Ovde rešavamo zadatke iz MML Book

MML-book Chapter 2. Linear Algebra, Exercises

Koristićemo R i matlib biblioteku.

library(matlib)

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2.1

  1. Jeste Abelova grupa.
  • Zatvorenost: Neka su \(x, y \in \mathbb{R}\backslash \{ -1 \}\). Pretpostavimo da je \(a \star b = -1 \;\Leftrightarrow\; a b + a + b = -1 \;\Leftrightarrow\; a ( b + 1 ) = -1 ( b + 1 )\).
    Poslednja jednakost je moguća samo ako je \(a = -1\) ili \(b = -1\), dakle, zatvorenost važi.
  • Asocijativnost: \(( a \star b ) \star c = (a b + a + b) \star c = a b c + a c + b c + a b + a + b + c\) \(a \star ( b \star c ) = a \star (b c + b + c) = a b c + a b + a c + a + b c + b + c\)
    Vidimo da asocijativnost važi.
  • Neutralni element je \(0\): \(0 \star b = b\), \(a \star 0 = a\).
  • Inverzni element: \(a \star a' = a a' + a + a' = 0 \;\Leftrightarrow\; a' ( a + 1) = -a \;\Leftrightarrow\; a' = -a / ( a + 1 )\), što može da se izračuna za sve elemente iz \(\mathbb{R}\backslash \{ -1 \}\).
  • Komutativnost: očigledna.
  1. Rešenja su: \(x \in \{ -3, 1 \}\).
  • \(3 \star x \star x = 3 \star ( x^2 + 2 x ) = 3 x^2 + 6 x + 3 + x^2 + 2 x = 4 x^2 + 8 x + 3 = 15 \;\Leftrightarrow\;\)
    \(4 x^2 + 8 x - 12 = 0 \;\Leftrightarrow\; x \in \{ -3, 1 \}\).

2.2

Za \(k = 0, 1, \ldots, n-1,\; k \in \overline{k}\). Skup \(\mathbb{Z}_n = \{ \overline{0}, \overline{1}, \ldots, \overline{n-1} \}\) predstavlja ostatke pri deljenju sa \(n\). Operacije \(\oplus\) i \(\otimes\) su sabiranje i množenje po modulu \(n\).

  1. Jeste Abelova grupa:
  • Zatvorenost: Pošto za svako \(k \in \mathbb{Z}\) jedini mogući ostaci pri deljenu sa \(n\) su predstavljeni u \(\mathbb{Z}_n\), zatvorenost je očigledna.
  • Asocijativnost sledi iz asocijativnosti sabiranja:
    \(( \overline{a} \oplus \overline{b} ) \oplus \overline{c} = ( \overline{a + b} ) \oplus \overline{c} = \overline{( a + b ) + c} = \overline{ a + ( b + c )} = \overline{a} \oplus ( \overline{b + c} ) = \overline{a} \oplus ( \overline{b} \oplus \overline{c} )\)
  • Neutralni element je \(\overline{0}\): \(\overline{0} \oplus \overline{k} = \overline{0 + k} = \overline{k} = \overline{k + 0} = \overline{k} \oplus \overline{0}\).
  • Inverzni element: za \(\overline{k}\), inverzni element je \(\overline{n - k}\), jer \(\overline{k} \oplus \overline{n - k} = \overline{k + n - k} = \overline{n} = \overline{0}\).
  • Komutativnost: očigledna.
#########
#  2.2  #
#########

# b
matrix(1:4,ncol=1) %*% matrix(1:4,nrow=1) %% 5
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]    2    4    1    3
## [3,]    3    1    4    2
## [4,]    4    3    2    1

  1. Očigledno je zatvorena operacija, asocijativnost i komutativnost slede, kao i kod \(\oplus\) iz asocijativnosti i komutativnosti množenja u skupu \(\mathbb{Z}\). Isto, neutralni element za množenje je \(\overline{1}\).
    Iz tablice se vidi da svaki element ima inverzni: \(1^{-1} = 1, 2^{-1} = 3, 3^{-1} = 2, 4^{-1} = 4\).

  2. U skupu \(\mathbb{Z}_8\) važi \(\overline{4} \otimes \overline{2} = \overline{4 \cdot 2} = \overline{8} = \overline{0}\), tako da \(( \mathbb{Z}_8 \backslash \{ 0 \}, \otimes )\) nije ni grupoid.

