TODO

reverse of 2x2 matrix d & -b \\ -c & a

Лемма о фальшивом разложении определителя

https://en.wikipedia.org/wiki/Gramian_matrix
https://en.wikipedia.org/wiki/Vandermonde_matrix

https://ru.wikipedia.org/wiki/Сопряжённый_оператор, https://en.wikipedia.org/wiki/Hermitian_adjoint
https://en.wikipedia.org/wiki/Adjugate_matrix

Effect of multiplying a matrix by a diagonal matrix – https://solitaryroad.com/c108.html

Я правильно понимаю, что между билинейными формами и двойственными пространствами тоже есть связь?
Да, связь есть. Главный факт такой: если есть билинейная форма b: V x U -> R, то она дает два линейных отображения V -> U^* по правилу v -> b(v, -) и U -> V^* по правилу u -> b( -, u). Если билинейная форма не вырождена (у нее нет левого и правого ядер), то эти отображения являются изоморфизмами (в некотором смысле естественными). В частном случае скалярного произведения (-,-) : V x V -> R это дает изоморфизм V -> V^* по правилу v -> (v, -). То есть любая линейная функция на Евклидовом пространстве – это скалярное произведение с некоторым вектором.

По поводу задач на двойственные пространства.
Можно глянуть в Кострикина следующие задачи: 36.9-36.17, 36.19, 36.20. Можно и 36.18 поглядеть, но там надо догадаться до ответа.
Для 36.9 и 36.10 понадобится определение сопряженного базиса. Базис e_1,…,e_n из V и базис f_1,…,f_n из V^* сопряжены, если верно: f_i(e_j) = 1, если i = j и f_i(e_j) = 0, если i != j.

Есть принцип двойственности для векторных пространств. О соответствии различных характеристик у V и V^*. Если тебе интересно, то можешь поглядеть на мои задачи для очников прошлого года на эту тему. Их можно найти тут: https://yadi.sk/d/rWjfAGc33N9gsX
Смотреть надо на семинар 15 задача 10 и семинар 18 задачи: 6,7 и 8.
Задачи эти не то чтобы сильно простые.
Простые задачи я постараюсь подобрать из Кострикина. Но это не сегодня.
На квадратичные формы в Кострикине тоже есть задачи. Можно просто открыть раздел на квадратичные формы и попытать начать решать по-порядку. Но я как буду подбирать задачи на двойственное пространство, загляну и туда. Постараюсь завтра это сделать вместе с подбором домашнего задания по новой теме.

set of invertible matrices is open and dense
Где-то прочитал, что важно всегда помнить, что множество обратимых матриц открыто и плотно.
Да, я про это рассказывал, когда объяснял принцип продолжения по непрерывности. Даже набирал листочек по поводу метода решения задач этим принципом. Он есть среди материалов курса.

напоминаю про задачу о написании матрицы с заданным характеристическим многочленом.
..
смежный вопрос. пусть дано реккурентное соотношение порядка n, например числа фибоначи a_{n+2}=a_{n+1}+a_n. заметить что (a_n, a_{n-1}) связанны матричным преобразованием. (к предыдущему — в случае общего соотношения можно написать его характеристический многочлен). в терминах линейно алгебры научиться возводить матрицу в большую степень:
1) в базисе из собственных векторов
2) используя теорему гамильтона-кэли и выразив остаток от деления многочлена t^n на характеристический многочлен, это можно сделать зная собственные числа
https://en.wikipedia.org/wiki/Cayley–Hamilton_theorem#n-th_Power_of_matrix

