#linear algebra

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I'm prolly gonna do some math for the first time in month, in like 14 mins or so

Wish me luck I'll post results

(the equations only work on my blog)

A rotation in 2 dimensions about the origin can be represented with a rotation matrix where θ is the angle of rotation with respect to the x axis:

$$\begin{bmatrix} cos( \theta) & -sin( \theta ) \\\ sin( \theta ) & cos( \theta ) \end{bmatrix}$$

A 2 dimensional rotation leaves a 0 dimensional point fixed as seen in the picture below where O is said 0 dimensional point:

(image courtesy of Wikipedia)

A 3 dimensional rotation about the an axis is similar to a 2 dimensional rotation matrix with the only difference being that the fixed axis has a 1 in its diagonal entry; the 3d rotation matrix across the x axis is given below:

$$\begin{bmatrix} 1 & 0 & 0 \\\ 0 & cos( \theta) & -sin( \theta ) \\\ 0 & sin( \theta ) & cos( \theta ) \end{bmatrix}$$

More generally, any square matrix with a determinant equal to 1 and an inverse that is equal to its transpose can be seen as a rotation matrix (including the identity matrix which is just a zero rotation).

For a simple rotation in any n dimensional space, there is a (n-2) dimensional hyperplane that remains stationary. This stationary hyperplane is the origin point in 2 dimensions, the axis of rotation in 3 dimensions, and the "plane of rotation" in 4 dimensions.

A simple 4D rotation is a rotation around a given plane defined by two of the 4 orthogonal axes in the 4D space; the rotation matrix around the xy plane is given below:

(the equations only work on my blog)

A polyhedron that is topologically identical to a sphere obeys the equation given below:

$$V-E+F=2$$

The Platonic solids all obey the equation because the boundary of a sphere can be molded into the shape of all 5 of the Platonic solids. More generally, any shape that can be derived by molding a sphere in a continuous and reversable fashion obeys the formula.

This formula along with some basic geometry and linear algebra can be used to show that there are only 5 regular convex polyhedra.

A Platonic solid is defined as a polyhedra that has identical vertices, edges, and faces where said faces are all convex regular polygons and said polygons are strictly convex meaning the faces cannot be coplanar and they cannot intersect one another.

The definition above means that a Platonic solid can be defined by which polygon makes up its faces and how many polygons are met at a vertex.

The Schläfli symbol {p,q} gives the shape of its faces {p} where p is the number of sides and {q} is the number of faces (for regular convex polyhedra) joined at each vertex.

The polyhedron formula given above can be used alongside other properties of the faces, edges, and vertices to create a system of linear equations with with only 5 valid solutions.

Every face of a Platonic solid is joined by p edges. If one were to break off each of the faces of a Platonic solid the number of faces F would stay the same but the number of edges E would double to 2E (as all edges connect to two faces). After separating the faces from one another, one can see that the new number of edges 2E is p times greater than the number of faces F. Therefore, the expression pF is equal to 2E.

$$pF = 2E$$

By again separating the faces of a Platonic solid, one can see that the number of vertices increases by q as q faces are joined at each vertex. This new number of vertices qV is the same number as 2E as the dislocated faces have the same number of vertices and edges as they are regular convex polygons. This means that 2E and qV are equal and both are also equal to pF.

$$2E = qV$$

We thus have three equations in terms of V, E, and F and p and q: the original Euler's equation and the two equations that have just been derived.

$$V-E+F=2$$

$$2E - pF = 0$$

$$qV - 2E = 0$$

The above equations can be thought of as a system of linear equations and can thus be rewritten as a matrix equation.

$$\begin{pmatrix} 1 & -1 & 1 \\\ 0 & 2 & -p \\\ q & -2 & 0 \end{pmatrix} \begin{pmatrix} V \\\ E \\\ F \end{pmatrix} = \begin{pmatrix} 2 \\\ 0 \\\ 0 \end{pmatrix}$$

The matrix equation can be solved by taking the inverse of the square matrix multiplied on the left for both sides; the inverse cancels out the one on the lefthand side of the equation:

$$\begin{pmatrix} V \\\ E \\\ F \end{pmatrix} = \frac{1}{2p-pq+2q} \begin{pmatrix} 2p & 2 & 2-p \\\ pq & q & -p \\\ 2q & q-2 & -2 \end{pmatrix} \begin{pmatrix} 2 \\\ 0 \\\ 0 \end{pmatrix}$$