  3. \(( \mathbb{Z}_n \backslash \{ 0 \}, \otimes )\) je Abelova grupa \(\Leftrightarrow\) n je prost broj. Dokaz:
    Komutativnost, asocijativnost i neutralnost \(\overline{1}\) za \(\otimes\) proizilaze iz osobina grupoida \(( \mathbb{Z}\backslash \{ 0 \}, \cdot )\).
    Ako \(n\) nije prost broj, \(n = m \ k\), onda \(( \mathbb{Z}_n \backslash \{ 0 \}, \otimes )\) nije grupoid jer \(\overline{m} \otimes \overline{k} = \overline{n} = \overline{0}\).
    Neka je \(n\) prost broj i neka \(\overline{m} \in \mathbb{Z}_n \backslash \{ 0 \}\) i neka \(m \in \{ 0, 1, \ldots, n - 1 \}\). Brojevi \(n\) i \(m\) su uzajamno prosti, na osnovu Bezuove teoreme (Bézout’s identity) postoje \(x, y \in \mathbb{Z}\) tako da \(x \ n + y \ m = 1\). Onda je \(\overline{y} = \overline{m}^{-1}\), jer \[ \overline{y} \otimes \overline{m} = \overline{y \ m} = \overline{- x \ n + 1} = \overline{1} = \overline{m \ y} = \overline{m} \otimes \overline{y}.\]

2.3

Ako u grupoidu sa neutralnim elementom za podskup važi zatvorenost operacije i nalaženja inverznog elemeta, onda je taj podskup u odnosu na restrikciju operacije grupa. Zatvorenost operacije: \[\begin{bmatrix} 1 & x_{1} & y_{1}\\ 0 & 1 & z_{1}\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & x_{2} & y_{2}\\ 0 & 1 & z_{2}\\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & x_{1} + x_{2} & x_{1} \ z_{2} + y_{1} + y_{2}\\ 0 & 1 & z_{1}+z_{2}\\ 0 & 0 & 1 \end{bmatrix}\] Zatvorenost traženja inverznog elementa:
\[\begin{bmatrix} 1 & x & y\\ 0 & 1 & z\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -x & x \ z - y \\ 0 & 1 & -z \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\\end{bmatrix} = \begin{bmatrix} 1 & -x & x \ z - y \\ 0 & 1 & -z \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & x & y\\ 0 & 1 & z\\ 0 & 0 & 1 \end{bmatrix}\] Komutativnost ne važi

2.4

Pomnožiti matrice, ako je moguće.

#########
#  2.4  #
#########

# a
# Ne mogu se pomnožiti

# b
A = matrix(c(1:9), ncol = 3, byrow = T);
B = matrix(c(1,1,0,0,1,1,1,0,1), ncol = 3, byrow = T); B
##      [,1] [,2] [,3]
## [1,]    1    1    0
## [2,]    0    1    1
## [3,]    1    0    1
A%*%B
##      [,1] [,2] [,3]
## [1,]    4    3    5
## [2,]   10    9   11
## [3,]   16   15   17
# c
B%*%A
##      [,1] [,2] [,3]
## [1,]    5    7    9
## [2,]   11   13   15
## [3,]    8   10   12
# d
A = matrix(c(1,2,1,2,4,1,-1,-4), ncol = 4, byrow = T); A
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    1    2
## [2,]    4    1   -1   -4
B = matrix(c(0,3,1,-1,2,1,5,2), ncol=2, byrow = T); B
##      [,1] [,2]
## [1,]    0    3
## [2,]    1   -1
## [3,]    2    1
## [4,]    5    2
A%*%B
##      [,1] [,2]
## [1,]   14    6
## [2,]  -21    2
# e
B%*%A
##      [,1] [,2] [,3] [,4]
## [1,]   12    3   -3  -12
## [2,]   -3    1    2    6
## [3,]    6    5    1    0
## [4,]   13   12    3    2

2.5

#########
#  2.5  #
#########

# a

A = matrix(c(1,2,2,5,1,5,-1,2,-1,-7,1,-4,-1,-5,3,2),ncol=4); A
##      [,1] [,2] [,3] [,4]
## [1,]    1    1   -1   -1
## [2,]    2    5   -7   -5
## [3,]    2   -1    1    3
## [4,]    5    2   -4    2
b = c(1,-2,4,6); b
## [1]  1 -2  4  6
gaussianElimination(A,b)
##      [,1] [,2] [,3]       [,4]        [,5]
## [1,]    1    0    0  0.6666667  1.83333333
## [2,]    0    1    0 -2.6666667 -0.08333333
## [3,]    0    0    1 -1.0000000  0.75000000
## [4,]    0    0    0  0.0000000 -0.50000000
# Matrica $A$ i proširena matrica $A|b$ nemaju isti rang, sistem je nemoguć

