric positive-definite tensor U satisfying U² = C.
- In C’s orthonormal eigenvector frame, the square root keeps the eigenvectors and replaces each positive eigenvalue λᵢ with √λᵢ.
- Positive eigenvalues make real square roots possible, while negative eigenvalues would require complex values.
- For a tensor F with det(F) > 0, the right polar decomposition is F = RU, where U = √(FᵀF).
- The left polar decomposition is F = VR, where V = √(FFᵀ).
- Both U and V are symmetric positive-definite tensors, and R is a rotation with determinant +1.
- Invertibility of F and Fᵀ means a nonzero vector w cannot be mapped to zero by either tensor.
- The quadratic form w·(FᵀF)w equals (Fw)·(Fw), which is strictly positive for nonzero w.
- The same argument makes FFᵀ positive definite, and reversing the order under transposition shows both products are symmetric.
- Define R = FU⁻¹; because U² = FᵀF and U is symmetric, RᵀR simplifies to the identity.
- From F = RU and det(F) > 0 with det(U) > 0, it follows that det(R) > 0.
- An orthogonal real tensor has determinant ±1, so positive determinant makes R a rotation with det(R) = +1.
- Define Q = V⁻¹F to obtain the left factorization F = VQ, then compare it with F = RU.
- Wise uses S = QᵀV and the orthogonality of Q to relate R to S and U.
- The resulting identity U² = S², together with the uniqueness of the SPD square root, gives S = U and R = Q.
- The second-order tensor inner product is defined as S:D = tr(SᵀD).
- A frame with three basis vectors yields nine dyads eᵢ ⊗ eⱼ, which form a basis for second-order tensors.
- The dyads are orthonormal under the tensor inner product, with pairings represented by Kronecker deltas.
- Writing S and D in the dyadic basis reduces their inner product to the component sum Σᵢⱼ SᵢⱼDᵢⱼ.
- The component formula shows that the tensor inner product is symmetric in S and D.
- A symmetric tensor is orthogonal under this product to every skew-symmetric tensor, so S:D = S:sym(D) when S is symmetric.
- A fourth-order tensor maps a second-order tensor to another second-order tensor and must act linearly.
- In three dimensions, four indices each ranging from 1 to 3 give 3⁴ = 81 components.
- Addition and scalar multiplication are defined through the maps’ outputs, while multiplying fourth-order tensors means composing their actions.
- With an orthonormal frame, components Cᵢⱼₖₗ are defined by applying C to a dyad and taking the corresponding vector or tensor component.
- If U = C[T], the coordinate rule is Uᵢⱼ = CᵢⱼₖₗTₖₗ, with summation over k and l.
- This generalizes matrix-vector multiplication, replacing one summed index with two; the equivalent dyadic inner-product formula is previewed for the next class.
Transcript excerpt
0:05 All right. Well, welcome back to continue mechanics. This is class number four. We're still in chapter one talking about tensors and tensor algebra. We should finish up with this chapter by next time. So, that will be our fifth class. Got a lot of material to cover today. Fortunately or unfortunately, we're we're making good progress at good speed. Um, I'll try to slow down at the harder parts, but um things are going to start accumulating and things are going to be not necessarily get getting more complicated, but uh just uh the details are going to start piling up on each other. Today we're going to start talking about igen values, igen vectors
0:54 and principal invariance among other things. What is an igen value? So you've you've seen the definition for this before. An igen pair in fact for a second order tensor is a scalar and a vector. So the scalar well so let's back up a little bit. Let's take the tensor to be s. The scalar is labeled lambda and the vector is a unit vector labeled e. Those two things to be an igen pair have to satisfy this equation. See equals lambda E. So you can see what happens here. Um S takes this vector E and and maps it into a scalar multiple of E. So this just says that in this um direction I'm
1:44 simply stretching out this vector E. And since um since uh this operation is linear, this mat this this tensor is linear. This shows that every vector in this subspace along this line or along this direction E stays in this direction stays in this subspace. So it leaves this subspace invariant. this one-dimensional subspace here lambda is called the igen value of s and e is called the associated igen
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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.