Tensors

Tensors - Introduction

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1 - How do basis vector components change WRT change of basis?

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We define:

• old basis: {𝑒₁, 𝑒₂}
• new basis: {𝑒̃₁, 𝑒̃₂}

We define:

• forward transform: old basis → new basis

  𝑒̃1 = 𝑎·𝑒1+ 𝑏·𝑒2 𝑒̃2 = 𝑐·𝑒1+ 𝑑·𝑒2

  $\left[\begin{array}{cc}{e}_1& e_2\end{array}\right]\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]=\left[\begin{array}{cc}{\tilde{e}}_1& {\tilde{e}}_2\end{array}\right]$

• backward transform: new basis → old basis

  𝑒1 = 𝑎·𝑒̃1+ 𝑏·𝑒̃2 𝑒2 = 𝑐·𝑒̃1+ 𝑑·𝑒̃2

  $\left[\begin{array}{cc}{\tilde{e}}_1& {\tilde{e}}_2\end{array}\right]\left[\begin{array}{cc}e& g\\ f& h\end{array}\right]=\left[\begin{array}{cc}{e}_1& e_2\end{array}\right]$

Forward transform is the inverse of Backward transform

• $\left[\begin{array}{cc}{e}_1& e_2\end{array}\right]\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]=\left[\begin{array}{cc}{\tilde{e}}_1& {\tilde{e}}_2\end{array}\right]$
• $\left[\begin{array}{cc}{\tilde{e}}_1& {\tilde{e}}_2\end{array}\right]\left[\begin{array}{cc}e& g\\ f& h\end{array}\right]\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]=\left[\begin{array}{cc}{\tilde{e}}_1& {\tilde{e}}_2\end{array}\right]$

Thus

• ${\tilde{e}}_j={\sum }_{i=1}^n{F}_{ij}{e}_i$
• $e_j={\sum }_{i=1}^n{B}_{ij}{\tilde{e}}_i$

Confirming the inverse behavior of the above equations

• $e_j={\sum }_{i=1}^n{B}_{ij}{\tilde{e}}_i$
• $e_j={\sum }_{i=1}^n{B}_{ij}{\sum }_{k=1}^n{F}_{ki}{e}_k$
• $e_j={\sum }_{k=1}^n{\sum }_{i=1}^n{B}_{ij}{F}_{ki}{e}_k$
• $e_j={\sum }_{k=1}^n{𝛿}_{jk}{e}_k$

Thus 𝛿_𝑖𝑗 is the kronecker delta function:

• ${𝛿}_{jk}={\sum }_{i=1}^n{B}_{ij}{F}_{ki}$

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2 - How do vector components change WRT change of basis?

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• 𝑣 = 𝑣[1] · 𝑒₁ + 𝑣[2] · 𝑒₂
• 𝑣̃ = 𝑣̃[1] · 𝑒̃₁ + 𝑣̃[2] · 𝑒̃₂
• 𝑣 = 𝑣̃ are geometrically the same vector

$v=\tilde{v}={\sum }_{i=1}^{n}v[i]\cdot e_i$ $v=\tilde{v}={\sum }_{i=1}^n\tilde{v}[i]\cdot {\tilde{e}}_i$

$v=\tilde{v}={\sum }_{j=1}^{n}v[j]\cdot e_j$ $v=\tilde{v}={\sum }_{j=1}^{n}v[j]\cdot {\sum }_{i=1}^n{B}_{ij}{\tilde{e}}_i$ $v=\tilde{v}={\sum }_{i=1}^n{\sum }_{j=1}^n{B}_{ij}v[j]{\tilde{e}}_i$ Thus $\tilde{v}[i]={\sum }_{j=1}^n{B}_{ij}v[j]$	$v=\tilde{v}={\sum }_{j=1}^n\tilde{v}[j]\cdot {\tilde{e}}_j$ $v=\tilde{v}={\sum }_{j=1}^n\tilde{v}[j]\cdot {\sum }_{i=1}^n{F}_{ij}{e}_i$ $v=\tilde{v}={\sum }_{i=1}^n{\sum }_{j=1}^n{F}_{ij}\tilde{v}[j]e_i$ Thus $v[i]={\sum }_{j=1}^n{F}_{ij}\tilde{v}[j]$
Thus
𝐵 is used to transform vector components from old to new	𝐹 is used to transform vector components from new to old

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3 - Covector Introduction

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Covectors are functions 𝛼: 𝑉→ℝ that map a vector to a number and also obey the following rules:

• 𝛼(𝑣 + 𝑢) = 𝛼(𝑣) + 𝛼(𝑢)
• 𝛼(𝑛·𝑣) = 𝑛·𝛼(𝑣)

Covectors can also be viewed as elements of dual vector space 𝑉*:

• (𝑛·𝛼)(𝑣) = 𝑛·𝛼(𝑣)
• (𝛼+𝛽)(𝑣) = 𝛼(𝑣) + 𝛽(𝑣)

Covectors can be visualized as level sets

What does a covector measure when we write [2 1]? like 2 of what and 1 of what?

Covectors don’t live in the vector space 𝑉, thus we can’t use basis vectors in 𝑉 like {𝑒₁, 𝑒₂} to measure covectors

epsilon covectors 𝜀𝑖 are defined as: 𝜀1(𝑒1) = 1 𝜀1(𝑒2) = 0 𝜀2(𝑒1) = 0 𝜀2(𝑒2) = 1 Thus: 𝜀𝑖(𝑒𝑗) = 𝛿𝑖𝑗 # the Kronecker delta function

In other words, let: 𝐸 = [𝑒1|𝑒2] a matrix whose columns are the basis vectors {𝑒1, 𝑒2} 𝐸ˆ = [𝜀1|𝜀2] a matrix whose columns are the epsilon covectors {𝜀1, 𝜀2} The system of equations can be expressed as: ${\hat{E}}^{T}E=I$ $\left[\begin{array}{cc}{𝜀}_1^1& {𝜀}_2^1\\ {𝜀}_1^2& {𝜀}_2^2\end{array}\right]\left[\begin{array}{cc}{e}_{11}& e_{21}\\ e_{12}& e_{22}\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

epsilon covector 𝜀𝑖 consumes arbitrary vector 𝑣:
𝜀1(𝑣) = 𝜀1(𝑣[1]·𝑒1 + 𝑣[2]·𝑒2) 𝜀1(𝑣) = 𝑣[1] · 𝜀1(𝑒1) + 𝑣[2] · 𝜀1(𝑒2) 𝜀1(𝑣) = 𝑣[1]	𝜀2(𝑣) = 𝜀2(𝑣[1]·𝑒1 + 𝑣[2]·𝑒2) 𝜀2(𝑣) = 𝑣[1] · 𝜀2(𝑒1) + 𝑣[2] · 𝜀2(𝑒2) 𝜀2(𝑣) = 𝑣[2]

arbitrary covector 𝛼 consumes arbitrary vector 𝑣: 𝛼(𝑣) = 𝛼(𝑣[1]·𝑒1 + 𝑣[2]·𝑒2) 𝛼(𝑣) = 𝑣[1]·𝛼(𝑒1) + 𝑣[2]·𝛼(𝑒2) 𝛼(𝑣) = 𝜀1(𝑣)·𝛼(𝑒1) + 𝜀2(𝑣)·𝛼(𝑒2) 𝛼(𝑣) = 𝜀1(𝑣)·𝛼1 + 𝜀2(𝑣)·𝛼2 𝛼(𝑣) = 𝛼1·𝜀1(𝑣) + 𝛼2·𝜀2(𝑣) 𝛼(𝑣) = (𝛼1·𝜀1 + 𝛼2·𝜀2) (𝑣) Thus 𝛼 = 𝛼1·𝜀1 + 𝛼2·𝜀2 Thus any arbitrary covector 𝛼 can be expressed as a linear combination of basis epsilon covectors {𝜀1, 𝜀2}