получить формулу для чисел фибоначи.
посмотреть 4 параграф кострикина и получить ту же формулу через метод неопределенных коэффициентов
..
почитать про коммутирующие операторы, связь их собственных подпространств
*2017 3 7. a) частный случай вращательных, это составленные из 4х блоков. рассмотреть самый простой – диагональный случай блока, найти собственные числа.
b) записать вращение матрицы A в алгебраическом виде (поворот это два последовательных отражения в матрице). пусть S элементарное преобразование отражающее строки отн. середины. тогда S и исходная матрица коммутируют. S^2=id, у S два собственных под-ва. док. что на отрицательном собственные значения A равны нулю. значит ненулевые лежать в положительном (S-инвариантные)

Задача 8
За столом сидят старателей, перед каждым из которых лежит кучка золотого песка. Каждую минуту все старатели берут половину песка из левой кучки и половину из правой кучки и кладут себе. Опишите асимптотическое поведение количества песка в кучках для произвольного.
(лучше не гуглить решение, оно все равно плохое; найти симметрию в задаче, перевести на язык коммутирующих операторов)

кострикин 15.3 (найти собственные вектора матрицы)

http://www.mi-ras.ru/~podolskii/files/20162017/cw12_plus.pdf 10 задача. переписать это в терминах условных вероятностей и линейной алгебры (т.е. как марковский процесс), чтобы все решалось однотипно
http://wiki.cs.hse.ru/Дискретная_математика_1_2016/2017 там очень хорошие задачи и лекции. если будет тягостно с линейной алгеброй бери пачками задачи оттуда. первые 15 листков нужно в конечном счете прорешать

next time

двойственность — переход между представлением оболочкой и матрицей
двойственные пространства
теорема Гамильтона-Кэли

Systems of linear equations

Gaussian elimination

Elementary row operations are used to reduce a matrix to what is called triangular form.
Row switching, row multiplication, row addition.
These don’t change the solution set.

http://en.wikipedia.org/wiki/Elementary_matrix

Elementary row operations do not change the solution set:

$\boxed{\Rightarrow}$ let $x: \ Ax=0 \quad \Leftrightarrow \quad E \cdot Ax = E \cdot 0$,
then $(EA)x=0$,
then $x$ is in null set of $EA$

$\boxed{\Leftarrow}$ let $x: \ EAx=0$,
then $Ax = (E^{-1}E)Ax = E^{-1} (EA)x = E^{-1} 0 = 0$,
then $x$ is in null set of $A$

But they change the image, see what happens after row switching:

A new matrix represents a projection to completely another plane.

Elementary row operations do not change the kernel of a matrix (and hence do not change the solution set), but they do change the image.

Dually, elementary column operations do not change the image, but they do change the kernel.

Gauss-Jordan elimination vs Gaussian elimination

Usually one use Gauss-Jordan to invert a matrix and Gaussian for rank.

Both are $O(n^3)$ for an $n$ by $n$ full rank matrix, but order of magnitude of Gauss-Jordan elimination is $n^3$, whereas that for Gaussian elimination is $\frac{2}{3}n^3$

Row echelon form — result of Gaussian elimination:

Reduced row echelon form — result of Gauss-Jordan elimination:

and

TODO LOW: http://cstheory.stackexchange.com/questions/3921/what-is-the-actual-time-complexity-of-gaussian-elimination

Block matrices

https://en.wikipedia.org/wiki/Block_matrix

Solution set of Homogeneous systems

Every homogeneous system has at least the zero solution.

if $\operatorname{rank}(A) = n$ then there is only trivial solution: $\vec{x} = \vec{0}$
if $\operatorname{rank}(A) < n$ then there is a basis of $(n - \operatorname{rank}(A))$ linear independent solutions, general solution looks like $c_1\vec{x}^1+\ldots+c_{n-r}\vec{x}^{n-r}$

http://ru.wikipedia.org/wiki/Решение_систем_линейных_алгебраических_уравнений

if $x_1$, $x_2$ are solutions for $Ax=0$
then $x_1 + x_2$ is solution for the homogenous system $Ax=0$
$A (x_1 + x_2) = 0 - 0 = 0$

TODO: proof of theorem about solution set of homogeneous system of linear equations

Elementary row operations

блочное преобразование матриц

элементарные преобразования столбцов
(или строк)

Умножение на следующее:
E -B
0 E

есть преобразования строк.

