Matrices, Scalars, Vectors and Vector Calculus 01

Let us imagine that we have a system of coordinates $ {S} $ and a system of coordinates $ {S'} $ that is rotated relatively to $ {S} $. Let us consider a point $ {P} $ that has coordinates $ {(x_1,x_2,x_3)} $ on $ {S} $ and coordinates $ {(x'_1,x'_2,x'_3)} $ on $ {S'} $.
In general it is obvious that $ {x'_1=x'_1(x_1,x_2,x_3)} $, $ {x'_2=x'_2(x_1,x_2,x_3)} $ and that $ {x'_3=x'_3(x_1,x_2,x_3)} $.
Since the transformation from $ {S} $ to $ {S'} $ is just a rotation we can assume that the transformation is linear. Hence we can write explicitly
$ {\begin{aligned} x'_1 &= \lambda _{11}x_1+ \lambda _{12}x_2 +\lambda _{13}x_3 \\ x'_2 &= \lambda _{21}x_1+ \lambda _{22}x_2 +\lambda _{23}x_3 \\ x'_3 &= \lambda _{31}x_1+ \lambda _{32}x_2 +\lambda _{33}x_3 \end{aligned}} $
Another way to write the three previous equations in a more compact way is:

$ \displaystyle x'_i=\sum_{j=1}^3 \lambda_{ij}x_j $

In case you don't see how the previous equation is a more compact way of writing the first equations I'll just lay out the $ {i=1} $ case.

$ \displaystyle x'_1=\sum_{j=1}^3 \lambda_{1j}x_j $

Now all that we have to do is to sum from $ {j=1} $ to $ {j=3} $ and we get the first equation. For the other two a similar reasoning applies.
If we want to make a transformation from $ {S'} $ to $ {S} $ the inverse transformation is

$ \displaystyle x_i=\sum_{j=1}^3 \lambda_{ji}x'_j $

The previous notation suggests that the $ {\lambda} $ indexes can be arranged in a form of a matrix:

$ \displaystyle \lambda= \left(\begin{array}{ccc} \lambda_{11} & \lambda_{12} & \lambda_{13} \\ \lambda_{21} & \lambda_{22} & \lambda_{23} \\ \lambda_{31} & \lambda_{32} & \lambda_{33} \end{array} \right) $

In the literature the previous matrix has the name of rotation matrix or transformation matrix.

— 1. Properties of the rotation matrix —

For the transformation $ {x'_i=x'_i(x_i)} $

$ \displaystyle \sum_j \lambda_{ij}\lambda_{kj}=\delta_{ik} $

Where $ {\delta_{ik}} $ is a matrix known as Kronecker delta and its definition is

$ \displaystyle \delta_{ik}=\begin{cases} 0 \quad i\neq k\\ 1 \quad i=k \end{cases} $

For the inverse transformation $ {x_i=x_i(x'_i)} $ it is

$ \displaystyle \sum_i \lambda_{ij}\lambda_{ik}=\delta_{jk} $

The previous relationships are called orthogonality relationships.

— 2. Matrix operations, definitions and properties —

Let us represent the coordinates of a point $ {P} $ by a column vector

$ \displaystyle x = \left(\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array}\right) $

Using the usual notation of linear algebra we can write the transformation equations as $ {x'=\mathbf{\lambda} x} $
Where we define the matrix product, $ {\mathbf{AB}=\mathbf{C}} $, to be possible only when the number of columns of $ {\mathbf{A}} $ is equal to the number of rows of $ {\mathbf{B}} $
The way to calculate a specific element of the matrix $ {\mathbf{C}} $, we will denote this element by the symbol $ {\mathbf{C}_{ij}} $ is,

$ \displaystyle \mathbf{C}_{ij}=[\mathbf{AB}]_{ij}=\sum_k A_{ik}B_{kj} $

Given the definition of a matrix product it should be clear that in general one has $ {\mathbf{AB} \neq \mathbf{BA}} $
As an example let us look into;