# b

A = matrix(c(1,1,2,-1,-1,1,-1,2,0,0,0,0,0,-3,1,-2,1,0,-1,-1), ncol = 5); A
##      [,1] [,2] [,3] [,4] [,5]
## [1,]    1   -1    0    0    1
## [2,]    1    1    0   -3    0
## [3,]    2   -1    0    1   -1
## [4,]   -1    2    0   -2   -1
b = c(3,6,5,-1); b
## [1]  3  6  5 -1
gaussianElimination(A,b) -> Ab1 ; Ab1
##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    0    0    0   -1    3
## [2,]    0    1    0    0   -2    0
## [3,]    0    0    0    1   -1   -1
## [4,]    0    0    0    0    0    0
Ab2=rbind(Ab1[1:2,],-(1:6==3),Ab1[3,],-(1:6==5)); Ab2
##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    0    0    0   -1    3
## [2,]    0    1    0    0   -2    0
## [3,]    0    0   -1    0    0    0
## [4,]    0    0    0    1   -1   -1
## [5,]    0    0    0    0   -1    0

Rešenje sistema su svi vektori \({\boldsymbol x} = [3,0,0,-1,0]^T + x_3 [0,0,1,0,0]^T + x_5 [ 1,2,0,1,1 ]^T\), \(x_3, x_5 \in \mathbb{R}\).

2.6

#########
#  2.6  #
#########

A=matrix(c(0,1,0,0,1,0,0,0,0,1,1,0,0,1,0,0,0,1), ncol = 6, byrow = TRUE); A
##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    0    1    0    0    1    0
## [2,]    0    0    0    1    1    0
## [3,]    0    1    0    0    0    1
b=c(2,-1,2); b
## [1]  2 -1  2
gaussianElimination(A,b) -> Ab1 ; Ab1
##      [,1] [,2] [,3] [,4] [,5] [,6] [,7]
## [1,]    0    1    0    0    0    1    2
## [2,]    0    0    0    1    0    1   -1
## [3,]    0    0    0    0    1   -1    0
Ab2=rbind(-(1:7==1),Ab1[1,],-(1:7==3),Ab1[2:3,],-(1:7==6)); Ab2
##      [,1] [,2] [,3] [,4] [,5] [,6] [,7]
## [1,]   -1    0    0    0    0    0    0
## [2,]    0    1    0    0    0    1    2
## [3,]    0    0   -1    0    0    0    0
## [4,]    0    0    0    1    0    1   -1
## [5,]    0    0    0    0    1   -1    0
## [6,]    0    0    0    0    0   -1    0

Rešenje sistema su svi vektori \({\boldsymbol x} = [0,2,0,-1,0,0]^T + x_1 [1,0,0,0,0,0]^T + x_3 [0,0,1,0,0,0]^T + x_6 [0,1,0,1,-1,-1]^T\), \(x_1, x_3, x_6 \in \mathbb{R}\).

2.7

#########
#  2.7  #
#########

A = matrix(c(6,4,3,6,0,9,0,8,0), ncol = 3, byrow = TRUE); A
##      [,1] [,2] [,3]
## [1,]    6    4    3
## [2,]    6    0    9
## [3,]    0    8    0
A = A - 12 * diag(3); A
##      [,1] [,2] [,3]
## [1,]   -6    4    3
## [2,]    6  -12    9
## [3,]    0    8  -12
A = rbind(A,c(1,1,1)); A
##      [,1] [,2] [,3]
## [1,]   -6    4    3
## [2,]    6  -12    9
## [3,]    0    8  -12
## [4,]    1    1    1
b = c(0,0,0,1); b
## [1] 0 0 0 1
gaussianElimination(A,b) -> Ab1 ; Ab1
##      [,1] [,2] [,3]  [,4]
## [1,]    1    0    0 0.375
## [2,]    0    1    0 0.375
## [3,]    0    0    1 0.250
## [4,]    0    0    0 0.000

Rešenje: \({\boldsymbol x} = [0.375,0.375,0.250]^T = [3/8, 3/8, 2/8]^T\).