справа – столбцов

Problems

https://www.dropbox.com/home/hse_math?preview=linear_algebra+01+--+Seminar1_problems.pdf

Example:

$\exists A \quad A^n = 0$:

bilinear forms with char(F) not equal 2

bilinear function is a generalization of vector dot product
main purpose of bilinear theory is to make matrix simpler by choosing a good basis, so it’s good to know properties that do not depend on basis: kernel, etc

bilinear function $\alpha: V \times V \mapsto F$ is linear in each of its arguments

everywhere it is assumed that $\operatorname{char}(F) \neq 2$

characteristic of field

TODO: what is characteristic of F? http://ru.wikipedia.org/wiki/Характеристика_кольца,

if exists, the characteristic $\operatorname{char}(R)$ is the smallest $n$ such that $\forall r \ : \ nr =0$

if doesn’t exist, then $\operatorname{char}(R) = 0$
when $\operatorname{char}(R) = 1$, the ring is trivial set: {0, 1}

$\operatorname{char} \mathbb{Z} = \operatorname{char} \mathbb{Q} = \operatorname{char} \mathbb{R} = 0$

$\operatorname{char} \mathbb{Z}_n = n$

$\operatorname{char}(\mathbb{F}_{p^m}) = p$

TODO: why characteristic of GF(p^m)=p?

dim of kernel of bilinear form

$\ker\alpha = { y : \alpha(x,y) = 0 }$

$\ker\alpha = { y \ : \ \alpha(e_i,y) = 0, \ i = 1, \ldots, n }$
this gives a system of linear equations, which matrix is $A$
then
$\operatorname{dim}\ker\alpha = n - \operatorname{rank} A$

examples of bilinear forms

example. vector dot product is a bilinear function
example. $\alpha (f, g) = \int_a^b f(x)g(x)dx$ is a bilinear function in $C[a,b]$

TODO: add more examples of bilinear forms

matrix representation of bilinear form

$x = \sum x_i e_i$, $y = \sum y_j e_j$

, where $a_{ij} = \alpha(e_i, e_j)$

$A = (a_{ij})$ is matrix representation in basis ${e_1, \ldots, e_n }$

$\alpha(x, y) = X^\top A \ Y$, where X and Y are rows of coordinates

in a given basis each bilinear map can be uniquely represented by a matrix

examples of matrices of bilinear forms

$f : F^3 \mapsto F$
$f (x, y) = x_1 y_2 + x_1 y_3 - 2 x_3 y_1$

$a_{11} = f ( < 1, 0, 0 > , < 1, 0, 0 > ) = 1 \cdot 0 + 1 \cdot 0 - 2 \cdot 0 \cdot 1 = 0$
$a_{12} = f ( < 1, 0, 0 > , < 0, 1, 0 > ) = 1 \cdot 1 + 1 \cdot 0 - 2 \cdot 0 \cdot 0 = 1$
$a_{12} = f ( < 1, 0, 0 > , < 0, 0, 1 > ) = 1 \cdot 0 + 1 \cdot 1 - 2 \cdot 0 \cdot 0 = 1$
$\ldots$
$a_{31} = f ( < 0, 0, 1 > , < 1, 0, 0 > ) = 0 \cdot 0 + 0 \cdot 1 - 2 \cdot 1 \cdot 1 = 2$
$\ldots$
$a_{33} = f ( < 0, 0, 1 > , < 0, 0, 1 > ) = 0 \cdot 0 + 0 \cdot 1 - 2 \cdot 1 \cdot 0 = 0$

change of basis

$(e’_1, \ldots, e’_n) = (e_1, \ldots, e_n) \cdot C$
$X = CX’$, $Y = CY’$
$\alpha(x, y) = (CX’)^\top \ A \ (CY’) = (X’)^\top C^\top A \ C \ Y’$
then in new basis $A = C^\top A \ C$