$ \displaystyle \mathbf{A}=\left( \begin{array}{cc} 2 & 1\\ -1 & 3 \end{array}\right) ;\quad \mathbf{B}=\left( \begin{array}{cc} -1 & 2\\ 4 & -2 \end{array}\right) $

With

$ \displaystyle \mathbf{AB}=\left( \begin{array}{cc} 2\times (-1)+1\times 4 & 2\times 2+1\times (-2)\\ -1\times (-1)+3\times 4 & -1\times 2+3\times (-2) \end{array}\right)=\left( \begin{array}{cc} 2 & 2\\ 13 & -8 \end{array}\right) $

and

$ \displaystyle \mathbf{BA}=\left( \begin{array}{cc} -4 & 5\\ 10 & -2 \end{array}\right) $

We'll say that $ {\lambda^T} $ is the transposed of $ {\lambda} $ and calculate the matrix elements of the transposed matrix by $ {\lambda_{ij}^T=\lambda_{ji}} $. In a more pedestrian way one can say that in order to obtain the transpose of a given matrix one needs only to exchange its rows and columns.
For a given matrix $ {\mathbf{A}} $ it exists another matrix $ {\mathbf{U}} $ such as $ {\mathbf{AU}=\mathbf{UA}=\mathbf{A}} $. The matrix $ {\mathbf{U}} $ is said to be the unit matrix and usually one can represent it by $ {\mathbf{U}=\mathbf{1}} $.
If $ {\mathbf{AB}=\mathbf{BA}=\mathbf{1}} $, then $ {\mathbf{A}} $ and $ {\mathbf{B}} $ are said to be the inverse of each other and $ {\mathbf{B}=\mathbf{A}^{-1}} $, $ {\mathbf{A}=\mathbf{B}^{-1}} $.
Now for the rotation matrices it is
$ {\begin{aligned} \lambda \lambda ^T &= \left( \begin{array}{cc} \lambda_{11} & \lambda_{12}\\ \lambda_{21} & \lambda_{22} \end{array}\right)\left( \begin{array}{cc} \lambda_{11} & \lambda_{21}\\ \lambda_{12} & \lambda_{22} \end{array}\right) \\ &= \left( \begin{array}{cc} \lambda_{11}^2+\lambda_{22}^2 & \lambda_{11}\lambda_{21}+\lambda_{12}\lambda_{22}\\ \lambda_{21}\lambda_{11}+\lambda_{22}\lambda_{12} & \lambda_{21}^2+\lambda_{22}^2 \end{array}\right)\\ &=\left( \begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)\\ &= \mathbf{1} \end{aligned}} $
Where the second last equality follows from what we've seen in section 1.
Thus $ {\lambda ^T=\lambda ^{-1}} $.
Just to finish up this section let me just mention that even though, in general, matrix multiplication isn't commutative it still is associative. Thus $ {(\mathbf{AB})\mathbf{C}=\mathbf{A}(\mathbf{BC})} $. Also matrix addition has just the definition one would expect. Namely $ {C_{ij}=A_{ij}+B_{ij}} $.
If one inverts all three axes at the same time the matrix that we get is the so called inversion matrix and it is

$ \displaystyle \left( \begin{array}{ccc} -1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & -1 \end{array}\right) $

Since it can be shown that rotation matrices always have their determinant equal to $ {1} $ and that the inversion matrix has a $ {-1} $ determinant we know that there isn't any continuous transformation that maps a rotation into an inversion.

— 3. Vectors and Scalars —

In Physics quantities are either scalars or vectors (they can also be tensors but since they aren't needed right away I'll just pretend that they don't exist for the time being). These two entities are defined according to their transformation properties.
Let $ {\lambda} $ be a coordinate transformation, $ {\displaystyle\sum_j\lambda_{ij}\lambda_{kj}=\delta_{ij}} $, if it is:

• $ {\displaystyle\sum_j\lambda_{ij}\varphi=\varphi} $ then $ {\varphi} $ is said to be a scalar.
• $ {\displaystyle\sum_j\lambda_{ij}A_j=A'_i} $ for $ {A_1} $, $ {A_2} $ and $ {A_3} $ then $ {(A_1,A_2,A_3)} $ is said to be a vector.