The epsilon covectors 𝜀1 and 𝜀2 form the basis vectors of the dual vector space 𝑉*: 𝜀1 and 𝜀2 are linearly independent 1. proof 1, 𝜀2} from the dual vector space 𝑉* is linearly independent if: scalars {𝑎1, 𝑎2} must be ALL zero such that: 𝑎1𝜀1 + 𝑎2𝜀2 = 𝜀0 # where 𝜀0is the null functional (i.e. zero vector in 𝑉*) We need to prove the above statement. The expression 𝑎1𝜀1 + 𝑎2𝜀2 = 𝜀0 naturally leads to (𝑎1𝜀1 + 𝑎2𝜀2)(𝑣) = 𝜀0(𝑣); for all 𝑣∊𝑉. Where we have both sides “consume” an arbitrary vector 𝑣∊𝑉: (𝑎1𝜀1 + 𝑎2𝜀2)(𝑣) = 𝜀0(𝑣); for all 𝑣∊𝑉 (𝑎1𝜀1 + 𝑎2𝜀2)(𝑣) = 0; for all 𝑣∊𝑉 # by assumption that 𝜀0(𝑣) = 0 for all 𝑣∊𝑉 𝑎1𝜀1(𝑣) + 𝑎2𝜀2(𝑣) = 0; for all 𝑣∊𝑉 # because they are linear transformations Thus the original statement that we need to prove now becomes: scalars {𝑎1, 𝑎2} must be ALL zero such that: 𝑎1𝜀1(𝑣) + 𝑎2𝜀2(𝑣) = 0; for all 𝑣∊𝑉 Since the new statement is true for every vector 𝑣∊𝑉, it must also be true for the basis vectors {𝑒1, 𝑒2}: that the scalars {𝑎1, 𝑎2} must be ALL zero such that: 𝑎1𝜀1(𝑒𝑖) + 𝑎2𝜀2(𝑒𝑖) = 0; for all 𝑒𝑖∊{𝑒1, 𝑒2} Evaluating the above statement for each basis vector yields: that the scalars {𝑎1, 𝑎2} must be ALL zero such that: for 𝑒𝑖=𝑒1: 𝑎1𝜀1(𝑒1) + 𝑎2𝜀2(𝑒1) = 0 𝑎1*1 + 𝑎2*0 = 0 𝑎1 = 0 for 𝑒𝑖=𝑒1: 𝑎1𝜀1(𝑒2) + 𝑎2𝜀2(𝑒2) = 0 𝑎1*0 + 𝑎2*1 = 0 𝑎2 = 0 Thus, 𝑎1 and 𝑎2 must equal to 0. A set of covectors {𝜀 every arbitrary covector 𝛼∊𝑉* can be expressed as a linear combination of {𝜀1, 𝜀2} see above

The components of covector 𝛼 can be extracted as follows: 𝛼 = 𝛼1·𝜀1 + 𝛼2·𝜀2 𝛼(𝑒1) = 𝛼1 𝛼(𝑒2) = 𝛼2 Thus: the components of an arbitrary covector (e.g. [2 1]) are measured by the number of covector/level-set lines that the basis vectors {𝑒1, 𝑒2} pierces

RECAP:

• starting with basis vectors 𝑒_𝑖 of a vector space 𝑉
• you can derive the epsilon covectors 𝜀^𝑖 as so: 𝜀^𝑖(𝑒_𝑗) = 𝛿_𝑖𝑗
• in which the epsilon covectors form a dual basis of the dual vector space 𝑉*

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4 - How do covector components change WRT change of basis?

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Let’s define:

𝜀1(𝑒1) = 1 = 𝜀̃1(𝑒̃1) 𝜀1(𝑒2) = 0 = 𝜀̃1(𝑒̃2) 𝜀2(𝑒1) = 0 = 𝜀̃2(𝑒̃1) 𝜀2(𝑒2) = 1 = 𝜀̃2(𝑒̃2)

𝜀𝑖(𝑒𝑗) = 𝛿𝑖𝑗 = 𝜀̃𝑖(𝑒̃𝑗)

Thus

𝛼 = 𝛼̃ = 𝛼1·𝜀1 + 𝛼2·𝜀2	𝛼 = 𝛼̃ = 𝛼̃1·𝜀̃1 + 𝛼̃2·𝜀̃2
𝛼(𝑒1) = 𝛼1 𝛼(𝑒2) = 𝛼2	𝛼(𝑒̃1) = 𝛼̃1 𝛼(𝑒̃2) = 𝛼̃2

Thus

𝛼 = 𝛼1·𝜀1 + 𝛼2·𝜀2 𝛼 = 𝛼(𝑒1)·𝜀1 + 𝛼(𝑒2)·𝜀2

𝛼 = 𝛼̃1·𝜀̃1 + 𝛼̃2·𝜀̃2 𝛼 = 𝛼(𝑒̃1)·𝜀̃1 + 𝛼(𝑒̃2)·𝜀̃2

Thus

$𝛼={\sum }_{j=1}^n{𝛼}_j\cdot {𝜀}^j$ $𝛼={\sum }_{j=1}^n𝛼(e_j)\cdot {𝜀}^j$ $𝛼={\sum }_{j=1}^n𝛼({\sum }_{i=1}^n{B}_{ij}{\tilde{e}}_i)\cdot {𝜀}^j$ $𝛼={\sum }_{j=1}^n{\sum }_{i=1}^n{B}_{ij}\cdot 𝛼({\tilde{e}}_i)\cdot {𝜀}^j$ $𝛼={\sum }_{j=1}^n{\sum }_{i=1}^n{B}_{ij}\cdot {\widetilde{𝛼}}_i\cdot {𝜀}^j$	$𝛼={\sum }_{j=1}^n{\widetilde{𝛼}}_j\cdot {\widetilde{𝜀}}^j$ $𝛼={\sum }_{j=1}^n𝛼({\tilde{e}}_j)\cdot {\widetilde{𝜀}}^j$ $𝛼={\sum }_{j=1}^n𝛼({\sum }_{i=1}^n{F}_{ij}{e}_i)\cdot {\widetilde{𝜀}}^j$ $𝛼={\sum }_{j=1}^n{\sum }_{i=1}^n{F}_{ij}\cdot 𝛼(e_i)\cdot {\widetilde{𝜀}}^j$ $𝛼={\sum }_{j=1}^n{\sum }_{i=1}^n{F}_{ij}\cdot {𝛼}_i\cdot {\widetilde{𝜀}}^j$
${𝛼}_j={\sum }_{i=1}^n{B}_{ij}\cdot {\widetilde{𝛼}}_i$	${𝛼}_j={\sum }_{i=1}^n{F}_{ij}\cdot {𝛼}_i$

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5 - How do dual basis covectors change WRT change of basis?

