Tensors

a tensor is an object that is invariant under a change of coordinate/basis and has COMPONENTS that change in a special predictable way under a change of coordinates/basis
a tensor is a collection of vectors and covectors combined together using the tensor product
tensors can take several different forms (e.g. scalars, vectors, covectors, linear maps, bilinear maps, multilinear maps, etc)

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
  
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backward transform: new basis → old basis
- 𝑒₁ = 𝑎·𝑒̃₁+ 𝑏·𝑒̃₂
  𝑒₂ = 𝑐·𝑒̃₁+ 𝑑·𝑒̃₂
  
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Forward transform is the inverse of Backward transform

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Thus

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Confirming the inverse behavior of the above equations

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Thus 𝛿_𝑖𝑗 is the kronecker delta function:

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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

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Thus
𝐵 is used to transform vector components from old to new	𝐹 is used to transform vector components from new to old

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 𝜀^𝑖 are defined as:

𝜀¹(𝑒₁) = 1
𝜀¹(𝑒₂) = 0
𝜀²(𝑒₁) = 0
𝜀²(𝑒₂) = 1

Thus:

𝜀^𝑖(𝑒_𝑗) = 𝛿_𝑖𝑗 # 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:

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epsilon covector 𝜀^𝑖 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
  A set of covectors {𝜀¹, 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(𝑒_𝑖) + 𝑎₂𝜀²(𝑒_𝑖) = 0; for all 𝑒_𝑖∊{𝑒₁, 𝑒₂}
  Evaluating the above statement for each basis vector yields:
  - that the scalars {𝑎₁, 𝑎₂} must be ALL zero such that:
  - for 𝑒_𝑖=𝑒₁:
    𝑎₁𝜀¹(𝑒₁) + 𝑎₂𝜀²(𝑒₁) = 0
    𝑎₁*1 + 𝑎₂*0 = 0
    𝑎₁ = 0
  - for 𝑒_𝑖=𝑒₁:
    𝑎₁𝜀¹(𝑒₂) + 𝑎₂𝜀²(𝑒₂) = 0
    𝑎₁*0 + 𝑎₂*1 = 0
    𝑎₂ = 0
  Thus, 𝑎₁ and 𝑎₂ must equal to 0.
every arbitrary covector 𝛼∊𝑉* can be expressed as a linear combination of {𝜀¹, 𝜀²}
1. 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*

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)

𝜀^𝑖(𝑒_𝑗) = 𝛿_𝑖𝑗 = 𝜀̃^𝑖(𝑒̃_𝑗)

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

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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

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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:

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Basis of Linear Maps

{𝑒₁𝜀¹, 𝑒₁𝜀², 𝑒₂𝜀¹, 𝑒₂𝜀²} is one possible basis for linear map 𝐿: ℝ²→ℝ² where:

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Thus

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𝐿 = 𝐿^𝑖_𝑗 𝑒_𝑖𝜀^𝑗 # Einstein's notation

Thus: linear maps can be written as linear combinations of vector-covector pairs

How is a Basis 𝑒_𝑖𝜀^𝑗 a Linear Map (That Eats a Vector and Outputs a Vector)?

Given:

𝐿 = 𝐿^𝑖_𝑗 𝑒_𝑖𝜀^𝑗
𝑣 = 𝑣^𝑘𝑒_𝑘

Let 𝐿 eat a vector 𝑣:

𝐿(𝑣) = 𝐿(𝑣)
𝐿(𝑣) = 𝐿^𝑖_𝑗𝑒_𝑖𝜀^𝑗(𝑣^𝑘𝑒_𝑘)
𝐿(𝑣) = 𝐿^𝑖_𝑗𝑣^𝑘𝑒_𝑖𝜀^𝑗(𝑒_𝑘)
𝐿(𝑣) = 𝐿^𝑖_𝑗𝑣^𝑘𝑒_𝑖𝛿^𝑗_𝑘
𝐿(𝑣) = 𝐿^𝑖_𝑗𝑣^𝑗𝑒_𝑖
𝐿(𝑣) = 𝑤^𝑖𝑒_𝑖 # 𝐿^𝑖_𝑗𝑣^𝑗is a scalar, thus replace with 𝑤^𝑖= 𝐿^𝑖_𝑗𝑣^𝑗
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 ℝ^𝑚 spanned by column vectors
row space of 𝐴
- is the subspace of ℝ^𝑛 spanned by row vectors

The Rank-Nullity Theorem

Given a linear transformation 𝐿 from ℝ^𝑛 to ℝ^𝑚, then:

dim ker(𝐿) + dim range(𝐿) = 𝑛

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

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Detailed

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Thus
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Simple (using vector-covector pairs)