symmetric and skew-symmetric bilinear forms

symmetric: $\alpha(y, x) = \alpha(x, y)$
skew-symmetric: $\alpha(y, x) = - \alpha(x, y)$
there are also non-symmetric and non-skew-symmetric bilinear functions

bilinear function is symmetric $\iff$ $A^\top = A$
bilinear function is skew-symmetric $\iff$ $A^\top = - A$

this property is independent of basis

symmetric matrix:

skew-symmetric matrix always have zeros on its diagonal:

Theorem:
if $\operatorname{char}(F) \neq 2$, then vector space of all bilinear forms is direct sum of subspaces: of symmetric bilinear forms and skew-symmetric bilinear forms
Proof:
if $\alpha$ is symmetric and skew-symmetric simultaneously, then $\alpha(x, y) = \alpha(y, x) = - \alpha(x, y) \ \Rightarrow \ 2\alpha(x, y) = 0 \ \Rightarrow \ \alpha(x, y) = 0$

i.e. every bilinear form can be represented as a sum of symmetric and skew-symmetric bilinear forms:
$\alpha(x,y) \ = \ \frac{1}{2}{ \alpha(x,y) + \alpha(y,x) } \ + \ \frac{1}{2}{ \alpha(x,y) - \alpha(y,x) }$
or in for their matrices: $A \ = \ \frac{1}{2}(A + A^\top) \ + \ \frac{1}{2}(A - A^\top)$

but if $\operatorname{char}(F) = 2$ the theorem does not work:
$F = {0, 1}$, then a skew-symmetric form is the same as a symmetric form: $\alpha(-1, x) = - \alpha(1, x) = \alpha(1, x)$
and, for example, $\begin{pmatrix}0 & 1 \ 0 & 1\end{pmatrix}$ can’t be represented as a sum of two symmetric matrices

TODO: do i need alternating forms which are f(v,v) = 0?
If the characteristic of F is not 2 then the converse is also true: every skew-symmetric form is alternating.
If, however, char(F) = 2 then a skew-symmetric form is the same as a symmetric form and there exist alternating forms which are not symmetric/skew-symmetric.
A bilinear form is alternating if and only if its coordinate matrix is skew-symmetric and the diagonal entries are all zero (which follows from skew-symmetry when char(F) ≠ 2).

TODO: in bilinear forms section, what if $\operatorname{char} F = 2$ +? https://docs.google.com/a/egotv.ru/viewer?a=v&q=cache:2cBYrJa0egoJ:www.math.uregina.ca/~bailey/pjc60/slides/JIHtalk.pdf+&hl=en&gl=ru&pid=bl&srcid=ADGEESiFkiK-RE_grFgDgATUbJigqZNg55FUUXogfSgVP5uy1H2ONLVOwNkV_F-DME8oZlYc8aakE46cDwGcaUamOKz39EEZYkQRbswF7-J-NPotAv4bf6Bc2OCHaeRtsxAEIHtNPZ3j&sig=AHIEtbTQgPTST7GrVjPGu_XgwEySzql5Vw

Quadratic forms

$q(x) = \sum a_{ij} \, x_i x_j \ : \ V \mapsto F$
$A = (a_{ij})$ is the symmetric matrix of the quadratic form

the matrix is symmetric when $\operatorname{char} F \neq 2$

examples:

any symmetric bilinear form b defines a quadratic form: $q(x) = \alpha(x,x)$

we can restore the bilinear function, this is called a polarization of a quadratic form: $\alpha(x,y)=\frac{1}{2}{q(x+y)-q(x)-q(y)} = x^\mathrm{T}Ay = y^\mathrm{T}Ax$
this is from $\alpha(x+y, x+y) = q(x) + q(y) + 2\alpha(x,y)$
these two processes are the inverses of one another $\alpha(x, -x)=\frac{1}{2}{q(0)-q(x)-q(-x)}$ $\Leftrightarrow$ $q(x) = \alpha(x,x) + \frac{1}{2}q(0)$. Substitute $x=0$ and get $q(0) = 0$, then $q(x) = \alpha(x,x)$