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We define:

• 𝜀^𝑖(𝑒_𝑗) = 𝛿_𝑖𝑗 = 𝜀̃^𝑖(𝑒̃_𝑗)

By definition of basis

${\widetilde{𝜀}}^i={\sum }_{j=1}^n{𝛼}_j\cdot {𝜀}^j$ ${\widetilde{𝜀}}^i={\sum }_{j=1}^n{\widetilde{𝜀}}^i(e_j)\cdot {𝜀}^j$ ${\widetilde{𝜀}}^i={\sum }_{j=1}^n{\widetilde{𝜀}}^i({\sum }_{k=1}^n{B}_{kj}\cdot {\tilde{e}}_k)\cdot {𝜀}^j$ ${\widetilde{𝜀}}^i={\sum }_{j=1}^n{\sum }_{k=1}^n{B}_{kj}\cdot {\widetilde{𝜀}}^i({\tilde{e}}_k)\cdot {𝜀}^j$ ${\widetilde{𝜀}}^i={\sum }_{j=1}^n{\sum }_{k=1}^n{B}_{kj}\cdot {𝛿}_{ik}\cdot {𝜀}^j$

${𝜀}^i={\sum }_{j=1}^n{\widetilde{𝛼}}_j\cdot {\widetilde{𝜀}}^j$ ${𝜀}^i={\sum }_{j=1}^n{𝜀}^i({\tilde{e}}_j)\cdot {\widetilde{𝜀}}^j$ ${\widetilde{𝜀}}^i={\sum }_{j=1}^n{𝜀}^i({\sum }_{k=1}^n{F}_{kj}\cdot e_k)\cdot {\widetilde{𝜀}}^j$ ${\widetilde{𝜀}}^i={\sum }_{j=1}^n{\sum }_{k=1}^n{F}_{kj}\cdot {𝜀}^i(e_k)\cdot {\widetilde{𝜀}}^j$ ${𝜀}^i={\sum }_{j=1}^n{\sum }_{k=1}^n{F}_{kj}\cdot {𝛿}_{ik}\cdot {\widetilde{𝜀}}^j$

${\widetilde{𝜀}}^i={\sum }_{j=1}^n{B}_{ij}\cdot {𝜀}^j$

${𝜀}^i={\sum }_{j=1}^n{F}_{ij}\cdot {\widetilde{𝜀}}^j$

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6 - Linear Maps Introduction

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A linear map 𝐿: maps vectors to vectors, 𝐿: 𝑉→𝑊 (e.g. 𝐿: 𝑉→𝑉) add inputs or the outputs, 𝐿(𝑣+𝑤) = 𝐿(𝑣) + 𝐿(𝑤) scale the inputs or outputs, 𝐿(𝓃·𝑣) = 𝓃·𝐿(𝑣) The Matrix Multiplication can be purely Derived from the Linearity Rules Above 𝑤 = 𝐿(𝑣) = 𝐿(𝑣1𝑒1 + 𝑣2𝑒2) 𝑤 = 𝐿(𝑣) = 𝑣1𝐿(𝑒1) + 𝑣2𝐿(𝑒2) 𝐿(𝑒1) = 𝐿11𝑒1 + 𝐿21𝑒2 𝐿(𝑒2) = 𝐿12𝑒1 + 𝐿22𝑒2 𝑤 = 𝐿(𝑣) = 𝑣1(𝐿11𝑒1 + 𝐿21𝑒2) + 𝑣2(𝐿12𝑒1 + 𝐿22𝑒2) 𝑤 = 𝐿(𝑣) = 𝑣1𝐿11𝑒1 + 𝑣1𝐿21𝑒2 + 𝑣2𝐿12𝑒1 + 𝑣2𝐿22𝑒2 𝑤 = 𝐿(𝑣) = (𝐿11𝑣1+ 𝐿12𝑣2)𝑒1  +  (𝐿21𝑣1+ 𝐿22𝑣2)𝑒2 𝑤 = 𝐿(𝑣) = (        𝑤1)𝑒1  +  (         𝑤2)𝑒2 Thus: 𝑤1 = 𝐿11𝑣1+ 𝐿12𝑣2 𝑤2 = 𝐿21𝑣1+ 𝐿22𝑣2 Thus: $\left[\begin{array}{c}{w}_1\\ w_2\end{array}\right]=\left[\begin{array}{cc}{L}_1^1& L_2^1\\ L_1^2& L_2^2\end{array}\right]\left[\begin{array}{c}{v}_1\\ v_2\end{array}\right]$ Basis of Linear Maps {𝑒1𝜀1,  𝑒1𝜀2,  𝑒2𝜀1,  𝑒2𝜀2} is one possible basis for linear map 𝐿: ℝ2→ℝ2 where: $e_1{𝜀}^1=\left[\begin{array}{c}1\\ 0\end{array}\right]\left[\begin{array}{cc}1& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$ $e_1{𝜀}^2=\left[\begin{array}{c}1\\ 0\end{array}\right]\left[\begin{array}{cc}0& 1\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ $e_2{𝜀}^1=\left[\begin{array}{c}0\\ 1\end{array}\right]\left[\begin{array}{cc}1& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 1& 0\end{array}\right]$ $e_2{𝜀}^2=\left[\begin{array}{c}0\\ 1\end{array}\right]\left[\begin{array}{cc}0& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$ Thus $L=L_1^1{e}_1{𝜀}^1+L_2^1{e}_1{𝜀}^2+L_1^2{e}_2{𝜀}^1+L_2^2{e}_2{𝜀}^2$ 𝐿 = 𝐿𝑖𝑗 𝑒𝑖𝜀𝑗 # Einstein’s notation Thus: linear maps can be written as linear combinations of vector-covector pairs How is a Basis 𝑒𝑖𝜀𝑗 a Linear Map (That Eats a Vector and Outputs a Vector)? Given: 𝐿 = 𝐿𝑖𝑗 𝑒𝑖𝜀𝑗 𝑣 = 𝑣𝑘𝑒𝑘 Let 𝐿 eat a vector 𝑣: 𝐿(𝑣) = 𝐿(𝑣) 𝐿(𝑣) = 𝐿𝑖𝑗𝑒𝑖𝜀𝑗(𝑣𝑘𝑒𝑘) 𝐿(𝑣) = 𝐿𝑖𝑗𝑣𝑘𝑒𝑖𝜀𝑗(𝑒𝑘) 𝐿(𝑣) = 𝐿𝑖𝑗𝑣𝑘𝑒𝑖𝛿𝑗𝑘 𝐿(𝑣) = 𝐿𝑖𝑗𝑣𝑗𝑒𝑖 𝐿(𝑣) = 𝑤𝑖𝑒𝑖 # 𝐿𝑖𝑗𝑣𝑗is a scalar, thus replace with 𝑤𝑖= 𝐿𝑖𝑗𝑣𝑗 𝐿(𝑣) = 𝑤 # which outputs a vector

Given a linear transformation 𝐿, the set of vectors for which 𝐿 vanishes is called the KERNEL of 𝐿 kernel(𝐿) = {𝑥∊𝑋 : 𝐿𝑥 = 0 in 𝑌} range (𝐿) = {𝑦∊𝑌 : ∃𝑥∊𝑋 such that 𝑇𝑥 = 𝑦} Given a linear transformation 𝐿: kernel of 𝐿 is a linear subspace of 𝑋 range  of 𝐿 is a linear subspace of 𝑌 The dimension of range(𝐿) is called the rank of 𝐿 (i.e. rank(𝑇)) Matrices: Null-Spaces, Column-Spaces, Row Spaces null-space of 𝐴 is the kernel of 𝐴 the kernel space is orthogonal to the row vectors/space column space of 𝐴 is the range of 𝐴 is the subspace of ℝ𝑚 spanned by column vectors row space of 𝐴 is the subspace of ℝ𝑛 spanned by row vectors The Rank-Nullity Theorem Given a linear transformation 𝐿 from ℝ𝑛 to ℝ𝑚, then: dim ker(𝐿) + dim range(𝐿) = 𝑛

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7 - How do linear transformations change WRT change of basis?