2.8

#########
#  2.8  #
#########

# a
A = matrix(c(2,3,4,3,4,5,4,5,6), ncol = 3, byrow = T); A
##      [,1] [,2] [,3]
## [1,]    2    3    4
## [2,]    3    4    5
## [3,]    4    5    6
A = cbind(A,diag(3)); A
##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    2    3    4    1    0    0
## [2,]    3    4    5    0    1    0
## [3,]    4    5    6    0    0    1
gaussianElimination(A) -> A1; A1
##      [,1] [,2] [,3] [,4] [,5] [,6]
## [1,]    1    0   -1    0   -5    4
## [2,]    0    1    2    0    4   -3
## [3,]    0    0    0    1   -2    1
# Nema inverznu, rang matrice je 2

# b
A = matrix(c(1,0,1,0,0,1,1,0,1,1,0,1,1,1,1,0), ncol = 4, byrow = T); A
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    1    0
## [2,]    0    1    1    0
## [3,]    1    1    0    1
## [4,]    1    1    1    0
A0 = cbind(A,diag(4)); A0
##      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
## [1,]    1    0    1    0    1    0    0    0
## [2,]    0    1    1    0    0    1    0    0
## [3,]    1    1    0    1    0    0    1    0
## [4,]    1    1    1    0    0    0    0    1
gaussianElimination(A0) -> A1; A1
##      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
## [1,]    1    0    0    0    0   -1    0    1
## [2,]    0    1    0    0   -1    0    0    1
## [3,]    0    0    1    0    1    1    0   -1
## [4,]    0    0    0    1    1    1    1   -2
A1 = A1[,5:8]; A1
##      [,1] [,2] [,3] [,4]
## [1,]    0   -1    0    1
## [2,]   -1    0    0    1
## [3,]    1    1    0   -1
## [4,]    1    1    1   -2
A %*% A1
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0    0
## [2,]    0    1    0    0
## [3,]    0    0    1    0
## [4,]    0    0    0    1
library(xtable)
print(xtable(A1, digits = 0), floating=FALSE, tabular.environment="bmatrix", hline.after=NULL, include.rownames=FALSE, include.colnames=FALSE)
## % latex table generated in R 4.5.0 by xtable 1.8-4 package
## % Thu Dec  4 21:29:39 2025
## \begin{bmatrix}{rrrr}
##   0 & -1 & 0 & 1 \\ 
##   -1 & 0 & 0 & 1 \\ 
##   1 & 1 & 0 & -1 \\ 
##   1 & 1 & 1 & -2 \\ 
##   \end{bmatrix}

Rešenje: \(A^{-1} = \begin{bmatrix} 0 & -1 & 0 & 1\\ -1 & 0 & 0 & 1\\ 1 & 1 & 0 & -1\\ 1 & 1 & 1 & -2 \end{bmatrix}\).

2.9

  1. Da. span \(( [1,1,1]^T, [0,1,-1]^T ) = \lambda [1,1,1]^T + \mu^3 [0,1,-1]^T, \lambda, \mu \in \mathbb{R}\)
  2. Ne. Nije grupa.
  3. Da ako i samo ako \(\gamma = 0\).
  4. Ne. (Homogenost!)

2.10

#########
#  2.10 #
#########

# a
A = matrix(c(2,-1,3,1,1,-2,3,-3,8), ncol = 3); A
##      [,1] [,2] [,3]
## [1,]    2    1    3
## [2,]   -1    1   -3
## [3,]    3   -2    8
gaussianElimination(A)
##      [,1] [,2] [,3]
## [1,]    1    0    2
## [2,]    0    1   -1
## [3,]    0    0    0
# b
A = matrix(c(1,2,1,0,0,1,1,0,1,1,1,0,0,1,1), ncol = 3); A
##      [,1] [,2] [,3]
## [1,]    1    1    1
## [2,]    2    1    0
## [3,]    1    0    0
## [4,]    0    1    1
## [5,]    0    1    1
gaussianElimination(A)
##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    0    1    0
## [3,]    0    0    1
## [4,]    0    0    0
## [5,]    0    0    0
  1. Linearno zavisni: \({\boldsymbol x}_3 = 2 {\boldsymbol x}_1 - {\boldsymbol x}_2\)
  2. Linearno nezavisni, rang \(([ {\boldsymbol x}_1, {\boldsymbol x}_2, {\boldsymbol x}_3]) = 3\).

2.11

#########
#  2.11 #
#########

A = matrix(c(1,1,1,1,2,3,2,-1,1), ncol = 3); A
##      [,1] [,2] [,3]
## [1,]    1    1    2
## [2,]    1    2   -1
## [3,]    1    3    1
y = c(1,-2,5)
gaussianElimination(A,y)->A1; A1
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0   -6
## [2,]    0    1    0    3
## [3,]    0    0    1    2

\({\boldsymbol y} = -6 {\boldsymbol x}_1 + 3 {\boldsymbol x}_2 + 2 {\boldsymbol x}_3\).

