Matrices and second order tensors

An anisotropic material responds differently in different directions. A scalar coefficient cannot encode that behavior, but a linear map can. A second-order tensor in this Euclidean setting is the coordinate-independent linear relation; a matrix is its representation in one chosen orthonormal frame. Changing the frame changes the entries, not the physical relation.

1 Linear systems and matrix maps

A matrix $A\in\mathbb R^{m\times n}$ maps $x$ to $Ax$, with $(Ax)_i=\sum_j a_{ij}x_j$. Composition gives matrix multiplication. Solving $Ax=b$ by Gaussian elimination uses reversible operations on the augmented matrix: swap rows, scale by nonzero constants and add row multiples. A contradictory zero-coefficient row gives no solution; otherwise free variables describe nonuniqueness.

For a square matrix, invertibility means every $b$ has a unique solution. A nonzero vector with $Az=0$ proves failure of uniqueness, since $A(x+z)=Ax$. This is physically important: an inverse reconstruction cannot distinguish two states separated by a null direction.

An eigenvector satisfies $Av=\lambda v$ with $v\ne0$. It describes a direction preserved by the map up to scale. Symmetric matrices have an orthonormal eigenbasis. One proof maximizes $v^TAv$ on the unit sphere, obtains an eigenvector by tangent differentiation, observes that its perpendicular complement is invariant under symmetry, and inducts. This gives principal directions for a symmetric material tensor.

2 Tensor components and coordinate changes

Let old and new orthonormal coordinates satisfy $x=Qx'$, where $Q^TQ=I$. If a linear relation is $y=Ax$, then $y'=Q^Ty$ and

$$y'=Q^TAQx',\qquad A'=Q^TAQ.$$

This is the transformation law for the linear-map tensor's components in orthonormal frames. An arbitrary array is not automatically a tensor unless its role and transformation law are specified. General nonorthonormal coordinate changes require distinguishing vector and covector components; the simple orthogonal formula is not being asserted for every type of tensor in arbitrary coordinates.

The dyadic product $a\otimes b$ is the map $v\mapsto a(b\cdot v)$, represented by $ab^T$. For an orthonormal eigenbasis $q_i$ of symmetric $A$,

$$A=\sum_i\lambda_i q_i\otimes q_i.$$

The identity is $\sum_iq_i\otimes q_i$. Trace and determinant are frame-invariant: $\operatorname{tr}(Q^TAQ)=\operatorname{tr}(AQQ^T)=\operatorname{tr}A$ by finite index sums, and determinants multiply to give $\det A'=\det A$. The sum of eigenvalues is the trace and their product the determinant.

3 Worked anisotropic transport model

Suppose the flux relation is $j=-K\nabla c$ with

$$K=\begin{pmatrix}3&1\\1&3\end{pmatrix}.$$

The unit directions $q_1=(1,1)/\sqrt2$ and $q_2=(1,-1)/\sqrt2$ have eigenvalues 4 and 2. In that principal frame, $K'=\operatorname{diag}(4,2)$. A concentration gradient $(1,0)$ produces $j=(-3,-1)$, which is not parallel to the gradient. Anisotropy changes direction as well as magnitude.

For every nonzero $g$, $g^TKg=4(g\cdot q_1)^2+2(g\cdot q_2)^2>0$, so $j\cdot\nabla c=-\nabla c^TK\nabla c\le0$. Under this constitutive model, flux has a down-gradient component. Positivity is a property to check, not a consequence of calling an array a transport tensor.

In a simplified continuum stress model, traction on a unit-normal plane is $t(n)=\sigma n$. It is the normal component $n^T\sigma n$ that gives signed normal traction; the remaining vector $\sigma n-(n^T\sigma n)n$ is tangential. For symmetric $\sigma$, an eigenvector direction has zero tangential traction. The physical law and sign conventions must be supplied by the application; this chapter studies their linear algebra.