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Detailed

$L(e_j)={\sum }_{i=1}^n{L}_j^i{e}_i$

${\tilde{e}}_j={\sum }_{i=1}^n{F}_j^i{e}_i$ $e_j={\sum }_{i=1}^n{B}_j^i{\tilde{e}}_i$

$L({\tilde{e}}_i)=L({\tilde{e}}_i)$ $L({\tilde{e}}_i)=L({\sum }_{k=1}^n{F}_i^k{e}_k)$ $L({\tilde{e}}_i)={\sum }_{k=1}^n{F}_i^{k}L(e_k)$ $L({\tilde{e}}_i)={\sum }_{k=1}^n{F}_i^k{\sum }_{j=1}^n{L}_k^j{e}_j$ $L({\tilde{e}}_i)={\sum }_{k=1}^n{\sum }_{j=1}^n{F}_i^k{L}_k^j{e}_j$ $L({\tilde{e}}_i)={\sum }_{k=1}^n{\sum }_{j=1}^n{F}_i^k{L}_k^j{\sum }_{t=1}^n{B}_j^t{\tilde{e}}_t$ $L({\tilde{e}}_i)={\sum }_{k=1}^n{\sum }_{j=1}^n{\sum }_{t=1}^n{F}_i^k{L}_k^j{B}_j^t{\tilde{e}}_t$ ${\sum }_{t=1}^{n}L{˜}_i^t{\tilde{e}}_t={\sum }_{k=1}^n{\sum }_{j=1}^n{\sum }_{t=1}^n{F}_i^k{L}_k^j{B}_j^t{\tilde{e}}_t$	Einstein’s notation: $L({\tilde{e}}_i)=L({\tilde{e}}_i)$ $L({\tilde{e}}_i)=L(F_i^j{e}_i)$ $L({\tilde{e}}_i)=F_i^{j}L(e_i)$ $L({\tilde{e}}_i)=F_i^j{L}_j^k{e}_k$ $L({\tilde{e}}_i)=F_i^j{L}_j^k{B}_k^t{\tilde{e}}_t$ $L{˜}_i^t{\tilde{e}}_t=F_i^j{L}_j^k{B}_k^t{\tilde{e}}_t$ $L{˜}_i^t=F_i^j{L}_j^k{B}_k^t$ $L{˜}_i^t=B_k^t{L}_j^k{F}_i^j$
Thus
$L{˜}_i^t={\sum }_{k=1}^n{\sum }_{j=1}^n{F}_i^k{L}_k^j{B}_j^t$ $L˜=BLF$

Simple (using vector-covector pairs)

Given:

Linear Map Definition	Basis Vectors	Basis Covectors
𝐿 = 𝐿˜𝑖𝑗 𝑒̃𝑖𝜀̃𝑗 𝐿 = 𝐿𝑖𝑗 𝑒𝑖𝜀𝑗	${\tilde{e}}_j=F_j^i{e}_i$ $e_j=B_j^i{\tilde{e}}_i$	${\widetilde{𝜀}}^i=B_j^i{𝜀}^j$ ${𝜀}^i=F_j^i{\widetilde{𝜀}}^j$

Then

Then start with the definition of linear map 𝐿: 𝐿 = 𝐿𝑖𝑗 𝑒𝑖𝜀𝑗 Next, transform all the basis vectors and basis covectors individually 𝐿 = 𝐿𝑖𝑗 𝐵𝑘𝑖𝑒̃𝑘𝐹𝑗𝑙𝜀̃𝑙 𝐿 = (𝐵𝑘𝑖𝐿𝑖𝑗𝐹𝑗𝑙)𝑒̃𝑘𝜀̃𝑙 𝐿 = (    𝐿˜𝑘𝑙)𝑒̃𝑘𝜀̃𝑙 Thus: 𝐿𝑘𝑙 = 𝐹𝑘𝑖𝐿˜𝑖𝑗𝐵𝑗𝑙 𝐿˜ = 𝐵𝐿𝐹

Then start with the definition of linear map 𝐿: 𝐿 = 𝐿˜𝑖𝑗 𝑒̃𝑖𝜀̃𝑗 Next, transform all the basis vectors and basis covectors individually 𝐿 = 𝐿˜𝑖𝑗 𝐹𝑘𝑖𝑒𝑘𝐵𝑗𝑙𝜀𝑙 𝐿 = (𝐹𝑘𝑖𝐿˜𝑖𝑗𝐵𝑗𝑙)𝑒𝑘𝜀𝑙 𝐿 = (    𝐿𝑘𝑙)𝑒𝑘𝜀𝑙 Thus: 𝐿𝑘𝑙 = 𝐹𝑘𝑖𝐿˜𝑖𝑗𝐵𝑗𝑙 𝐿 = 𝐹𝐿˜𝐵

Another Way

Given:

• basis 𝑒 = {𝑒₁, …, 𝑒_𝑛}
• basis 𝑓 = {𝑓₁, …, 𝑓_𝑛}
• matrix 𝐹 = [𝑓] = [𝑓₁ … 𝑓_𝑛] where 𝑓_𝑖 are columns expressed in 𝑒

Then:

• 𝑣_𝑒 = 𝐹 𝑣_𝑓
• 𝑣_𝑓 = 𝐹^-1 𝑣_𝑒

Hence if:

• 𝑇∊𝐿(ℝ^𝑛) and matrix 𝐴_𝑒 is a realization of transformation 𝑇 expressed in basis 𝑒

What is the realization matrix 𝐴_𝑓expressed in basis 𝑓?

• by definition:
  • 𝑦_𝑓 = 𝐴_𝑓𝑥_𝑓
  • 𝑦_𝑒 = 𝐴_𝑒𝑥_𝑒
• then:
  • 𝑦_𝑓 = 𝐹^-1𝑦_𝑒
  • 𝑦_𝑓 = 𝐹^-1𝐴_𝑒𝑥_𝑒
  • 𝑦_𝑓 = 𝐹^-1𝐴_𝑒𝐹𝑥_𝑓
  • 𝐴_𝑓𝑥_𝑓 = 𝐹^-1𝐴_𝑒𝐹𝑥_𝑓
• thus:
  • 𝐴_𝑓 = 𝐹^-1𝐴_𝑒𝐹

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8 - How do basis of a linear transformation change WRT change of basis?

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TODO

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9 - Metric Tensor Introduction

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Metric Tensors

• are invariant to change of basis
• measures: length & angle

Length

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• ||𝑣||² = 𝑣·𝑣
• ||𝑣||² = (𝑣¹𝑒₁+ 𝑣²𝑒₂)·(𝑣¹𝑒₁+ 𝑣²𝑒₂)
• ||𝑣||² = (𝑣¹·𝑣¹)(𝑒₁·𝑒₁) + 2(𝑣¹·𝑣²)(𝑒₂·𝑒₂) + (𝑣²·𝑣²)(𝑒₂·𝑒₂)

$g_{e_i}={\left[\begin{array}{cc}{e}_1\cdot e_1& e_1\cdot e_2\\ e_2\cdot e_1& e_2\cdot e_2\end{array}\right]}_{e_i}$	$g_{{\tilde{e}}_i}={\left[\begin{array}{cc}{\tilde{e}}_1\cdot {\tilde{e}}_1& {\tilde{e}}_1\cdot {\tilde{e}}_2\\ {\tilde{e}}_2\cdot {\tilde{e}}_1& {\tilde{e}}_2\cdot {\tilde{e}}_2\end{array}\right]}_{{\tilde{e}}_i}$
||𝑣||2 = 𝑣𝑖𝑣𝑗(𝑒𝑖·𝑒𝑗) = 𝑣𝑖𝑣𝑗𝑔𝑖𝑗 (𝑒𝑖·𝑒𝑗) = 𝑔𝑖𝑗	||𝑣||2 = 𝑣̃𝑖𝑣̃𝑗(𝑒̃𝑖·𝑒̃𝑗) = 𝑣̃𝑖𝑣̃𝑗𝑔̃𝑖𝑗 (𝑒̃𝑖·𝑒̃𝑗) = 𝑔̃𝑖𝑗