matrix of a polar bilinear form is equal to matrix of quadratic form
as a consequence, when $\operatorname{char} F \neq 2$, the theories of symmetric bilinear forms and of quadratic forms in n variables are essentially the same

orthogonal complement

$\operatorname{char} F \neq 2$
$\alpha(x, y)$ is a symmetric or skew-symmetric bilinear form
$x$ and $y$ are /orthogonal/ if $\alpha(x, y) = 0$
when bilinear form is skew-symmetric, every vector is orthogonal to itself

the orthogonal complement of a subspace $W$ of a vector space $V$ equipped with a bilinear form is the set $W^\bot$ of all vectors in $V$ that are orthogonal to every vector in $W$
$W^\bot = {x\in V \ : \ \forall y\in W \ \alpha(x, y) = 0 }$

$V^\bot = \operatorname{Ker}\alpha$

if $\alpha(x, y) \not\equiv 0$, then $\dim W + \dim {W^\bot} = \dim V$
and $(W^\bot)^\bot = W$

orthogonal basis for symmetric bilinear form

A basis is orthogonal with respect to the bilinear function when basis vectors are mutually orthogonal
When $\operatorname{char} F \neq 2$, for a symmetric bilinear form there is always an orthogonal basis
a basis is orthogonal $\iff$ the matrix representation is a diagonal matrix, then bilinear and quadratic function look like this: $\alpha(x, y) = a_1 x_1 y_1 + \ldots + a_n x_n y_n$, $q(x) = a_1 x_1^2 + \ldots + a_n x_n^2$

Gram–Schmidt process — orthonormalising a set of vectors

$\mathbf{b}_1$ and $\mathbf{b}_2$ are orthogonal:

$\mathbf{b}_2$ and $\mathbf{b}_3$ are orthogonal similarly:

etc

canonical, or normal form of quadratic function

If $F = \mathbb C$, coefficients $\alpha_i = 0, 1$, i.e. canonical quadratic function look like $q(x) = x_1^2 + \ldots + x_r^2$ after a suitable permutation of basis vectors, $r = \operatorname{rank} q$.

If $F = \mathbb R$, coefficients $\alpha_i = -1, 0, 1$, i.e. canonical quadratic function look like $q(x) = x_1^2 + \ldots + x_k^2 - x_{k+1}^2 - \ldots - x_{k+l}^2$ after a suitable permutation of basis vectors.
$k+l = \operatorname{rank} q$

a pair $(k,l)$ is called signature of $q$

law of inertia

$k$ and $l$ do not depend on a particular choice of diagonalizing basis

// it’s about invariant signature, the word “inertia” was used because no better word was found

$k$ is max dimention of a subspace on which $q(x)$ is positive definite
(same for $l$)

TODO: proof of law of inertia

positive definite and negative definite quadratic forms

$q(x) > 0$ for $x \neq 0$ is positive definite
$q(x) < 0$ for $x \neq 0$ is negative definite

bilinear form $\alpha(x, y)$ is positive definite if associated quadratic form $q(x)$ is positive definite (same for negative)
dot product $(x, y)$ of vectors is positive definite

normal form of positive definite quadratic function is $q(x) = x_1^2 + \ldots + x_k^2$

Sylvester’s criterion — is a quadratic form positive definite

Sylvester’s criterion states that a matrix $A$ is positive definite $\iff$ all the following matrices have a positive determinant:
the upper left 1-by-1 corner
the upper left 2-by-2 corner
…
the upper left n-by-n corner – $A$ itself






in another words, all leading principal minors are positive

if every $\Delta_k$ is non-zero then the negative index of inertia is equal to the number of sign changes in the sequence $\Delta_0=1, \Delta_1, \ldots, \Delta_n=\det A$

TODO: do i need $F = Z_p$?