2.12

#########
#  2.12 #
#########

A = matrix(c(1,1,-3,1,2,-1,0,-1,-1,1,-1,1), ncol = 3); A
##      [,1] [,2] [,3]
## [1,]    1    2   -1
## [2,]    1   -1    1
## [3,]   -3    0   -1
## [4,]    1   -1    1
gaussianElimination(A)
##      [,1] [,2]       [,3]
## [1,]    1    0  0.3333333
## [2,]    0    1 -0.6666667
## [3,]    0    0  0.0000000
## [4,]    0    0  0.0000000
B = matrix(c(-1,-2,2,1,2,-2,0,0,-3,6,-2,-1), ncol = 3); B
##      [,1] [,2] [,3]
## [1,]   -1    2   -3
## [2,]   -2   -2    6
## [3,]    2    0   -2
## [4,]    1    0   -1
gaussianElimination(B)
##      [,1] [,2] [,3]
## [1,]    1    0   -1
## [2,]    0    1   -2
## [3,]    0    0    0
## [4,]    0    0    0
C = cbind( A[,1:2], -B[,1:2]); C
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    1   -2
## [2,]    1   -1    2    2
## [3,]   -3    0   -2    0
## [4,]    1   -1   -1    0
gaussianElimination(C) -> C1; C1
##      [,1] [,2] [,3]       [,4]
## [1,]    1    0    0 -0.4444444
## [2,]    0    1    0 -1.1111111
## [3,]    0    0    1  0.6666667
## [4,]    0    0    0  0.0000000
C2 = rbind(C1[1:3,],-(1:4==4))

D=matrix(round(-9*C2[,4]),ncol = 1); D
##      [,1]
## [1,]    4
## [2,]   10
## [3,]   -6
## [4,]    9
C %*% D
##      [,1]
## [1,]    0
## [2,]    0
## [3,]    0
## [4,]    0
(C[,1:2] %*% D[1:2,])
##      [,1]
## [1,]   24
## [2,]   -6
## [3,]  -12
## [4,]   -6
(C[,1:2] %*% D[1:2,])/6
##      [,1]
## [1,]    4
## [2,]   -1
## [3,]   -2
## [4,]   -1
(C[,3:4] %*% -D[3:4,])
##      [,1]
## [1,]   24
## [2,]   -6
## [3,]  -12
## [4,]   -6
(C[,3:4] %*% -D[3:4,])/6
##      [,1]
## [1,]    4
## [2,]   -1
## [3,]   -2
## [4,]   -1

\(U_1 \cap U_2 = \mbox{span} ( [4, -1, -2, -1]^T )\)

2.13

#########
#  2.13 #
#########

A = matrix(c(1,1,2,1,0,-2,1,0,1,-1,3,1),ncol=3); A
##      [,1] [,2] [,3]
## [1,]    1    0    1
## [2,]    1   -2   -1
## [3,]    2    1    3
## [4,]    1    0    1
gaussianElimination(A)
##      [,1] [,2] [,3]
## [1,]    1    0    1
## [2,]    0    1    1
## [3,]    0    0    0
## [4,]    0    0    0
A=matrix(c(3,1,7,3,-3,2,-5,-1,0,3,2,2),ncol=3); A
##      [,1] [,2] [,3]
## [1,]    3   -3    0
## [2,]    1    2    3
## [3,]    7   -5    2
## [4,]    3   -1    2
gaussianElimination(A)
##      [,1] [,2] [,3]
## [1,]    1    0    1
## [2,]    0    1    1
## [3,]    0    0    0
## [4,]    0    0    0

\(U_1 \cap U_2 = \mbox{span}\: ( [ 1, 1, -1 ]^T)\).

2.14

#########
#  2.14 #
#########

A = matrix(c(1,1,2,1,0,-2,1,0,1,-1,3,1), ncol = 3); A
##      [,1] [,2] [,3]
## [1,]    1    0    1
## [2,]    1   -2   -1
## [3,]    2    1    3
## [4,]    1    0    1
gaussianElimination(A)
##      [,1] [,2] [,3]
## [1,]    1    0    1
## [2,]    0    1    1
## [3,]    0    0    0
## [4,]    0    0    0
B = matrix(c(3,1,7,3,-3,2,-5,-1,0,3,2,2),ncol=3,byrow = F); B
##      [,1] [,2] [,3]
## [1,]    3   -3    0
## [2,]    1    2    3
## [3,]    7   -5    2
## [4,]    3   -1    2
gaussianElimination(B)
##      [,1] [,2] [,3]
## [1,]    1    0    1
## [2,]    0    1    1
## [3,]    0    0    0
## [4,]    0    0    0
C = cbind( A[,1:2],-B[,1:2]); C
##      [,1] [,2] [,3] [,4]
## [1,]    1    0   -3    3
## [2,]    1   -2   -1   -2
## [3,]    2    1   -7    5
## [4,]    1    0   -3    1
gaussianElimination(C)
##      [,1] [,2] [,3] [,4]
## [1,]    1    0   -3    0
## [2,]    0    1   -1    0
## [3,]    0    0    0    1
## [4,]    0    0    0    0
D=matrix(c(3,1,1,0), ncol = 1); D
##      [,1]
## [1,]    3
## [2,]    1
## [3,]    1
## [4,]    0
C %*% D
##      [,1]
## [1,]    0
## [2,]    0
## [3,]    0
## [4,]    0
C[,1:2] %*% D[1:2,]
##      [,1]
## [1,]    3
## [2,]    1
## [3,]    7
## [4,]    3