Angles

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Say we have Unit Vectors

metric-tensor-angles.drawio

𝑒̃1 = 𝑒1 𝑒̃2 = 𝑐𝑜𝑠(𝜃)·𝑒1 + 𝑠𝑖𝑛(𝜃)·𝑒2 three possible combinations: 𝑒̃1·𝑒̃1 = 𝑒1·𝑒1= 1 𝑒̃1·𝑒̃2 = 𝑒1· (𝑐𝑜𝑠(𝜃)·𝑒1 + 𝑠𝑖𝑛(𝜃)·𝑒2) = 𝑐𝑜𝑠(𝜃)(𝑒1·𝑒1) + 𝑠𝑖𝑛(𝜃)(𝑒1·𝑒2) = 𝑐𝑜𝑠(𝜃) 𝑒̃2·𝑒̃2 = (𝑐𝑜𝑠(𝜃)·𝑒1 + 𝑠𝑖𝑛(𝜃)·𝑒2) · (𝑐𝑜𝑠(𝜃)·𝑒1 + 𝑠𝑖𝑛(𝜃)·𝑒2) = 𝑐𝑜𝑠(𝜃)2(𝑒1·𝑒1) + 𝑠𝑖𝑛(𝜃)2(𝑒2·𝑒2) + 2·𝑠𝑖𝑛(𝜃)·𝑐𝑜𝑠(𝜃)(𝑒1·𝑒2) = 𝑐𝑜𝑠(𝜃)2 + 𝑠𝑖𝑛(𝜃)2 = 1

Thus:

𝑒1·𝑒1 = 1 𝑒1·𝑒2 = 0 𝑒2·𝑒2 = 1	𝑒̃1·𝑒̃1 = 1 𝑒̃1·𝑒̃2 = 𝑐𝑜𝑠(𝜃) 𝑒̃2·𝑒̃2 = 1
$g_{e_i}={\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]}_{e_i}$	$g_{{\tilde{e}}_i}={\left[\begin{array}{cc}1& cos(𝜃)\\ cos(𝜃)& 1\end{array}\right]}_{{\tilde{e}}_i}$

Say We Have Arbitrary Vectors

metric-tensor-angles-2.drawio

First, define 2 new basis vectors {𝑒̃1, 𝑒̃2}: 𝑒̃1·𝑒̃1 = 1 𝑒̃1·𝑒̃2 = 𝑐𝑜𝑠(𝜃) 𝑒̃2·𝑒̃2 = 1 The metric tensor (𝑣·𝑤) is computed as: (𝑣·𝑤) = (𝑎·𝑒̃1)·(𝑏·𝑒̃2) = 𝑎𝑏·(𝑒̃1·𝑒̃2) = 𝑎𝑏·𝑐𝑜𝑠(𝜃) = ||𝑣||·||𝑤||·𝑐𝑜𝑠(𝜃) Thus: $\frac{v\cdot w}{||v||||w||}=cos(𝜃)$

The angle between 2 vectors can be computed entirely by the metric tensor

• (𝑣·𝑤) = (𝑣¹·𝑒₁ + 𝑣²·𝑒₂) · (𝑤¹·𝑒₁ + 𝑤²·𝑒₂)
  • = (𝑣¹·𝑒₁ + 𝑣²·𝑒₂) · (𝑤¹·𝑒₁ + 𝑤²·𝑒₂)
  • = 𝑣¹𝑤¹(𝑒₁·𝑒₁) + 𝑣¹𝑤²(𝑒₁·𝑒₂) + 𝑣²𝑤¹(𝑒₂·𝑒₁) + 𝑣²𝑤²(𝑒₂·𝑒₂)
  • = 𝑣¹𝑤¹𝑔₁₁ + 𝑣¹𝑤²𝑔₁₂ + 𝑣²𝑤¹𝑔₂₁ + 𝑣²𝑤²𝑔₂₂
  • = 𝑣^𝑖𝑤^𝑗𝑔_𝑖𝑗

Lengths & Angles Summary

• 𝑣·𝑣  = ||𝑣||²                 = 𝑣^𝑖𝑣^𝑗 𝑔_𝑖𝑗
• 𝑤·𝑤 = ||𝑤||²                = 𝑤^𝑖𝑤^𝑗 𝑔_𝑖𝑗
• 𝑣·𝑤  = ||𝑣|| ||𝑤|| 𝑐𝑜𝑠(𝜃) = 𝑣^𝑖𝑤^𝑗 𝑔_𝑖𝑗
• $\frac{v\cdot w}{||v||||w||}=cos(𝜃)$

Components of a Metric Tensor

• 𝑔_𝑖𝑗 = 𝑒_𝑖·𝑒_𝑗 = 𝑒_𝑗·𝑒_𝑖= 𝑔_𝑗𝑖

Thus the metric tensor is a symmetric matrix

Metric Tensor Algebraic Properties

• 𝑔: 𝑉⨯𝑉 → ℝ
• multiplication
  • 𝑎·(𝑣^𝑖𝑤^𝑗𝑔_𝑖𝑗) = (𝑎·𝑣^𝑖)𝑤^𝑗𝑔_𝑖𝑗 = 𝑣^𝑖(𝑎·𝑤^𝑗)𝑔_𝑖𝑗
  • 𝑎·𝑔(𝑣,𝑤) = 𝑔(𝑎·𝑣,𝑤) = 𝑔(𝑣,𝑎·𝑤) # simplified
• addition
  • 𝑔(𝑣+𝑢,𝑤) = 𝑔(𝑣,𝑤) + 𝑔(𝑢,𝑤)
  • 𝑔(𝑣,𝑤+𝑢) = 𝑔(𝑣,𝑤) + g(𝑣,𝑢)
• symmetric
  • 𝑔(𝑣,𝑤) = 𝑔(𝑤,𝑣)
• positive definite
  • 𝑔(𝑣,𝑣) = ||𝑣||² ≥ 0

Metric Tensors are a type of bilinear form with 2 additional properties:

• symmetric
• positive definite

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10 - How do Metric Tensor Components Change WRT Change of Basis

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HOW METRIC TENSOR COMPONENTS CHANGE WRT CHANGE OF BASIS

• 𝑔_𝑘𝑙= 𝑒_𝑘·𝑒_𝑙
• 𝑔̃_𝑖𝑗 = 𝑒̃_𝑖·𝑒̃_𝑗

𝑔̃𝑖𝑗 = 𝑒̃𝑖·𝑒̃𝑗 𝑔̃𝑖𝑗 = (𝐹𝑘𝑖 𝑒𝑘)·(𝐹𝑙𝑗 𝑒𝑙) 𝑔̃𝑖𝑗 = 𝐹𝑘𝑖𝐹𝑙𝑗 (𝑒𝑘·𝑒𝑙) 𝑔̃𝑖𝑗 = 𝐹𝑘𝑖𝐹𝑙𝑗 𝑔𝑘𝑙

𝑔𝑖𝑗 = 𝑒𝑖·𝑒𝑗 𝑔𝑖𝑗 = (𝐵𝑘𝑖 𝑒̃𝑘)·(𝐵𝑙𝑗 𝑒̃𝑙) 𝑔𝑖𝑗 = 𝐵𝑘𝑖𝐵𝑙𝑗 (𝑒̃𝑘·𝑒̃𝑙) 𝑔𝑖𝑗 = 𝐵𝑘𝑖𝐵𝑙𝑗 𝑔̃𝑘𝑙

CONFIRM THE SQUARED LENGTH OF A VECTOR REMAINS THE SAME WRT CHANGE OF BASIS

verify the following two statements are equivalent:

• ||𝑣||² = 𝑣^𝑖𝑣^𝑗 𝑔_𝑖𝑗
• ||𝑣||² = 𝑣̃^𝑖𝑣̃^𝑗 𝑔̃_𝑖𝑗

proof:

• ||𝑣||² = 𝑣̃^𝑖𝑣̃^𝑗 𝑔̃_𝑖𝑗
• ||𝑣||² = (𝐵^𝑖_𝑎 𝑣^𝑎) (𝐵^𝑗_𝑏 𝑣^𝑏) (𝐹^𝑘_𝑖 𝐹^𝑙_𝑗 𝑔_𝑘𝑙)
• ||𝑣||² = 𝑣^𝑎𝑣^𝑏𝑔_𝑘𝑙(𝐵^𝑖_𝑎𝐹^𝑘_𝑖 𝐵^𝑗_𝑏𝐹^𝑙_𝑗)
• ||𝑣||² = 𝑣^𝑎𝑣^𝑏𝑔_𝑘𝑙(𝛿_𝑎𝑘 𝛿_𝑏𝑙) # 𝐵𝐹 simplifies to Kronecker delta function
• ||𝑣||² = 𝑣^𝑘𝑣^𝑙𝑔_𝑘𝑙
• ||𝑣||² = 𝑣^𝑖𝑣^𝑗𝑔_𝑖𝑗

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11 - How do Basis of a Metric Tensor Change WRT Change of Basis

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TODO

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12 - Bilinear Forms Introduction

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METRIC TENSOR ALGEBRAIC PROPERTIES

• 𝑔: 𝑉⨯𝑉 → ℝ
• 𝑎·𝑔(𝑣,𝑤) = 𝑔(𝑎·𝑣,𝑤) = 𝑔(𝑣,𝑎·𝑤) # simplified
• 𝑔(𝑣+𝑢,𝑤) = 𝑔(𝑣,𝑤) + 𝑔(𝑢,𝑤)
• 𝑔(𝑣,𝑤+𝑢) = 𝑔(𝑣,𝑤) + g(𝑣,𝑢)
• 𝑔(𝑣,𝑤) = 𝑔(𝑤,𝑣)
• 𝑔(𝑣,𝑣) = ||𝑣||² ≥ 0

BILINEAR FORM DEFINITION

• 𝐵: 𝑉⨯𝑉 → ℝ
• 𝑎·𝐵(𝑣,𝑤) = 𝐵(𝑎·𝑣,𝑤) = 𝐵(𝑣,𝑎·𝑤)
• 𝐵(𝑣+𝑢,𝑤) = 𝐵(𝑣,𝑤) + 𝐵(𝑢,𝑤)
• 𝐵(𝑣,𝑤+𝑢) = 𝐵(𝑣,𝑤) + 𝐵(𝑣,𝑢)

thus bilinear forms are (0,2)-tensors:

• ${\tilde{\mathcal{B}}}_{ij}=F_i^k{F}_j^l{\mathcal{B}}_{kl}$
• ${\mathcal{B}}_{ij}=B_i^k{B}_j^l{\tilde{\mathcal{B}}}_{kl}$

FORMS are functions that take vectors as input and output a field

• the linear form takes in 1 vector
• the bilinear forms take in 2 vectors

BILINEAR FORMS

• are linear combinations of covector-covector pairs
  • $\mathcal{B}={\mathcal{B}}_{ij}{𝜀}^i{𝜀}^j$

How do Bilinear Form components Change WRT change of Basis?

see: Tensor - 13 - How do Bilinear Form Components Change WRT Change of Basis

Why a row of rows?

• $\left[\begin{array}{cc}{𝛼}_1& {𝛼}_2\end{array}\right]\otimes \left[\begin{array}{cc}{𝛽}_1& {𝛽}_2\end{array}\right]$
• $\left[\begin{array}{cc}\left[\begin{array}{cc}{𝛼}_1& {𝛼}_2\end{array}\right]{𝛽}_1& \left[\begin{array}{cc}{𝛼}_1& {𝛼}_2\end{array}\right]{𝛽}_2\end{array}\right]$
• $\left[\begin{array}{cc}\left[\begin{array}{cc}{𝛼}_1{𝛽}_1& {𝛼}_2{𝛽}_1\end{array}\right]& \left[\begin{array}{cc}{𝛼}_1{𝛽}_2& {𝛼}_2{𝛽}_2\end{array}\right]\end{array}\right]$

Bilinear Forms V⨯𝑉 → ℝ:

• $\mathcal{B}(v,w)\to {\mathcal{B}}_{ij}{v}^i{w}^j=\left[\begin{array}{cc}{v}^1& v^2\end{array}\right]\left[\begin{array}{cc}{\mathcal{B}}_{11}& {\mathcal{B}}_{12}\\ {\mathcal{B}}_{21}& {\mathcal{B}}_{22}\end{array}\right]\left[\begin{array}{c}{w}^1\\ w^2\end{array}\right]$
• $\mathcal{B}(v,w)\to \left[\begin{array}{c}\left[\begin{array}{cc}{\mathcal{B}}_{11}& {\mathcal{B}}_{12}\end{array}\right]\left[\begin{array}{cc}{\mathcal{B}}_{21}& {\mathcal{B}}_{22}\end{array}\right]\end{array}\right]\left[\begin{array}{c}{v}^1\\ v^2\end{array}\right]\left[\begin{array}{c}{w}^1\\ w^2\end{array}\right]$
• $\mathcal{B}(v,w)\to \left[\begin{array}{c}\left[\begin{array}{cc}{\mathcal{B}}_{11}& {\mathcal{B}}_{12}\end{array}\right]v^1+\left[\begin{array}{cc}{\mathcal{B}}_{21}& {\mathcal{B}}_{22}\end{array}\right]v^2\end{array}\right]\left[\begin{array}{c}{w}^1\\ w^2\end{array}\right]$
• $\mathcal{B}(v,w)\to \left[\begin{array}{c}\left[\begin{array}{cc}{\mathcal{B}}_{11}{v}^1& {\mathcal{B}}_{12}{v}^1\end{array}\right]+\left[\begin{array}{cc}{\mathcal{B}}_{21}{v}^2& {\mathcal{B}}_{22}{v}^2\end{array}\right]\end{array}\right]\left[\begin{array}{c}{w}^1\\ w^2\end{array}\right]$
• $\mathcal{B}(v,w)\to \left[\begin{array}{cc}{\mathcal{B}}_{11}{v}^1+{\mathcal{B}}_{21}{v}^2& {\mathcal{B}}_{12}{v}^1+{\mathcal{B}}_{22}{v}^2\end{array}\right]\left[\begin{array}{c}{w}^1\\ w^2\end{array}\right]$
• $\mathcal{B}(v,w)\to {\mathcal{B}}_{11}{v}^1{w}^1+{\mathcal{B}}_{21}{v}^2{w}^1+{\mathcal{B}}_{12}{v}^1{w}^2+{\mathcal{B}}_{22}{v}^2{w}^2$
• $\mathcal{B}(v,w)\to {\mathcal{B}}_{ij}{v}^i{w}^j$

Basis of a Bilinear Form

{𝜀¹𝜀¹,  𝜀¹𝜀²,  𝜀²𝜀¹,  𝜀²𝜀²} is one possible basis for bilinear form 𝐵: ℝ²→ℝ where:

• ${𝜀}^1{𝜀}^1=\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{cc}1& 0\end{array}\right]=\left[\begin{array}{cc}\left[\begin{array}{cc}1& 0\end{array}\right]1& \left[\begin{array}{cc}1& 0\end{array}\right]0\end{array}\right]=\left[\begin{array}{cc}\left[\begin{array}{cc}1& 0\end{array}\right]& \left[\begin{array}{cc}0& 0\end{array}\right]\end{array}\right]$
• ${𝜀}^1{𝜀}^2=\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{cc}0& 1\end{array}\right]=\left[\begin{array}{cc}\left[\begin{array}{cc}1& 0\end{array}\right]0& \left[\begin{array}{cc}1& 0\end{array}\right]1\end{array}\right]=\left[\begin{array}{cc}\left[\begin{array}{cc}0& 0\end{array}\right]& \left[\begin{array}{cc}1& 0\end{array}\right]\end{array}\right]$
• ${𝜀}^2{𝜀}^1=\left[\begin{array}{cc}0& 1\end{array}\right]\left[\begin{array}{cc}1& 0\end{array}\right]=\left[\begin{array}{cc}\left[\begin{array}{cc}0& 1\end{array}\right]1& \left[\begin{array}{cc}0& 1\end{array}\right]0\end{array}\right]=\left[\begin{array}{cc}\left[\begin{array}{cc}0& 1\end{array}\right]& \left[\begin{array}{cc}0& 0\end{array}\right]\end{array}\right]$

Thus

• 𝐵 = 𝐵¹₁𝜀¹𝜀¹+ 𝐵¹₂𝜀¹𝜀² + 𝐵²₁𝜀²𝜀¹ + 𝐵²₂𝜀²𝜀²
• 𝐵 = 𝐵^𝑖_𝑗 𝜀^𝑖𝜀^𝑗 # Einstein’s notation

Thus: bilinear forms can be written as linear combinations of covector-covector pairs

How is a Basis 𝜀^𝑖𝜀^𝑗 a Bilinear Form (That Eats 2 Vectors and Outputs a Scalar)?

Given: 𝐵 = 𝐵𝑖𝑗 𝜀𝑖𝜀𝑗 𝑣 = 𝑣𝑘𝑒𝑘 𝑤 = 𝑤𝑙𝑒𝑙 Let 𝐵 eat a vectors 𝑣 and 𝑤: 𝐵(𝑣,𝑤) = 𝐵(𝑣,𝑤) 𝐵(𝑣,𝑤) = 𝐵𝑖𝑗 𝜀𝑖𝜀𝑗(𝑣𝑘𝑒𝑘𝑤𝑙𝑒𝑙) 𝐵(𝑣,𝑤) = 𝐵𝑖𝑗𝑣𝑘𝑤𝑙𝜀𝑖(𝑒𝑘) 𝜀𝑗(𝑒𝑙) 𝐵(𝑣,𝑤) = 𝐵𝑖𝑗𝑣𝑘𝑤𝑙𝛿𝑖𝑘𝛿𝑗𝑙 𝐵(𝑣,𝑤) = 𝐵𝑖𝑗𝑣𝑖𝑤𝑗# 𝐵𝑖𝑗𝑣𝑖𝑤𝑗is a scalar

$B(v,w)=\left[\begin{array}{cc}\left[\begin{array}{cc}{B}_1^1& B_2^1\end{array}\right]& \left[\begin{array}{cc}{B}_1^2& B_2^2\end{array}\right]\end{array}\right]\left[\begin{array}{c}{v}^1\\ v^2\end{array}\right]\left[\begin{array}{c}{w}^1\\ w^2\end{array}\right]$ $B(v,w)=\left[\begin{array}{c}\left[\begin{array}{cc}{B}_1^1& B_2^1\end{array}\right]v^1+\left[\begin{array}{cc}{B}_1^2& B_2^2\end{array}\right]v^2\end{array}\right]\left[\begin{array}{c}{w}^1\\ w^2\end{array}\right]$ $B(v,w)=\left[\begin{array}{c}\left[\begin{array}{cc}{B}_1^1{v}^1& B_2^1{v}^1\end{array}\right]+\left[\begin{array}{cc}{B}_1^2{v}^2& B_2^2{v}^2\end{array}\right]\end{array}\right]\left[\begin{array}{c}{w}^1\\ w^2\end{array}\right]$ $B(v,w)=\left[\begin{array}{cc}{B}_1^1{v}^1+B_1^2{v}^2& B_2^1{v}^1+B_2^2{v}^2\end{array}\right]\left[\begin{array}{c}{w}^1\\ w^2\end{array}\right]$ $B(v,w)=(B_1^1{v}^1+B_1^2{v}^2)w^1+(B_2^1{v}^1+B_2^2{v}^2)w^2$ $B(v,w)=B_1^1{v}^1{w}^1+B_1^2{v}^2{w}^1+B_2^1{v}^1{w}^2+B_2^2{v}^2{w}^2$ Which is a scalar $B(v,w)=B_j^i{v}^i{w}^j$

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13 - How do Bilinear Form Components Change WRT Change of Basis

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Detailed

Given

${\tilde{e}}_j=F_j^i{e}_i$ $e_j=B_j^i{\tilde{e}}_i$

${\widetilde{𝜀}}_i=B_j^i{𝜀}_j$ ${𝜀}_i=F_j^i{\widetilde{𝜀}}_j$

Then

$\mathcal{B}={\mathcal{B}}_{kl}{𝜀}^k{𝜀}^l$ $\mathcal{B}={\mathcal{B}}_{kl}(F_i^k{\widetilde{𝜀}}^i)(F_j^l{\widetilde{𝜀}}^j)$ $\mathcal{B}=(F_i^k{F}_j^l{\mathcal{B}}_{kl}){\widetilde{𝜀}}^i{\widetilde{𝜀}}^j$ $\mathcal{B}=({\tilde{\mathcal{B}}}_{ij}){\widetilde{𝜀}}^i{\widetilde{𝜀}}^j$ Thus $\tilde{\mathcal{B}}=F_i^k{F}_j^l{\mathcal{B}}_{kl}$

$\mathcal{B}={\tilde{\mathcal{B}}}_{kl}{\widetilde{𝜀}}^k{\widetilde{𝜀}}^l$ $\mathcal{B}={\tilde{\mathcal{B}}}_{kl}(B_i^k{𝜀}^i)(B_j^l{𝜀}^j)$ $\mathcal{B}=(B_i^k{B}_j^l{\tilde{\mathcal{B}}}_{kl}){𝜀}^i{𝜀}^j$ $\mathcal{B}={\mathcal{B}}_{ij}{B}_i^k{𝜀}^i{𝜀}^j$ Thus ${\mathcal{B}}_{ij}=B_i^k{B}_j^l{\tilde{\mathcal{B}}}_{kl}$

The bilinear map consumes 2 vectors and outputs a scalar

Given

• $\mathcal{B}={\mathcal{B}}_{ij}{𝜀}^i{𝜀}^j$
• $v=v^k{e}_k$
• $w=w^l{e}_l$

Then

• $s=\mathcal{B}(v,w)$
• $s={\mathcal{B}}_{ij}{𝜀}^i{𝜀}^j(v^k{e}_k,w^l{e}_l)$
• $s={\mathcal{B}}_{ij}{𝜀}^i(v^k{e}_k){𝜀}^j(w^l{e}_l)$
• $s={\mathcal{B}}_{ij}{v}^k{w}^l{𝜀}^i(e_k){𝜀}^j(e_l)$
• $s={\mathcal{B}}_{ij}{v}^k{w}^l{𝛿}_k^i{𝛿}_l^j$
• $s={\mathcal{B}}_{ij}{v}^i{w}^j$

Simple (using covector-covector pairs)