4.3 Symmetry gives orthogonal eigenvectors

For real symmetric $A$ and real vectors $u,v$, if $Au=\lambda u$, $Av=\mu v$, then

$$\lambda u^Tv=(Au)^Tv=u^TA^Tv=u^TAv=\mu u^Tv.$$

Distinct eigenvalues therefore have orthogonal eigenspaces. The stronger spectral theorem states that every real symmetric $n\times n$ matrix, where $n\ge1$, has an orthonormal eigenbasis, hence $A=QDQ^T$ with $Q^TQ=I$ and real diagonal $D$.

Here is a proof using earlier chapters. The continuous function $x^TAx$ attains a maximum on the unit sphere, which is closed and bounded. At a maximizing unit vector $v$, the regular constraint $x^Tx=1$ has nonzero gradient $2v$. The multiplier condition gives $2Av=2\lambda v$, so $v$ is a real unit eigenvector. Its perpendicular subspace is invariant: if $v^Tw=0$, then $v^TAw=(Av)^Tw=\lambda v^Tw=0$. Restrict $A$ to that $(n-1)$-dimensional subspace in an orthonormal basis; the restriction remains symmetric. Induction, beginning with dimension one, supplies an orthonormal eigenbasis there. Together with $v$, it proves the theorem. An orthonormal basis for the perpendicular subspace is constructed by the Gram–Schmidt method.

Closing the maximizing-vector and orthogonal-complement steps

The maximizing-vector step can also be checked without a general multiplier theorem. If $v$ is a unit maximizer and $w\cdot v=0$, the curve $q(t)=(v+tw)/\|v+tw\|$ lies on the sphere and has $q'(0)=w$. Differentiating $q(t)^TAq(t)$ at its maximum gives $2w^TAv=0$. Decompose $Av=(v^TAv)v+w_0$ with $w_0\perp v$ and choose $w=w_0$; then $\|w_0\|^2=0$. Thus $Av=(v^TAv)v$, a real eigenpair. In dimension one the conclusion is immediate.

To construct the complement basis used in the induction, extend $v$ to a linear basis by adding standard coordinate vectors outside the current span. Successively subtract the projections onto previously obtained orthonormal vectors and divide the nonzero remainder by its norm. The remainder cannot vanish because the starting list is linearly independent. The resulting vectors after $v$ form an orthonormal basis of $v^\perp$. The restriction matrix has entries $u_i^TAu_j=u_j^TAu_i$, so is symmetric. This supplies the induction's construction and hypothesis, including the empty complement in dimension one.

4 Graded practice

E1 Foundation. Solve $\begin{pmatrix}2&1\\1&3\end{pmatrix}x=(4,7)^T$.

E2 Tensor components. For $A=\operatorname{diag}(4,1)$ and $Q=\frac1{\sqrt2}\begin{pmatrix}1&-1\\1&1\end{pmatrix}$, compute $Q^TAQ$ and check its trace and determinant.

E3 Traction. For $\sigma=\begin{pmatrix}5&2\\2&1\end{pmatrix}$ and $n=(1,0)$, find traction and its normal and tangential parts.

E4 Proof. Show $a\otimes a$ is symmetric positive semidefinite and identify its rank when $a\ne0$.

5 Full solutions

S1. From $2x+y=4$ and $x+3y=7$, substitute $y=4-2x$ to get $-5x=-5$. Thus $x=1$, $y=2$, and direct substitution checks both equations.

S2. $Q^TAQ=\begin{pmatrix}5/2&-3/2\\-3/2&5/2\end{pmatrix}$. Trace is 5 and determinant is $25/4-9/4=4$, matching the original. Off-diagonal components can appear purely because axes rotate.

S3. $t=\sigma n=(5,2)$. The normal component vector is $(5,0)$ with scalar 5; the tangential component is $(0,2)$. A coordinate-axis plane need not be a principal plane.

S4. Its matrix is $aa^T$, which equals its transpose. For any $v$, $v^Taa^Tv=(a^Tv)^2\ge0$. Its image is the line spanned by $a$, and it maps $a$ to $\|a\|^2a\ne0$, so rank is 1.

Summary

Matrices represent linear maps and support inverse problems. Second-order tensor components transform with the coordinate frame; the map itself does not. Symmetry supplies principal directions, while positivity gives meaningful directional inequalities in transport models. Always state the tensor type, coordinates and physical convention.