The matrix equation above can again be simplified by multiplying the inverse matrix with the vector:

$$\begin{pmatrix} V \\\ E \\\ F \end{pmatrix} = \frac{1}{2p-pq+2q} \begin{pmatrix} 4p \\\ 2pq \\\ 4q \end{pmatrix} $$

The numbers p and q must be integers greater than 2 and the denominator of the fraction 2p-pq+2q must be greater than zero. Meaning that the possible values of p and q are bounded by the following inequalities:

$$2 < p < \frac{2q}{q-2}$$

$$2 < q < \frac{2p}{p-2}$$

There are only 5 solutions to the inequalities above and they each correspond to a solution of the system of linear equations which produces the Platonic solids:

{3,3} - tetrahedron

{3,4} - octahedron

{3,5} - icosahedron

{4,3} - hexahedron

(the equations only work on my blog directly)

Diagonalizing a matrix is a pretty well known transformation in linear algebra where a matrix A can be rewritten as a similarity transformation of a diagonal matrix.

$$AP = PD$$

$$A = PDP^{-1}$$

The diagonal matrix D has the eigenvalues as its diagonal entries, and the matrix P thus has eigenvectors as its columns:

$$(A - \lambda_n I)p_n = 0$$

Nonetheless, not all choices of A can be diagonalized. The fundamental issue that can arise is whether the matrix of the eigenvectors P is invertible or not. Diagonalization requires the inverse of P to exist and if all possible choices of P are uninvertible, then the matrix A cannot be transformed into a diagonal matrix through a similarity relation.

However, there are ways of creating a "pseudo-diagonal" matrix J where all diagonal entries are the eigenvalues of A (counting multiplicities) and the entries above the diagonal or the "superdiagonal entries" are all either zero or one. An example is given below:

$$J = \begin{pmatrix} \lambda_1 & 1 & 0\\\ 0 & \lambda_1 & 0\\\ 0 & 0 & \lambda_2 \end{pmatrix}$$

These "pseudo-diagonal" matrices are called Jordan normal forms. Every Jordan matrix can be though of as a diagonal matrix D added to a nilpotent matrix N which only has nonzero entries on the superdiagonal; the nonzero entries are always 1:

$$D+N = \begin{pmatrix} \lambda_1 & 0 & 0\\\ 0 & \lambda_1 & 0\\\ 0 & 0 & \lambda_2 \end{pmatrix} + \begin{pmatrix} 0 & 1 & 0\\\ 0 & 0 & 0\\\ 0 & 0 & 0 \end{pmatrix}$$

The ones in the superdiaongal are only ever positioned over a repeated eigenvalue as seen in the example above where a one is placed over the first λ. The ones always appear in Jordan "blocks"; they are submatrices where the diagonal is always the same eigenvalue and the superdiagonal is always one. A Jordan block of size three is given below:

$$\begin{pmatrix} \lambda & 1 & 0 \\\ 0 & \lambda & 1 \\\ 0 & 0 & \lambda \end{pmatrix}$$

The Jordan matrix as a whole is just the direct sum of all Jordan blocks.

A Jordan block of size one is just the eigenvalue on its own; a diagonal matrix can be thought of as a special case where all of the Jordan blocks are of size 1 thus having no ones in the superdiagonal.

Each eigenvalue has two types of multiplicities:

μ = the algebraic multiplicity (the number of times the eigenvalue repeats)

γ = the geometric multiplicity (the number of Jordan blocks for a specific eigenvalue)

The algebraic multiplicity is the number of times said eigenvalue acts as a root to the characteristic polynomial:

$$\Vert A- \lambda I \Vert = 0$$

The geometric multiplicity is a bit more of a hassle to derive. The geometric multiplicity is defined by the equation below where n is the number of columns and rows and the rank of a matrix is the number of linearly independent rows or columns:

$$\gamma = n - rank(A- \lambda I)$$

The geometric multiplicity is always either less than or equal to the algebraic multiplicity:

$$1 \le \gamma \le \mu \le n$$

If an eigenvalue has an algebraic multiplicity of three and a geometric multiplicity of two then the three repeated roots will have two Jordan blocks associated with them. The only possible way this could work is to have a Jordan block of size one and size two as that is the only way to break up three into two parts:

$$\begin{pmatrix} \lambda & 1 & 0 \\\ 0 & \lambda & 0 \\\ 0 & 0 & \lambda \end{pmatrix}$$

Nevertheless, the geometric multiplicity only gives the number of Jordan blocks; it says nothing about the size of those blocks. The example above is trivial as breaking up the number three into two smaller whole numbers can only be one or two.