TODO: do i need skew-symmetric bilinear functions? symplectic basic?

Eigenvalues and Eigenvectors

http://setosa.io/ev/eigenvectors-and-eigenvalues/

http://math.stackexchange.com/questions/243533/how-to-intuitively-understand-eigenvalue-and-eigenvector
http://math.stackexchange.com/questions/36815/a-simple-explanation-of-eigenvectors-and-eigenvalues-with-big-picture-ideas-of
http://www.quora.com/What-is-the-best-way-to-intuitively-explain-what-eigenvectors-and-eigenvalues-are
https://georgemdallas.wordpress.com/2013/10/30/principal-component-analysis-4-dummies-eigenvectors-eigenvalues-and-dimension-reduction/
http://anothermathgeek.hubpages.com/hub/What-the-Heck-are-Eigenvalues-and-Eigenvectors

“Eigen” is German for “peculiar to”
some authors call these characteristic values and vectors

[[file:.images/algebra.org.20121108-1456-402JmH.screenshot.png]]

operator $\mathcal A$ has a scalar /eigenvalue/ $\lambda$ if there is a nonzero /eigenvector/ $x$ such that $Ax = \lambda x$

if $\lambda$ is an eigenvalue of $A$ then $\lambda^k$ is an eigenvalue of $A^k$: $Ax = \lambda x \ \Rightarrow \ A^k x = \lambda^k x$
$A^k x = A^{k-1} \lambda x = \lambda A^{k-1} x = \cdots = \lambda^k x$

if $x$ is an eigenvector, then $kx$ is an eigenvector too: $A(kx) = \lambda (kx)$

eigenspace

/eigenspace/ associated with an eigenvalue $V^{\lambda} = \left{ \ x \ : \ A x = \lambda x \ \right}$ consists of $\boldsymbol 0$ and all eigenvectors associated with the eigenvalue
eigenspace is a subspace since nonempty and $Ax = \lambda x, \ Ay = \lambda y \ \Rightarrow \ A(\alpha x + \beta y) = \lambda (\alpha x + \beta y)$

eigenspaces are linearly independent
i.e. eigenvectors corresponding to different eigenvalues are linearly independent

$\dim V^{\lambda}$ is /geometric multiplicity/ of an eigenvalue $\lambda$

characteristic polynomial

The eigenvalues of a matrix A are precisely the solutions $\lambda$ to the equation
$\det(A - \lambda I) = 0$

$Ax = \lambda x \ \Leftrightarrow \ Ax - \lambda I x = 0 \ \Leftrightarrow \ (A - \lambda I) x = 0$
If there exists an inverse $(A - \lambda I)^{-1}$ then both sides can be left-multiplied by it to obtain $x=0$.
Therefore, if $\lambda$ is such that $(A - \lambda I)$ is invertible, $\lambda$ cannot be an eigenvalue.
If it is not invertible, $\lambda$ is an eigenvalue.

/characteristic polynomial/ $\chi_{\mathcal A}(t) = \det(tI - A)$ ( $= (-1)^n \, \det(A-tI)$ )
The solutions of the /characteristic equation/ $\chi_A(t) = 0$ are precisely the eigenvalues.