Given:

Linear Map Definition	Basis Vectors	Basis Covectors
𝐿 = 𝐿˜^𝑖_𝑗 𝑒̃_𝑖𝜀̃^𝑗 𝐿 = 𝐿^𝑖_𝑗 𝑒_𝑖𝜀^𝑗	Loading Loading	Loading Loading

Then

Then start with the definition of linear map 𝐿:

𝐿 = 𝐿^𝑖_𝑗 𝑒_𝑖𝜀^𝑗

Next, transform all the basis vectors and basis covectors individually

𝐿 = 𝐿^𝑖_𝑗 𝐵^𝑘_𝑖𝑒̃_𝑘𝐹^𝑗_𝑙𝜀̃^𝑙
𝐿 = (𝐵^𝑘_𝑖𝐿^𝑖_𝑗𝐹^𝑗_𝑙)𝑒̃_𝑘𝜀̃^𝑙
𝐿 = ( 𝐿˜^𝑘_𝑙)𝑒̃_𝑘𝜀̃^𝑙

Thus:

𝐿^𝑘_𝑙 = 𝐹^𝑘_𝑖𝐿˜^𝑖_𝑗𝐵^𝑗_𝑙
𝐿˜ = 𝐵𝐿𝐹

Then start with the definition of linear map 𝐿:

𝐿 = 𝐿˜^𝑖_𝑗 𝑒̃_𝑖𝜀̃^𝑗

Next, transform all the basis vectors and basis covectors individually

𝐿 = 𝐿˜^𝑖_𝑗 𝐹^𝑘_𝑖𝑒_𝑘𝐵^𝑗_𝑙𝜀^𝑙
𝐿 = (𝐹^𝑘_𝑖𝐿˜^𝑖_𝑗𝐵^𝑗_𝑙)𝑒_𝑘𝜀^𝑙
𝐿 = ( 𝐿^𝑘_𝑙)𝑒_𝑘𝜀^𝑙

Thus:

𝐿^𝑘_𝑙 = 𝐹^𝑘_𝑖𝐿˜^𝑖_𝑗𝐵^𝑗_𝑙
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

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

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TODO

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)

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\|\|𝑣\|\|² = 𝑣^𝑖𝑣^𝑗(𝑒_𝑖·𝑒_𝑗) = 𝑣^𝑖𝑣^𝑗𝑔_𝑖𝑗 (𝑒_𝑖·𝑒_𝑗) = 𝑔_𝑖𝑗	\|\|𝑣\|\|² = 𝑣̃^𝑖𝑣̃^𝑗(𝑒̃_𝑖·𝑒̃_𝑗) = 𝑣̃^𝑖𝑣̃^𝑗𝑔̃_𝑖𝑗 (𝑒̃_𝑖·𝑒̃_𝑗) = 𝑔̃_𝑖𝑗

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
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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:

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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
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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

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

𝑔̃_𝑖𝑗 = 𝑒̃_𝑖·𝑒̃_𝑗
𝑔̃_𝑖𝑗 = (𝐹^𝑘_𝑖 𝑒_𝑘)·(𝐹^𝑙_𝑗 𝑒_𝑙)
𝑔̃_𝑖𝑗 = 𝐹^𝑘_𝑖𝐹^𝑙_𝑗 (𝑒_𝑘·𝑒_𝑙)
𝑔̃_𝑖𝑗 = 𝐹^𝑘_𝑖𝐹^𝑙_𝑗 𝑔_𝑘𝑙

𝑔_𝑖𝑗 = 𝑒_𝑖·𝑒_𝑗
𝑔_𝑖𝑗 = (𝐵^𝑘_𝑖 𝑒̃_𝑘)·(𝐵^𝑙_𝑗 𝑒̃_𝑙)
𝑔_𝑖𝑗 = 𝐵^𝑘_𝑖𝐵^𝑙_𝑗 (𝑒̃_𝑘·𝑒̃_𝑙)
𝑔_𝑖𝑗 = 𝐵^𝑘_𝑖𝐵^𝑙_𝑗 𝑔̃_𝑘𝑙

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

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

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TODO

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:

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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
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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?