Given

Bilinear Map Definition	Basis Covectors
$\mathcal{B}={\tilde{\mathcal{B}}}_{ij}{\widetilde{𝜀}}^i{\widetilde{𝜀}}^j$ $\mathcal{B}={\mathcal{B}}_{ij}{𝜀}^i{𝜀}^j$	${\widetilde{𝜀}}^i=B_j^i{𝜀}^j$ ${𝜀}^i=F_j^i{\widetilde{𝜀}}^j$

Then

Then start with the definition of bilinear form 𝐵: $\mathcal{B}={\mathcal{B}}_{ij}{𝜀}^i{𝜀}^j$ Next, transform all the basis vectors and basis covectors individually $\mathcal{B}={\mathcal{B}}_{ij}{F}_k^i{\widetilde{𝜀}}^k{F}_l^j{\widetilde{𝜀}}^l$ $\mathcal{B}={\mathcal{B}}_{ij}{F}_k^i{F}_l^j{\widetilde{𝜀}}^k{\widetilde{𝜀}}^l$ Thus: ${\tilde{\mathcal{B}}}_{kl}={\mathcal{B}}_{ij}{F}_k^i{F}_l^j$

Then start with the definition of bilinear form 𝐵: $\mathcal{B}={\tilde{\mathcal{B}}}_{ij}{\widetilde{𝜀}}^i{\widetilde{𝜀}}^j$ Next, transform all the basis vectors and basis covectors individually $\mathcal{B}={\tilde{\mathcal{B}}}_{ij}{B}_k^i{𝜀}^k{B}_l^j{𝜀}^l$ $\mathcal{B}={\tilde{\mathcal{B}}}_{ij}{B}_k^i{B}_l^j{𝜀}^k{𝜀}^l$ Thus ${\mathcal{B}}_{kl}={\tilde{\mathcal{B}}}_{ij}{B}_k^i{B}_l^j$

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Tensors - Types

(m,n)-tensor

• m = number of contravariant indices (top of 𝑇)
• n = number of covariant indices (bottom of 𝑇)

For example, a (3,3)-tensor 𝑇 is denoted as:

• $T=T_{rst}^{ijk}\text{ is short for }{T}_{rst}^{ijk}{e}_i{e}_j{e}_k{𝜀}^r{𝜀}^s{𝜀}^t$
• $T={\tilde{T}}_{rst}^{ijk}\text{ is short for }{\tilde{T}}_{rst}^{ijk}{\tilde{e}}_i{\tilde{e}}_j{\tilde{e}}_k{\widetilde{𝜀}}^r{\widetilde{𝜀}}^s{\widetilde{𝜀}}^t$

How do components of a (3,3)-tensor 𝑇 change WRT change of basis?

• $T={\tilde{T}}_{abc}^{xyz}=B_i^x{B}_j^y{B}_k^z{T}_{rst}^{ijk}{F}_a^r{F}_b^s{F}_c^t$

  how it's computed

  Prerequisite knowledge:

  How Basis Vectors 𝑒𝑗 Change WRT Change of Basis	How Dual Basis Covectors 𝜀𝑖 Change WRT Change of Basis
  ${\tilde{e}}_j=F_j^i{e}_i$ $e_j=B_j^i{\tilde{e}}_i$	${\widetilde{𝜀}}^i=B_j^i{𝜀}^j$ ${𝜀}^i=F_j^i{\widetilde{𝜀}}^j$

  Start with the definition:

  • $T=T_{rst}^{ijk}{e}_i{e}_j{e}_k{𝜀}^r{𝜀}^s{𝜀}^t$

  Next, transform all the basis vectors and basis covectors individually and resolve:

  • $T=T_{rst}^{ijk}{B}_i^x{\tilde{e}}_x{B}_j^y{\tilde{e}}_y{B}_k^z{\tilde{e}}_z{F}_a^r{\widetilde{𝜀}}^a{F}_b^s{\widetilde{𝜀}}^b{F}_c^t{\widetilde{𝜀}}^c$
  • $T=(B_i^x{B}_j^y{B}_k^z{T}_{rst}^{ijk}{F}_a^r{F}_b^s{F}_c^t){\tilde{e}}_x{\tilde{e}}_y{\tilde{e}}_z{\widetilde{𝜀}}^a{\widetilde{𝜀}}^b{\widetilde{𝜀}}^c$
  • $T=({\tilde{T}}_{abc}^{xyz}){\tilde{e}}_x{\tilde{e}}_y{\tilde{e}}_z{\widetilde{𝜀}}^a{\widetilde{𝜀}}^b{\widetilde{𝜀}}^c$

  Thus

  • ${\tilde{T}}_{abc}^{xyz}=B_i^x{B}_j^y{B}_k^z{T}_{rst}^{ijk}{F}_a^r{F}_b^s{F}_c^t$

• $T=T_{abc}^{xyz}=F_i^x{F}_j^y{F}_k^z{\tilde{T}}_{rst}^{ijk}{B}_a^r{B}_b^s{B}_c^t$

  how it's computed

  Prerequisite knowledge:

  How Basis Vectors 𝑒𝑗 Change WRT Change of Basis	How Dual Basis Covectors 𝜀𝑖 Change WRT Change of Basis
  ${\tilde{e}}_j=F_j^i{e}_i$ $e_j=B_j^i{\tilde{e}}_i$	${\widetilde{𝜀}}^i=B_j^i{𝜀}^j$ ${𝜀}^i=F_j^i{\widetilde{𝜀}}^j$

  Start with the definition:

  • $T={\tilde{T}}_{rst}^{ijk}{\tilde{e}}_i{\tilde{e}}_j{\tilde{e}}_k{\widetilde{𝜀}}^r{\widetilde{𝜀}}^s{\widetilde{𝜀}}^t$

  Next, transform all the basis vectors and basis covectors individually and resolve:

  • $T={\tilde{T}}_{rst}^{ijk}{F}_i^x{e}_x{F}_j^y{e}_y{F}_k^z{e}_z{B}_a^r{𝜀}^a{B}_b^s{𝜀}^b{B}_c^t{𝜀}^c$
  • $T=(F_i^x{F}_j^y{F}_k^z{\tilde{T}}_{rst}^{ijk}{B}_a^r{B}_b^s{B}_c^t)e_x{e}_y{e}_z{𝜀}^a{𝜀}^b{𝜀}^c$
  • $T=(T_{rst}^{ijk})e_x{e}_y{e}_z{𝜀}^a{𝜀}^b{𝜀}^c$

  Thus

  • $T_{abc}^{xyz}=F_i^x{F}_j^y{F}_k^z{\tilde{T}}_{rst}^{ijk}{B}_a^r{B}_b^s{B}_c^t$

Resources

• NASA’s - An Introduction to Tensors for Students of Physics and Engineering

• Brian Keng's Tensor Introduction

  Original Article

  • https://bjlkeng.io/posts/tensors-tensors-tensors/

  Article Copy

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• YouTube - FloatHeadPhysics - Intuitive Tensors
• YouTube - Mu Prime Math - A Concrete Introduction to Tensor Products
• YouTube - EigenChris - Tensors for Beginners
• YouTube - EigenChris - Tensor Calculus

• Old Stuff

  Tensor Ranks

  Tensor Rank	Name	Description	# of components
  tensor of rank 0	scalar	magnitude (no direction)	1
  tensor of rank 1	vector	each component describes the magnitude in a particular direction magnitude in the x-direction magnitude in the y-direction magnitude in the z-direction	3 - for 3 dimensions
  tensor of rank 2	dyad	each direction is described with a vector x direction is described with a vector y direction is described with a vector z direction is described with a vector	9 - for 3 components for each of the 3 dimensions
  tensor of rank 3	triad		27

  Einstein’s Theory of Relativity required a tensor of rank 4 (x,y,z,t); thus 4*4*4*4=256 components to describe the Theory of Relativity

  Tensor - Represented as a Matrix

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