Matrix & Tensor

Notes Maths

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Matrices

Vector space

Definition

A set of elements on which vector addition and scalar multiplication are defined
Closed under these operations
Includes identity $\mathbf{0}$ for vector addition

Note:

2D spaces $\mathbb{R}^2 \neq \mathbb{C}$ , since $\mathbb{C}$ has a single point of $\infty$ .

Span

The set of all vectors that are linear combinations of any vectors of a set $\mathbb{S}$ .

Linearly independent

None of the vectors is a linear combination of others, or

$x_1\mathbf{e}_1+x_2\mathbf{e}_2+x_3\mathbf{e}_3+...=0 \iff x_1=x_2=x_3=...=0$

Basis

A set that spans $\mathbb{V}$ and are linearly independent.

The number of elements is the dimension of $\mathbb{V}$ .
Any vector can be expressed as a linear combination of the basis set in a unique way.
Can have infinite dimensions - e.g. Fourier series

Change of bases

Primed / unprimed opposite to lecture notes convention.

Idea: the same vector has different components in different bases.

$\mathbf{e}_j'=\mathbf{e}_i\mathbf{R}_{ij}$

where $\mathbf{R}_{ij}$ is the $i$ th component of $\mathbf{e}_j'$ in the unprimed basis, i.e.

$\mathbf{R}_{ij}= \begin{pmatrix} \uparrow & \uparrow & \uparrow\\ \mathbf{e_1'} & \mathbf{e_2'} &\mathbf{e_3'}\\ \downarrow &\downarrow &\downarrow \end{pmatrix}$

Check by reversing the argument

$\begin{pmatrix} \uparrow & \uparrow & \uparrow\\ \mathbf{e_1'} & \mathbf{e_2'} &\mathbf{e_3'}\\ \downarrow &\downarrow &\downarrow \end{pmatrix} \begin{pmatrix} \uparrow \\ \mathbf{e'}\\ \downarrow \end{pmatrix} =\begin{pmatrix} \uparrow \\ \mathbf{e'}\,\mathrm{in}\,\mathbf{e}\\ \downarrow \end{pmatrix}$

Now,

$\mathbf{x}=x_i\mathbf{e}_i=x_j'\mathbf{e}_j'=x_j'\mathbf{e}_i\mathbf{R}_{ij}\\ \implies x_i=\mathbf{R}_{ij}x_j'$

Inverse

Define the transformations

$\mathbf{e}_j'=\mathbf{e}_i\mathbf{R}_{ij} \\ \mathbf{e}_j=\mathbf{e}_i'\mathbf{S}_{ij}$

Sub in,

$\mathbf{e}_j=\mathbf{e}_k\mathbf{R}_{ki}\mathbf{S}_{ij}$

True iff $\mathbf{R}_{ki}\mathbf{S}_{ij}=\delta_{kj}$ , similarly, $\mathbf{S}_{ki}\mathbf{R}_{ij}=\delta_{kj}$ . Hence, inverse transform is represented by inverse matrix.

Transforming matrices

$\begin{align} \mathbf{A}\mathbf{x}=\mathbf{e}_i\mathbf{A}_{ij}x_j&=\mathbf{e}_i'\mathbf{A'}_{ij}x_j'\\ \mathbf{e}_i\mathbf{A}_{ij}x_j&=\mathbf{e}_k\mathbf{R}_{ki}\mathbf{A'}_{ij}x_j'\\ \mathbf{A}\mathbf{x}&=\mathbf{R}\mathbf{A'}\mathbf{R}^{-1}\mathbf{x'}\\ \mathbf{A}&=(\mathbf{R}\mathbf{A'})\mathbf{R}^{-1}\\ \end{align}$

Linear Operation

$\mathbf{A}(\alpha\mathbf{x}+\mathbf{y})=\alpha\mathbf{A}\mathbf{x}+\mathbf{A}\mathbf{y}$

Without reference to any bases.

Inner Product

$<\!\!x|y\!\!>=x_i^*y_i$

Linear in 2nd argument
Anti-linear in 1st argument
Hermitian: $<\!\!x|y\!\!>^*=<y|x>$
Positive-definite: $<\!\!x|x\!\!>\geq0$

Inner Product and bases

$<\!\!x|y\!\!>=<\!\!x_i\mathbf{e}_i|y_j\mathbf{e}_j\!\!>=x_i^*y_j<\!\!\mathbf{e}_i|\mathbf{e}_j\!\!>=x_i^*y_j\mathbf{G}_{ij}$

Metric coefficient: $\mathbf{G}_{ij}$ , hermitian

Cauchy-Schwartz

$|<\!\!\mathbf{x}|\mathbf{y}\!\!>|\leq||\mathbf{x}||\,||\mathbf{y}||$

Proof by considering $<\!\!\mathbf{x}-\alpha\mathbf{y}|\mathbf{x}-\alpha\mathbf{y}\!\!>$ , then choose appropriate phase of $\alpha$ , so that $\alpha<\!\!\mathbf{x}|\mathbf{y}\!\!>$ is real and positive; appropriate length of $\alpha=\frac{|\mathbf{x}|}{|\mathbf{y}|}$ .