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Bilinear Forms V⨯𝑉 → ℝ:

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Basis of a Bilinear Form

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

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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 = 𝐵^𝑖_𝑗 𝜀^𝑖𝜀^𝑗
𝑣 = 𝑣^𝑘𝑒_𝑘
𝑤 = 𝑤^𝑙𝑒_𝑙

Let 𝐵 eat a vectors 𝑣 and 𝑤:

𝐵(𝑣,𝑤) = 𝐵(𝑣,𝑤)
𝐵(𝑣,𝑤) = 𝐵^𝑖_𝑗 𝜀^𝑖𝜀^𝑗(𝑣^𝑘𝑒_𝑘𝑤^𝑙𝑒_𝑙)
𝐵(𝑣,𝑤) = 𝐵^𝑖_𝑗𝑣^𝑘𝑤^𝑙𝜀^𝑖(𝑒_𝑘) 𝜀^𝑗(𝑒_𝑙)
𝐵(𝑣,𝑤) = 𝐵^𝑖_𝑗𝑣^𝑘𝑤^𝑙𝛿^𝑖_𝑘𝛿^𝑗_𝑙
𝐵(𝑣,𝑤) = 𝐵^𝑖_𝑗𝑣^𝑖𝑤^𝑗# 𝐵^𝑖_𝑗𝑣^𝑖𝑤^𝑗is a scalar

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Which is a scalar

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

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Detailed

Given

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Then

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Thus

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Thus

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The bilinear map consumes 2 vectors and outputs a scalar

Given

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Then

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Simple (using covector-covector pairs)

Given

Bilinear Map Definition	Basis Covectors
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Then

Then start with the definition of bilinear form 𝐵:

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Next, transform all the basis vectors and basis covectors individually

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Thus:

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Then start with the definition of bilinear form 𝐵:

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Next, transform all the basis vectors and basis covectors individually

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Thus

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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:

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

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how it's computed
Prerequisite knowledge:
How Basis Vectors 𝑒_𝑗 Change WRT Change of Basis How Dual Basis Covectors 𝜀_𝑖 Change WRT Change of Basis

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Start with the definition:
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Next, transform all the basis vectors and basis covectors individually and resolve:
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Thus
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how it's computed
Prerequisite knowledge:
How Basis Vectors 𝑒_𝑗 Change WRT Change of Basis How Dual Basis Covectors 𝜀_𝑖 Change WRT Change of Basis

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Start with the definition:
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Next, transform all the basis vectors and basis covectors individually and resolve:
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Thus
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Name

Tensor Type

Longer
Syntax

Shorter
Syntax

How Components Change
WRT Change of Basis

Is an Element of

Available
Functions

Additional

basis vectors

covariant
(0,1)-tensor

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computation steps
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
    
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- backward transform: new basis → old basis
  - 𝑒₁ = 𝑎·𝑒̃₁+ 𝑏·𝑒̃₂
    𝑒₂ = 𝑐·𝑒̃₁+ 𝑑·𝑒̃₂
    
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Forward transform is the inverse of Backward transform
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Thus
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Confirming the inverse behavior of the above equations
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Thus 𝛿_𝑖𝑗 is the kronecker delta function:
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N/A

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basis vector components are covariant with the change of basis

vectors

contravariant
(1,0)-tensor

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computation steps

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

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Loading Loading Loading Thus Loading	Loading Loading Loading Thus Loading
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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vectors themselves are invariant to the change of basis
vector components are contravariant to the change of basis
column vectors are coordinate representations of vectors

dual basis covectors

contravariant
(1,0)-tensor

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computation steps
We define:
- 𝜀^𝑖(𝑒_𝑗) = 𝛿_𝑖𝑗 = 𝜀̃^𝑖(𝑒̃_𝑗)
By definition of basis
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N/A

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dual basis covector components are contravariant with the change of basis

covectors

covariant
(0,1)-tensor

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computation steps

Let's define:

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

𝜀^𝑖(𝑒_𝑗) = 𝛿_𝑖𝑗 = 𝜀̃^𝑖(𝑒̃_𝑗)

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

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covectors themselves are invariant to the change of basis
covector components are covariant to the change of basis
row vectors are coordinate representations of covectors

linear maps

(1,1)-tensor

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computation steps

Detailed

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Loading Loading Loading Loading Loading Loading Loading Loading	Einstein's notation: Loading Loading Loading Loading Loading Loading Loading Loading
Thus
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Simple (using vector-covector pairs)

Given:

Linear Map Definition	Basis Vectors	Basis Covectors
𝐿 = 𝐿˜^𝑖_𝑗 𝑒̃_𝑖𝜀̃^𝑗 𝐿 = 𝐿^𝑖_𝑗 𝑒_𝑖𝜀^𝑗	Loading Loading	Loading Loading

Then

Then start with the definition of linear map 𝐿:

𝐿 = 𝐿^𝑖_𝑗 𝑒_𝑖𝜀^𝑗

Next, transform all the basis vectors and basis covectors individually

𝐿 = 𝐿^𝑖_𝑗 𝐵^𝑘_𝑖𝑒̃_𝑘𝐹^𝑗_𝑙𝜀̃^𝑙
𝐿 = (𝐵^𝑘_𝑖𝐿^𝑖_𝑗𝐹^𝑗_𝑙)𝑒̃_𝑘𝜀̃^𝑙
𝐿 = ( 𝐿˜^𝑘_𝑙)𝑒̃_𝑘𝜀̃^𝑙

Thus:

𝐿^𝑘_𝑙 = 𝐹^𝑘_𝑖𝐿˜^𝑖_𝑗𝐵^𝑗_𝑙
𝐿˜ = 𝐵𝐿𝐹

Then start with the definition of linear map 𝐿:

𝐿 = 𝐿˜^𝑖_𝑗 𝑒̃_𝑖𝜀̃^𝑗

Next, transform all the basis vectors and basis covectors individually

𝐿 = 𝐿˜^𝑖_𝑗 𝐹^𝑘_𝑖𝑒_𝑘𝐵^𝑗_𝑙𝜀^𝑙
𝐿 = (𝐹^𝑘_𝑖𝐿˜^𝑖_𝑗𝐵^𝑗_𝑙)𝑒_𝑘𝜀^𝑙
𝐿 = ( 𝐿^𝑘_𝑙)𝑒_𝑘𝜀^𝑙

Thus:

𝐿^𝑘_𝑙 = 𝐹^𝑘_𝑖𝐿˜^𝑖_𝑗𝐵^𝑗_𝑙
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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linear transformations themselves are invariant to the change of basis
linear transformation components are both contravariant and covariant to the change of basis
matrices are coordinate representations of linear maps

bilinear forms

(0,2)-tensor

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computation steps
TODO

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metric tensors

(0,2)-tensor

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computation steps

HOW METRIC TENSOR COMPONENTS CHANGE WRT CHANGE OF BASIS

𝑔_𝑘𝑙= 𝑒_𝑘·𝑒_𝑙
𝑔̃_𝑖𝑗 = 𝑒̃_𝑖·𝑒̃_𝑗

𝑔̃_𝑖𝑗 = 𝑒̃_𝑖·𝑒̃_𝑗
𝑔̃_𝑖𝑗 = (𝐹^𝑘_𝑖 𝑒_𝑘)·(𝐹^𝑙_𝑗 𝑒_𝑙)
𝑔̃_𝑖𝑗 = 𝐹^𝑘_𝑖𝐹^𝑙_𝑗 (𝑒_𝑘·𝑒_𝑙)
𝑔̃_𝑖𝑗 = 𝐹^𝑘_𝑖𝐹^𝑙_𝑗 𝑔_𝑘𝑙

𝑔_𝑖𝑗 = 𝑒_𝑖·𝑒_𝑗
𝑔_𝑖𝑗 = (𝐵^𝑘_𝑖 𝑒̃_𝑘)·(𝐵^𝑙_𝑗 𝑒̃_𝑙)
𝑔_𝑖𝑗 = 𝐵^𝑘_𝑖𝐵^𝑙_𝑗 (𝑒̃_𝑘·𝑒̃_𝑙)
𝑔_𝑖𝑗 = 𝐵^𝑘_𝑖𝐵^𝑙_𝑗 𝑔̃_𝑘𝑙

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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lengths and angles themselves are invariant to the change of basis
metric tensor components are covariant to the change of basis
symmetric matrices are coordinate representations of metric tensors

metric tensors are a special type of bilinear forms

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
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

Tensors

Tensors - Introduction

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

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

3 - Covector Introduction

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

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

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

6 - Linear Maps Introduction

The Matrix Multiplication can be purely Derived from the Linearity Rules Above

Basis of Linear Maps

How is a Basis 𝑒𝑖𝜀𝑗 a Linear Map (That Eats a Vector and Outputs a Vector)?

Matrices: Null-Spaces, Column-Spaces, Row Spaces

The Rank-Nullity Theorem

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

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

9 - Metric Tensor Introduction

Length

Angles

Say we have Unit Vectors

Say We Have Arbitrary Vectors

Lengths & Angles Summary

Components of a Metric Tensor

Metric Tensor Algebraic Properties

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

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

12 - Bilinear Forms Introduction

How do Bilinear Form components Change WRT change of Basis?

Why a row of rows?

Basis of a Bilinear Form

How is a Basis 𝜀𝑖𝜀𝑗 a Bilinear Form (That Eats 2 Vectors and Outputs a Scalar)?

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

The bilinear map consumes 2 vectors and outputs a scalar

Tensors - Types

Resources

Original Article

Article Copy

Tensor Ranks

Tensor - Represented as a Matrix

How is a Basis 𝑒_𝑖𝜀^𝑗 a Linear Map (That Eats a Vector and Outputs a Vector)?

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