— 3.1. Operations between scalars and vectors —

I think that most people in here already know this but in the interest of a modicum of self containment I'll just enumerate some properties of scalars and vectors.

1. $ {\vec{A}+\vec{B}=\vec{B}+\vec{A}} $
2. $ {\vec{A}+(\vec{B}+\vec{C})=(\vec{A}+\vec{B})+\vec{C}} $
3. $ {\varphi+\psi=\psi+\varphi} $
4. $ {\varphi+(\psi+\xi)=(\varphi+\psi)+\xi} $
5. $ {\xi \vec{A}= \vec{B}} $ is a vector.
6. $ {\xi \varphi=\psi} $ is a scalar.

As an example we will show the second proposition 5 and the reader has to show the veracity of the last proposition.
In order to show that $ {\xi \vec{A}= \vec{B}} $ is a vector we have to show that it transforms like a vector.
$ {\begin{aligned} B'_i &= \displaystyle\sum_j \lambda_{ij}B_j\\ &= \displaystyle\sum_j \lambda_{ij}\xi A_j\\ &= \xi\displaystyle\sum_j \lambda_{ij} A_j\\ &= \xi A'_i \end{aligned}} $
Hence $ {\xi A} $ transforms like a vector.

— 4. Vector "products" —

The operations between scalars are pretty much well know by everybody, hence we won't take a look at them, but maybe it is best for us to take a look at two operations between vectors that are crucial for our future development.

— 4.1. Scalar product —

We can construct a scalar by using two vectors. This scalar is a measure of the projection of one vector into the other. Its definition is

$ \displaystyle \vec{A}.\cdot\vec{B}=\sum_i A_i B_i = AB\cos (A.B) $

For this operation deserve its name, one still has to prove that the result indeed is a scalar.
First one writes $ {A'_i=\displaystyle \sum_j\lambda_{ij}A_j} $ and $ {B'_i=\displaystyle \sum_k\lambda_{ik}B_k} $, where one changes the index of the second summation because we'll have to multiply the two quantities and that way the final result can be achieved much more easily.
Now it is
$ {\begin{aligned} \vec{A}'\cdot \vec{B}' &= \displaystyle\sum_i A'_i B'_i \\ &= \displaystyle \sum_i \left(\sum_j\lambda_{ij}A_j\right)\left( \sum_k\lambda_{ik}B_k \right)\\ &= \displaystyle \sum_j \sum_k \left( \sum_i \lambda_{ij}\lambda_{ik} \right)A_j B_k\\ &= \displaystyle \sum_j \left(\sum_k \delta_{jk}A_jB_k \right)\\ &= \displaystyle \sum_j A_j B_j \\ &= \vec{A}\cdot \vec{B} \end{aligned}} $
Hence $ {\vec{A}\cdot \vec{B}} $ is a scalar.

— 4.2. Vector product —

First we have to introduce the permutation symbol $ {\varepsilon_{ijk}} $. Its definition is $ {\varepsilon_{ijk}=0} $ if two or three of its indices are equal; $ {\varepsilon_{ijk}=1} $ if $ {i\,j\,k} $ is an even permutation of $ {123} $ (the even permutations are $ {123} $, $ {231} $ and $ {312} $); $ {\varepsilon_{ijk}=-1} $ if $ {i\,j\,k} $ is an odd permutation of $ {123} $ (the odd permutations $ {132} $, $ {321} $ and $ {213} $).
The vector product, $ {\vec{C}} $, of two vectors $ {\vec{A}} $ and $ {\vec{B}} $ is denoted by $ {\vec{C}=\vec{A}\times \vec{B}} $.
To calculate the components the components of the vector $ {\vec{C}} $ the following equation is to be used:

$ \displaystyle C_i=\sum_{j,k}\varepsilon_{ijk}A_j B_k $

Where $ {\displaystyle\sum_{j,k}} $ is shorthand notation for $ {\displaystyle\sum_j\sum_k} $.
As an example let us look into $ {C_1} $
$ {\begin{aligned} C_1 &= \sum_{j,k}\varepsilon_{1jk}A_j B_k\\ &= \varepsilon_{123}A_2 B_3+\varepsilon_{132}A_3 B_2\\ &= A_2B_3-A_3B_2 \end{aligned}} $
where we have used the definition of $ {\epsilon_{ijk}} $ throughout the reasoning.
One can also see that (this another exercise for the reader) $ {C_2=A_3B_1-A_1B_3} $ and that $ {C_3=A_1B_2-A_2B_1} $.
If one only wants to know the magnitude of $ {\vec{C}} $ the following equation should be used $ {C=AB\sin (A,B)} $.
After choosing the three axes that define our frame of reference one can choose as the basis of this space a set of three linearly independent vectors that have unit norm. These vectors are called unit vectors.
If we denote these vectors by $ {\vec{e}_i} $ any vector $ {\vec{A}} $ can be written as $ {\vec{A}=\displaystyle \sum _i \vec{e}_i A_i} $. We also have that $ {\vec{e}_i\cdot \vec{e}_j=\delta_{ij}} $ and $ {\vec{e}_i\times \vec{e}_j=\vec{e}_k} $. Another way to write the last equation is $ {\vec{e}_i\times \vec{e}_j=\vec{e}_k\varepsilon_{ijk}} $.

— 5. Vector differentiation with respect to a scalar —

Let $ {\varphi} $ be a scalar function of $ {s} $: $ {\varphi=\varphi(s)} $. Since both $ {\varphi} $ and $ {s} $ are scalars we know that their transformation equations are $ {\varphi=\varphi '} $ and $ {s=s'} $. Hence it also is $ {d\varphi=d\varphi '} $ and $ {ds=ds'} $
Thus it follows that for differentiation it is $ {d\varphi/ds=d\varphi'/ds'=(d\varphi/ds)'} $.
In order to define the derivative of a vector with respect to a scalar we will follow an analogous road.
We already know that it is $ {A'_i=\displaystyle \sum_j \lambda _{ij}A_j} $ hence
$ {\begin{aligned} \dfrac{dA'_i}{ds'} &= \dfrac{d}{ds'}\left( \displaystyle \sum_j \lambda _{ij}A_j \right)\\ &= \displaystyle \lambda _{ij}\dfrac{d A_j}{ds'}\\ &= \displaystyle \lambda _{ij}\dfrac{d A_j}{ds}\ \end{aligned}} $
where the last equality follows from the fact that $ {s} $ is a scalar.
From what we saw we can write

$ \displaystyle \frac{d A'_i}{ds'}= \left( \frac{d A_i}{ds} \right)'=\sum_j \lambda _{ij}\frac{d A_j}{ds} $

Hence $ {dA_j/ds} $ transforms like the coordinates of a vector which is the same as saying that $ {d\vec{A}/ds} $ is a vector.
The rules for differentiating vectors are:

• $ {\dfrac{d}{ds}(\vec{A}+\vec{B})= \dfrac{d\vec{A}}{ds}+\dfrac{d\vec{B}}{ds}} $
• $ {\dfrac{d}{ds}(\vec{A}\cdot\vec{B})= \vec{A}\cdot\dfrac{d\vec{B}}{ds}+\dfrac{d\vec{A}}{ds}\cdot \vec{B}} $
• $ {\dfrac{d}{ds}(\vec{A}\times\vec{B})= \vec{A}\times\dfrac{d\vec{B}}{ds}+\dfrac{d\vec{A}}{ds}\times \vec{B}} $
• $ {\dfrac{d}{ds}(\varphi\vec{A})= \varphi\dfrac{d\vec{A}}{ds}+\dfrac{d\varphi}{ds}\vec{A}} $

The proof of these rules isn't needed in order for us to develop any kind of special skills but if the reader isn't very used to this, then it is better for him to do them just to see how things happen.