# Matice v R for dummies pre predmet 1-PMA-215
# Autor: Radoslav Harman, KAMS, FMFI UK

# Vytvorenie matice
A <- matrix(c(1, 2, 3, 2, 3, 2), ncol=2); print(A)
A <- matrix(c(1, 2, 3, 2, 3, 2), ncol=2, byrow=TRUE); print(A)
B <- matrix(c(3, 4, 5, 1, 1, 2), ncol=2); print(B)

# Sucet matic a pozlozkovy (Hadamardov) sucin matic
print(A + B); print(A * B)

# Transpozicia matice
C<-t(B); print(C)

# Vyber diagonalnych prvkov matice
print(diag(A))

# Stopa matice
tr<-function(X){sum(diag(X))}; print(tr(C))

# Standardny sucin matic
print(A %*% C); print(C %*% A)

# Overme tvrdenie tr(AC)=tr(CA)
print(tr(A %*% C)); print(tr(C %*% A))

# Overme asociativnost sucinu matic
print((A %*% C) %*% B); print(A %*% (C %*% B))

# Vytvorenie diagonalnej a jednotkovej matice
D <- diag(c(2, 3, 4)); print(D)
I5 <- diag(5); print(I5)

# Zlepenie matic po stlpcoch a riadkoch
E <- cbind(D, B); print(E)
F <- rbind(D, C); print(F)

# Vytvorme nahodnu maticu
H <- matrix(sample(1:10, 25, replace=TRUE), ncol=5); print(H)

# Vypocitame inverziu matice H a hned to radsej overme
Hinv <- solve(H); print(Hinv)
print(H %*% Hinv); print(round(H %*% Hinv, 10))

# Vypocitajme QR rozklad matice
res <- qr(H); print(qr.Q(res)); print(qr.R(res))

# Dobry zvyk - overenie
print(qr.Q(res) %*% qr.R(res)) - H

# Nacitajme specialnu kniznicu
library(MASS)

# Vypocitajme zovseobecnenu inverziu matice A
print(ginv(A))

# Je to naozaj zovseobecnena inverzia?
G <- ginv(A); print(round( A %*% G %*% A, 10))

# Vyrobme si nejaku permutacnu maticu a pohrajme sa s nou
P <- I5[c(1, 3, 4, 2, 5),]; print(P); print(P %*% t(P))
print(H); print(P %*% H); print(H %*% P)

# Vyrobme si funkciu, ktora vrati rotacnu maticu typu 2x2
Ort2 <- function(x){matrix(c(cos(x), sin(x), -sin(x), cos(x)), ncol=2)}
print(Ort2(pi / 4)); U <- Ort2(pi / 4); print(U %*% t(U))

# Domcek
X<-matrix(c(0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0.5, 1.5, 0, 1, 0, 0, 1, 0), nrow=2)
plot(X[1,], X[2,], type="l", xlim=c(-2, 2), asp=1)

# Potocme si domcek
Y <- Ort2(pi / 6) %*% X; plot(Y[1,], Y[2,], type="l", xlim=c(-2, 2), asp=1)
Y <- Ort2(pi / 2) %*% X; plot(Y[1,], Y[2,], type="l", xlim=c(-2, 2), asp=1)

# Potocme si ho este viac
for(i in 0:1000){Y <- Ort2(2 * pi * (i/1000)) %*% X; plot(Y[1,], Y[2,], type="l", xlim=c(-3, 3), ylim=c(-3, 3), asp=1)}