\(U_1 \cap U_2 = \mbox{span}\: ( [ 3, 1, 7, 3 ]^T)\).

2.15

  1. \(F\) je podprostor jer jer skup rešenja homogenog sistema. \(G\) je podprostor jer je \(G = \mbox{span}\: ( [ 1, 1, 1 ]^T, [-1, 1, -3]^T )\).
  2. Neka \([ x, y, z ]^T \in F \cap G\). Onda \(x = a - b, y = a + b, z = a - 3 b\).
    \(x + y - z = 0 \;\Leftrightarrow\; a - b + a + b - ( a - 3 b ) = 0 \;\Leftrightarrow\; a = - 3 b \;\Leftrightarrow\; x = - 4 b, y = - 2 b, z = - 6 b\) Dakle, \(F \cap G = \mbox{span} ( [2, 1, 3]^T )\)
  3. Nađimo prvo bazu \(F\) koristeći \(-1\)-trik:
#########
#  2.15 #
#########

# c. 
A=rbind(c(1,1,-1),-(1:3==2),-(1:3==3)); A
##      [,1] [,2] [,3]
## [1,]    1    1   -1
## [2,]    0   -1    0
## [3,]    0    0   -1
B=cbind(c(1,1,1),c(-1,1,-3)); B
##      [,1] [,2]
## [1,]    1   -1
## [2,]    1    1
## [3,]    1   -3
C=cbind(A[,2:3],-B); C
##      [,1] [,2] [,3] [,4]
## [1,]    1   -1   -1    1
## [2,]   -1    0   -1   -1
## [3,]    0   -1   -1    3
gaussianElimination(C)->C1; C1
##      [,1] [,2] [,3] [,4]
## [1,]    1    0    0   -2
## [2,]    0    1    0   -6
## [3,]    0    0    1    3
C2 = rbind(C1[1:3,],-(1:4==4))
D = C2[,4]
C %*% D
##      [,1]
## [1,]    0
## [2,]    0
## [3,]    0
A[,2:3] %*% D[1:2]
##      [,1]
## [1,]    4
## [2,]    2
## [3,]    6
A[,2:3] %*% D[1:2]/2
##      [,1]
## [1,]    2
## [2,]    1
## [3,]    3

\(F \cap G = \mbox{span}\: ( [ 2, 1, 3 ]^T)\).

2.16

  1. Da. \(\forall f, g \in L^1( [ a, b ] ), \ \forall \lambda, \psi \in \mathbb{R}\int_a^b ( \lambda f ( x ) + \psi g ( x ) ) dx = \lambda \ \int_a^b f ( x ) dx + \psi \int_a^b g ( x ) dx\). Vektorski prostor \(L^1( [ a, b ] )\) je beskonačno-dimenzionalni vektorski prostor nad \(\mathbb{R}\).

  2. Da. \(\forall f, g \in C^1, \forall \lambda, \psi \in \mathbb{R}, \forall x \in \mathbb{R}\ ( \lambda f ( x ) + \psi g ( x ) )' = \lambda f' ( x ) + g' ( x )\). \(C^1\) i \(C^0\) su beskonačno-dimenzionalni vektorski prostori nad \(\mathbb{R}\).

  3. Ne. \(\Phi ( 0 ) = 1\).

  4. Da. Na osnovu distributivnosti množenja matrica i osobina množenja matrice skalarom.

  5. Da. Na osnovu distributivnosti množenja matrica i osobina množenja matrice skalarom. Ova transformacija predstavlja rotaciju za ugao \(\theta\) u \(\mathbb{R}^2\).