Hermitian Conjugate

$\mathbf{A}^\dagger=(\mathbf{A}^T)^*\\ \mathbf{AB}^\dagger=\mathbf{B}^\dagger\mathbf{A}^\dagger$

An operator is adjoint if $<\!\!\mathbf{A}^\dagger \mathbf{x}|\mathbf{y}\!\!>=<\!\!\mathbf{x}|\mathbf{A} \mathbf{y}\!\!>$ .

Special square matrices

Type	Relation
Real Symmetric	$A_{ij}=A_{ji}$
Real Antisymmetric	$A_{ij}=-A_{ji}$
Orthogonal	$\mathbf{A}^\mathrm{T}=\mathbf{A}^{-1}$
Hermitian	$\mathbf{A}^\dagger=\mathbf{A}$
Anti-hermitian	$\mathbf{A}^\dagger=-\mathbf{A}$
Unitary	$\mathbf{A}^\dagger=\mathbf{A}^{-1}$
Normal	$\mathbf{A}^\dagger\mathbf{A}=\mathbf{A}\mathbf{A}^\dagger$

Special results

If A is Hermitian then iA is anti-Hermitian
if A is Hermitian then exp(iA) is unitary

c.f. real number

Eigenvectors

Two things to remember:

$\mathbf{Ax}=\lambda\mathbf{x}\\ \mathbf{AS}=\mathbf{S\Lambda}$

Always find Eigenvalues first, by solving characteristic equation $\det{(\mathbf{A}-\lambda\mathbf{I})}=0$ .
Roots
1. n distinct roots - n linearly independent vectors
2. m repeated roots - may be 1 to m vectors corresponding to it, and any linear combinations of those are Eigenvectors.

Derivation of Eigenvalue, Eigenvector properties of special matrices.

Diagonalisation

Diagonalisation is also a similarity transformation, into the Eigenvector basis.

Only diagonalisable if the matrix has n independent eigenvectors.

Normal matrices can be diagonalised by unitary transformation. - Transformation between orthonormal bases is described by a unitary vector.

Tr and Det are most easily studied in the diagonal form, invariant under similarity transformations.

Forms

Quadratic form

Any homogeneous quadratic funtion is a quadratic form of a symmetric matrix.

$Q(\mathbf{x})=\mathbf{x}^T\mathbf{A}\mathbf{x}=A_{ij}x_i x_j$

By diagonalisation, we can get a $Q(\mathbf{x}')$ with no cross-terms.

The eigenvectors of A defines the principal axes of the quadratic form.
Positive (semi-)definite means all eigenvalues are larger than (or equal to) 0.

Quadratic surface

$Q(\mathbf{x})=k$

Ellipsoids - $\lambda$ same sign
Hyperboloids - $\lambda$ different signs
$\lambda_1=\lambda_2=\lambda_3$ - sphere
$\lambda_1=\lambda_2$ - revolution about z’ (axis of symmetry)
$\lambda_3=0$ - conic section translated along z’

Hermitian Form

Similar idea for complex vector space.

$H(\mathbf{x})=\mathbf{x}^\dagger\mathbf{H}\mathbf{x}=\sum\lambda_i|x_i'|^2$

Stationary property

Define Rayleigh quotient

$\lambda(\mathbf{x})=\frac{\mathbf{x}^\dagger\mathbf{A}\mathbf{x}}{\mathbf{x}^\dagger\mathbf{x}}$

$\lambda$ is a scalar
$\lambda(\alpha\mathbf{x})=\lambda(\mathbf{x})$
When $\mathbf{x}$ is eigenvector, $\lambda$ is eigenvalue.

Rayleigh-Ritz

Eigenvalues of a matrix is the stationery values of the Rayleigh quotient.

Proof by considering $\lambda(\mathbf{x}+\delta \mathbf{x})-\lambda(\mathbf{x})$ .