If an eigenvalue had an algebraic multiplicity of 4 and a geometric multiplicity of 2, there are two possible situations for how to split up the four roots into two Jordan blocks; the possible combinations are 1 and 3 or 2 and 2:

$$\begin{pmatrix} \lambda & 1 & 0 & 0 \\\ 0 & \lambda & 0 & 0\\\ 0 & 0 & \lambda & 1 \\\ 0 & 0 & 0 & \lambda \end{pmatrix}$$

or

$$\begin{pmatrix} \lambda & 1 & 0 & 0 \\\ 0 & \lambda & 1 & 0\\\ 0 & 0 & \lambda & 0 \\\ 0 & 0 & 0 & \lambda \end{pmatrix}$$

because

$$4 = 2+2 = 3+1$$

The two instances above are not equal to each other and different choices of the original matrix A will require one or the other. To determine which case applies, one must use the equation below which gives the number of Jordan blocks which are of size j (where the rank of a matrix is the number of linearly independent columns or rows):

$$rank((A- \lambda I)^{j+1}) - 2rank((A- \lambda I)^j) + rank((A- \lambda I)^{j-1})$$

The rank of a matrix can be determined by converting it into its row echelon form and counting the number of nonzero rows.

For diagonalization, the modal matrix P was composed of eigenvectors which each corresponded to the eigenvalues. A Jordan normal form requires a more generalized version of eigenvectors.

For any Jordan block of size i there will be i different columns in the modal matrix P. The equations for each of the columns is given below where p sub n is the nth column and λ is the eigenvalue of the Jordan block:

$$(A-\lambda I)p_1 = 0$$

$$(A-\lambda I)p_2 = p_1$$

$$\cdots$$

$$(A-\lambda I)p_i = p_{i-1}$$

The equations above can be alternatively solved by the equation below where k is the kth column of the i columns:

$$(A-\lambda I)^k p_k = 0$$

The generalized eigenvectors are the vectors that follow the equation above with k in this instance being the smallest integer where the equation equals zero; in other words, the equation below must also hold for an eigenvector of rank k:

$$(A-\lambda I)^{k-1} p_{k-1} \neq 0$$

In order to derive the Jordan matrix, you must follow the following steps:

Find the solutions to the characteristic polynomial for the eigenvalues (including repeated roots)

Determine the geometric multiplicity of each eigenvalue that is repeated

Find the size of the Jordan blocks by using the equation below (where the equation is for the number of Jordan blocks of size j)

$$rank((A- \lambda I)^{j+1}) - 2rank((A- \lambda I)^j) + rank((A- \lambda I)^{j-1})$$

Find the generalized eigenvectors for each of the Jordan blocks to put in the modal matrix P.

Find the inverse of P

$$A = P(D+N)P^{-1} = PJP^{-1}$$

While Jordan matrices are a bit more complicated than diagonal matrices, the Jordan matrices do have similar properties to diagonal matrices that are of use.

Every Jordan matrix can be written as a sum of a diagonal matrix D and a nilpotent matrix N where only the superdiagonal entries ever have a 1.

Every nilpotent matrix has the two main properties below:

$$DN = DN$$

$$\exists k \in \Bbb{N} \; \vert \; N^k = 0 \neq N^{k-1}$$

Because the diagonal matrix and nilpotent matrix commute and the nilpotent matrix eventually has a power which equals zero, the Jordan matrix raised to an nth power can be rewritten as a finite sum (essentially a binomial expansion).

$$J^n = \sum_{i=0}^k \binom{n}{i} D^{n-i} N^i$$

The number k is the smallest number where the nilpotent raised to said power equals zero.

This equation can thus be used to compute any matrix A raised to an nth power:

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