2.17

#########
#  2.17 #
#########

A=matrix(c(3,2,1,1,1,1,1,-3,0,2,3,1), ncol = 3, byrow = T); A
##      [,1] [,2] [,3]
## [1,]    3    2    1
## [2,]    1    1    1
## [3,]    1   -3    0
## [4,]    2    3    1
gaussianElimination(A)->A1; A1
##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    0    1    0
## [3,]    0    0    1
## [4,]    0    0    0

2.19

#########
#  2.19 #
#########

A=matrix(c(1,1,1,1,-1,1,0,0,1), ncol = 3); A
##      [,1] [,2] [,3]
## [1,]    1    1    0
## [2,]    1   -1    0
## [3,]    1    1    1
B=matrix(c(1,1,1,1,2,1,1,0,0), ncol = 3); B
##      [,1] [,2] [,3]
## [1,]    1    1    1
## [2,]    1    2    0
## [3,]    1    1    0
inv(B) %*% A %*% B
##      [,1] [,2] [,3]
## [1,]    6    9    1
## [2,]   -3   -5    0
## [3,]   -1   -1    0

\(\tilde{A}_{\Phi} = \begin{bmatrix} 6 & 9 & 1 \\ -3 & -5 & 0 \\ -1 & -1 & 0 \\ \end{bmatrix}\)

2.20

Neka su \(V\) i \(W\) konačnodimenzionalni vektorski prostori. Neka su \(B = ( b_1, \ldots, b_n )\) uređena baza od \(V\) i \(C = ( c_1, \ldots, c_m )\) uređena baza od \(W\). Neka je \(A_{\Phi}\) matrica linearne transformacije \(\Phi : V \rightarrow W\) u odnosu na baze \(B\) i \(C\).

Neka su \(\tilde{B} = ( \tilde{b}_1, \ldots, \tilde{b}_n )\) uređena baza od \(V\) i \(\tilde{C} = ( \tilde{c}_1, \ldots, \tilde{c}_m )\) uređena baza od \(W\). Neka su matrice \(S = [ \tilde{b}_1, \ldots, \tilde{b}_n ]\) i \(T = [ \tilde{c}_1, \ldots, \tilde{c}_m ]\).

Onda se matrica linearne transformacije \(\Phi\) u odnosu na baze \(\tilde{B}\) i \(\tilde{C}\) dobija formulom \(\tilde{A}_{\Phi} = T^{-1} A_{\Phi} S\).

#########
#  2.20 #
#########

# a
B1 = matrix(c(2,1,-1,-1), ncol = 2); B1
##      [,1] [,2]
## [1,]    2   -1
## [2,]    1   -1
gaussianElimination(B1)
##      [,1] [,2]
## [1,]    1    0
## [2,]    0    1
B2 = matrix(c(2,-2,1,1), ncol = 2); B2
##      [,1] [,2]
## [1,]    2    1
## [2,]   -2    1
gaussianElimination(B2)
##      [,1] [,2]
## [1,]    1    0
## [2,]    0    1

Vidimo da je \(\mbox{rk} ( [ {\boldsymbol b}_1, {\boldsymbol b}_2 ] ) = \mbox{rk} ( [ {\boldsymbol b}'_1, {\boldsymbol b}'_2 ]) = 2\), znači da su i \(B\) i \(B'\) baze od \(\mathbb{R}^2\).

Neka je \(\mbox{id}_{\mathbb{R}^2}\) identična linearna transformacija. \(\mbox{id}_{\mathbb{R}^2} : {\boldsymbol x} \mapsto {\boldsymbol x}\).

Onda su matrice \([ {\boldsymbol b}_1, {\boldsymbol b}_2 ]\) i \([ {\boldsymbol b}'_1, {\boldsymbol b}'_2 ]\) matrice transformacije \(\mbox{id}_{\mathbb{R}^2}\) u odnosu na standardnu bazu i redom bazu \(B\) i \(B'\).

Vektor \({\boldsymbol x}\) izražen u bazi \(B\) kao, recimo, \({\boldsymbol x} = 2 {\boldsymbol b}_1 + 3 {\boldsymbol b}_2 = \begin{bmatrix} 2 \\ 3 \\ \end{bmatrix}_B\) se u standardnoj bazi dobija kao \([ {\boldsymbol b}_1, {\boldsymbol b}_2 ] \begin{bmatrix} 2 \\ 3 \\ \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & -1 \\ \end{bmatrix} \begin{bmatrix} 2 \\ 3 \\ \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}\).