Cartesian Tensor

Use the transformation matrix $L$ :

$L_{ij}=\mathbf{e}_i'\cdot\mathbf{e}_j\\ \mathbf{L}= \begin{pmatrix} \leftarrow & \mathbf{e_1'} & \rightarrow\\ \leftarrow & \mathbf{e_2'} &\rightarrow\\ \leftarrow &\mathbf{e_3'} &\rightarrow \end{pmatrix} =\begin{pmatrix} \uparrow & \uparrow & \uparrow\\ ... & \mathbf{e_i}\,\mathrm{in}\,\mathbf{e_i}' &...\\ \downarrow &\downarrow &\downarrow \end{pmatrix} \\$

So the transformation law for a vector is (only this form makes sense for me…)

$v_i'=L_{ij}v_j= \begin{pmatrix} \uparrow & \uparrow & \uparrow\\ ... & \mathbf{e_i}\,\mathrm{in}\,\mathbf{e_i}' &...\\ \downarrow &\downarrow &\downarrow \end{pmatrix} \begin{pmatrix} \uparrow\\ \mathbf{v}\\\downarrow \end{pmatrix}$

Definition in terms of transformation laws

Vector: a set of coefficents $v_i$ , defined wrt orthonormal basis $\mathbf{e}_i$ , such that $v_i'$ wrt another basis $\mathbf{e}_i'$ are given by $v_i'=L_{ij}v_j$ .

Axial vectors: $v_i'=\det(\mathbf{L})L_{ij}v_j$

Tensors: $T_{ijk...}'=L_{ia}L_{jb}L_{kc}...T_{abc...}$

Pseudo-tensor: $T_{ijk...}'=\det(\mathbf{L})L_{ia}L_{jb}L_{kc}...T_{abc...}$

Examples

Everything is checked by the transformation law above.

Delta is a 2nd order tensor
Epsilon is a 3rd order pseudo tensor
Inertia tensor: $I_{ij}=\int_V\rho(\mathbf{x})(x_kx_k\delta_{ij}-x_ix_i)\,dV$

Derive from the angular momentum definition.

Operations

Addition
Outer product
Inner product (contraction)

Note: Cross-product is defined using the epsilon symbol, so any cross product is a pseudo-tensor.

Symmetric and antisymmetric properties are invariant.
- Contraction of symmetric & antisymmetric indices $S_{ijk...}A_{ijr...} = 0$ .

Proof by using symmetry properties once and relabel once.

Second-order tensors

Two new things

Antisymmetric (3D)

Only three degrees of freedom, hence can be represented by a vector.

$A_{ij}=\epsilon_{ijk}\omega_k$

The vector $\omega$ is a dual vector, defined by

$\omega_k=\frac{1}{2}\epsilon_{klm}A_{lm}$

Symmetric (3D)

Can be further decomposed, uniquely, into a traceless symmetric matrix and identity.

$S=\tilde{S}+\frac{\mathrm{Tr}(S)}{3}I$

Isotropic tensors

Tensors or pseudo-tensors with the same components in all frames.

Scalars are isotropic
No non-zero rank 1 isotropic tensor
Rank 2 isotropic tensors are $\lambda\delta_{ij}$

Proof by (wlog) considering a rotation $\pi/2$ by z-axis and by y-axis.

Rank 3 isotropic tensors are $\lambda\epsilon_{ijk}$
Rank 4 isotropic tensors are $\lambda\delta_{ij}\delta_{kl}+\mu\delta_{ik}\delta_{jl}+\nu\delta_{il}\delta_{jk}$

Applications

Since invariant, we can pick the most convenient frame (e.g eigenvector frame).
Integrals

Integral 1

$X_i=\int_{r\leq a}x_i\rho(r)\,dV$

Relabel the integration variables as primed. Transform $x_i'=R_{ij}x_j$ we find $X_i=X_i'$ . But only isotropic rank 1 tensor is $\mathbf{0}$ .

Integral 2

$K_{ij}=\int_{r\leq a}x_ix_j\rho(r)\,dV$

This is also isotropic, so $K_{ij}=\lambda\delta_{ij}$ .

$K_{ij}=\frac{1}{3}(\int_{r\leq a}r^2\rho(r)\,dV)\delta_{ij}$

Only need to do 1 integral.

Tensor fields

$\nabla$ , with components $\partial_i$ , is a vector in Cartesian only.

One idea (maybe) new

The derivative of a second-order tensor $\sigma_{ij}$ is a rank 3 tensor field $\partial_i\sigma_{jk}$ .

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Author：2LalA猪

Permalink：https://butteraddict.fun/2019/03/Matrix-Tensor/

Published on：30th March 2019, 4:19 pm

Updated on：9th June 2020, 3:24 pm

License：CC BY 4.0

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