Two similar matrices have the same characteristic polynomial: $\chi_{C^{-1}AC} (t)=\chi_{A}(t)$
/similar/: $C$ is invertible change of basis. $A’ = C^{-1}AC$. Then $\det(tI - A’) = \det(tC^{-1}IC - C^{-1}AC) = \det\left( C^{-1} (tI -A) C \right) = \det C^{-1} \det(tI - A) \det C = \det(tI - A)$

We can assume $\chi_{\mathcal A}(t) = \chi_A(t)$
characteristic polynominal does not depend on the choice of a basis

TODO: [#B] prove $\chi_A(t) = t^n - \operatorname{tr}A \ t^{n-1} + \ldots + (-1)^n \det A$
[[file:.images/algebra.org.20121113-2217-402pcW.screenshot.png]]

in practice

in practice the characteristic polynomial is used rarely

ironic thing is that the relationship between polynomial roots and eigenvalues is often exploited in the opposite direction
roots of a polynomial, one approach is to construct its companion matrix, and then find its eigenvalues
there are some methods that work by iteratively transforming the matrix in some way (e.g., Householder transformations, or Jacobi transformations)
this approach is used in the root finder in the Chebfun system, for example — it routinely finds roots of polynomials whose degrees are in the hundreds
in some sense, finding eigenvalues is easier than finding polynomial roots — certainly more high-quality numerical methods software is available to help out

examples of getting eigenvalues and eigenvectore

characteristic polynomial:

eigenvalues are 1, 2, 11

when we have eigenvalues, we can get corresponding eigenvectors from a system of linear equations:
$Ax = \lambda x$

another example

$\lambda_1 = 1, \ \lambda_2 = 6$

$y = 2x$

$\Rightarrow$ for an eigenvalue 6 the eigenvectors are $(a, 2a)$ for nonzero $a$

properties of eigenvalues

$f(x) = (x - \lambda_1)^{d_1} \ldots (x - \lambda_k)^{d_k}$

$\operatorname{tr}(A) = d_1 \lambda_1 + \cdots + d_k \lambda_k$

$\det A = \prod \lambda_i$
More generally, $\operatorname{tr}(A^k) = \sum \lambda_i^k$

matrix is invertible $\iff$ no zero eigenvalue

if $A$ is invertible, then eigenvalues of $A^{-1}$ are $\frac{1}{\lambda_1}, \ldots, \frac{1}{\lambda_n}$

TODO: [#B] http://en.wikipedia.org/wiki/Eigenvalue_algorithm
http://en.wikipedia.org/wiki/Divide-and-conquer_eigenvalue_algorithm

TODO: [#B] see also Complexification of vector space. Vinberg 222

diagonizable

diagonalizable matrices and maps are of interest because diagonal matrices are especially easy to handle: their eigenvalues and eigenvectors are known and one can raise a diagonal matrix to a power by simply raising the diagonal entries to that same power.

a square matrix is called /diagonalizable/ if it is similar to a diagonal matrix: $C^{-1} A C = D$
linear operator is called /diagonalizable/ if there exists a basis with respect to which the operator is represented by a diagonal matrix

linear operator is diagonalizable
$\iff$
the sum of the dimensions of its eigenspaces is equal to $\operatorname{dim}V$
$\iff$
there exists a basis of $V$ consisting of eigenvectors.

With respect to such a basis, operator will be represented by a diagonal matrix. The diagonal entries of this matrix are the eigenvalues.

[[.images/algebra.org.20121114-0007-402pjK.screenshot.png]]
Vinberg p226

$P^{-1}AP = D$
\begin{align} A^k &= (PDP^{-1})^k = (PDP^{-1}) \cdot (PDP^{-1}) \cdots (PDP^{-1}) \&= PD(P^{-1}P) D (P^{-1}P) \cdots (P^{-1}P) D P^{-1} \&= PD^kP^{-1} \end{align}

illustration

[[.images/20150601-1533-486w0z.image.jpg]]

[[.images/20150601-1534-486i-C.image.png]]

The vertical axis here is sum-of-eigenvalues. The horizontal axis is product-of-eigenvalues. So points on this “plane” could have come from a high-dimensional matrix repeating on a system. It’s helpful to know, due to theorems, that (1) the sum of diagonal entries = the sum of the eigenvalues, and (2) the product of diagonal entries = the determinant. So neither axis is super hard to compute.

superlearn

https://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process