Obrnuto, ako je \({\boldsymbol x} = \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}\) vektor u standardnoj bazi i želimo da ga prikažemo u bazi \(B\), onda dobijamo \({\boldsymbol x} = \begin{bmatrix} 2 \\ 3 \\ \end{bmatrix}_B\). Račun: \(\begin{bmatrix} 2 & -1 \\ 1 & -1 \\ \end{bmatrix}^{-1} \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 1 & -2 \\ \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ \end{bmatrix}\).

#########
#  2.20 #
#########

# b
P1=inv(B1) %*% B2; P1
##      [,1] [,2]
## [1,]    4    0
## [2,]    6   -1

Kad želimo da nađemo matricu promene iz baze \(B\) u bazu \(B'\), onda tražimo prikazivanje identične transformacije u odnosu na baze \(B\) i \(B'\): \(P_1 = \begin{bmatrix} 2 & -1 \\ 1 & -1 \\ \end{bmatrix}^{-1} \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 2 & 1 \\ -2 & 1 \\ \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 1 & -2 \\ \end{bmatrix} \begin{bmatrix} 2 & 1 \\ -2 & 1 \\ \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 6 & -1 \\ \end{bmatrix}\)

#########
#  2.20 #
#########

# c
P2 = matrix(c(1,2,-1,0,-1,2,1,0,-1), ncol = 3); P2
##      [,1] [,2] [,3]
## [1,]    1    0    1
## [2,]    2   -1    0
## [3,]   -1    2   -1
det(P2)
## [1] 4

Neka je matrica \(P_2 = [ {\boldsymbol c}_1, {\boldsymbol c}_2 , {\boldsymbol c}_3 ]\).

Pošto je \(\mbox{det} P_2 \neq 0\), \(C\) jeste baza, jer je punog ranga. Upravo matrica \(P_2\) je matrica promene iz baze \(C\) u standardnu bazu \(C'\).

#########
#  2.20 #
#########

# d
AB = matrix(c(1,1,1,-1,0,2,1,-1,1,3), ncol = 5); AB
##      [,1] [,2] [,3] [,4] [,5]
## [1,]    1    1    0    1    1
## [2,]    1   -1    2   -1    3
ABG = gaussianElimination(AB); ABG
##      [,1] [,2] [,3] [,4] [,5]
## [1,]    1    0    1    0    2
## [2,]    0    1   -1    1   -1
A = t(ABG[,3:5]); A
##      [,1] [,2]
## [1,]    1   -1
## [2,]    0    1
## [3,]    2   -1

\(\Phi ( {\boldsymbol b}_1 + {\boldsymbol b}_2 ) = \Phi ( {\boldsymbol b}_1 ) + \Phi ( {\boldsymbol b}_1 ) = \phantom{2 c_1 - } \; c_2 + c_3\)
\(\Phi ( {\boldsymbol b}_1 - {\boldsymbol b}_2 ) = \Phi ( {\boldsymbol b}_1 ) - \Phi ( {\boldsymbol b}_1 ) = 2 c_1 - c_2 + 3 c_3\)

Primenom Gausovih transformacija:
\(\Phi ( {\boldsymbol b}_1 ) = \phantom{-} c_1 \phantom{ + 3 c_2 } + 2 c_3\)
\(\Phi ( {\boldsymbol b}_2 ) = - c_1 + c_2 - \phantom{2} c_3\)

Dobijamo da je \(A_{\Phi} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \\ 2 & -1 \\ \end{bmatrix}\) matrica transformacija \(\phi\) u odnosu na uređene baze \(B = ( {\boldsymbol b}_1, {\boldsymbol b}_2 )\) i \(C = ( {\boldsymbol c}_1, {\boldsymbol c}_2, {\boldsymbol c}_3 )\).

#########
#  2.20 #
#########

# e
A1 = P2 %*% A%*% P1; A1
##      [,1] [,2]
## [1,]    0    2
## [2,]  -10    3
## [3,]   12   -4

Koristimo formulu \(\tilde{A}_{\Phi} = T^{-1} A_{\Phi} S\), gde je \(T^{-1} = P_2\) (dobijeno pod c) i \(S= P_1\) (dobijeno pod b).

#########
#  2.20 #
#########

# f
x = matrix(c(2,3), ncol=1); x
##      [,1]
## [1,]    2
## [2,]    3
# (i)
P1 %*% x
##      [,1]
## [1,]    8
## [2,]    9
# (ii)
A %*% P1 %*% x
##      [,1]
## [1,]   -1
## [2,]    9
## [3,]    7
# (iii)
P2 %*% A %*% P1 %*% x
##      [,1]
## [1,]    6
## [2,]  -11
## [3,]   12
# (iv)
A1 %*% x
##      [,1]
## [1,]    6
## [2,]  -11
## [3,]   12