import sglib
reload(sglib)
from sglib import *
# The following allows us to use 'k' as a "variable"
k=sy.Symbol('k')

a=col(1,2); b=col(3,4); c=col(-2,5)

#(1)
3*a

#(2)
6*b

#(3)
k*c

#(4)
a+b

#(5)
b.transpose()+c.transpose()

#(6)
a-c

#(7)
2*a+3*b

#(8)
3*b-2*c+a

#(9)
-2*b.transpose()+2*c.transpose()

#(10)
inner_product(a,b)

#(11)
inner_product(b,c)

#(12)
inner_product(c,a)

#(13)
Matrix([(1,2), (3,4)]) * col(1,-1) 

#(14)
Matrix([(0,1), (1,0)]) * col(2,3)

#(15)
# NOTE: Matrix is Identity Matrix, so we don't really need to work this!
a=sy.Symbol('a'); b=sy.Symbol('b') # Need a,b as variables
Matrix([(1,0), (0,1)]) * col(a,b)

z1=2+3*i; z2=1-i; z3=5*i

#(16)
z1+z2

#(17)
z1-z3

#(18)
z1+z2-z3

#(19)
(z1*z2).simplify()

#(20)
(z2*z3).simplify()

#(21)
z1.conjugate()

#(22)
z2.conjugate()

#(23)
abs(z1)

#(24)
abs(z2)

#(25)
abs(z3)

#(26)
Matrix([(1,2), (3,4)]).transpose()

#(27)
Matrix([(1,0,1), (0,2,3), (3,4,5)]).transpose()

#(28)
a=sy.Symbol('a'); b=sy.Symbol('b'); c=sy.Symbol('c'); d=sy.Symbol('d')
w=sy.Symbol('w'); x=sy.Symbol('x'); y=sy.Symbol('y'); z=sy.Symbol('z')
Matrix([(w,x),(y,z)]).transpose()

#(29)
Matrix([(1,2),(3,4)])*Matrix([(4,3),(2,1)])

#(30)
# 2nd one is Identity Matrix!
Matrix([(1,2),(3,4)])*Matrix([(1,0),(0,1)])

#(31)
Matrix([(a,b),(c,d)])*Matrix([(w,x),(y,z)])

#(32)
row(1,2)*col(4,3)

#(33)
col(1,2)*row(4,3)

#(34)
# Not the inverse
Matrix([(0,1),(1,0)])*Matrix([(1,0),(0,1)])

# Their product IS the Identity, so YES
Matrix([(0,1),(1,0)])*Matrix([(0,1),(1,0)])

# NO
Matrix([(0,1),(1,0)])*Matrix([(0,0),(0,0)])

#(35)
# We'll just ask for the inverse directly ...
# It turns out to be (c)
M = Matrix([(1,2),(3,4)])
Mi = M.inverse_ADJ()
Mi

# But, still we should CHECK it ...
M*Mi

#(36)
# (a) - NO
Matrix([(1,0),(0,0)]) * Matrix([(1,0),(0,0)])

# (b) - NO
Matrix([(1,0),(0,0)]) * Matrix([(0,0),(0,1)])

# (c) - NO
Matrix([(1,0),(0,0)]) * Matrix([(-1,0),(0,0)])

# So, its (d) - none of the above.
# Actually, this matrix has no inverse ...
Matrix([(1,0),(0,0)]).inverse_ADJ()

#
# Change of Basis Problems
#
b11 = 1/sqrt(2)*col(1,1); b12 = 1/sqrt(2)*col(1,-1)
b21 = 1/sqrt(5)*col(2,1); b22 = 1/sqrt(5)*col(-1,2)

#(37)
v = col(2,3)
v_b1 = col(inner_product(b11,v), inner_product(b12,v))
Print('$%s$' %myltx(v_b1))

#(38)
v = col(3,2)
v_b2 = col(inner_product(b21,v), inner_product(b22,v))
Print('$%s$' %myltx(v_b2))

#(39)
v = col(1,0)
v_b2 = col(inner_product(b21,v), inner_product(b22,v))
Print('$%s$' %myltx(v_b2))

#(40)
v = col(5,5)
v_b1 = col(inner_product(b11,v), inner_product(b12,v))
Print('$%s$' %myltx(v_b1))

M1 = Matrix([(1,2),(3,4)])
M2 = Matrix([(1,0),(0,0)])

M1*M2

M2*M1

v1 = col(a,b); v2=col(c,d)
v1,v2

#(52)
ip = inner_product
# Yes
v1 = col(1,0); v2 = col(0,1)
v1.norm(), v2.norm(), ip(v1,v2)

#(53) No - Orthogonal but not normalized
v1 = col(2,0); v2 = col(0,2)
v1.norm(), v2.norm(), ip(v1,v2)

#(54) Yes (the "blue" basis)
v1 = col(1,1)/sqrt(2); v2 = col(1,-1)/sqrt(2)
v1.norm(), v2.norm(), ip(v1,v2)

#(55) Yes
v1 = col(1,0); v2 = col(0,-1)
v1.norm(), v2.norm(), ip(v1,v2)

#(56) Yes
v1 = col(2,1)/sqrt(5); v2 = col(-1,2)/sqrt(5)
v1.norm(), v2.norm(), ip(v1,v2)

#(57) No - Normalized but not orthogonal
v1 = col(1,1)/sqrt(2); v2 = col(-1,2)/sqrt(5)
v1.norm(), v2.norm(), ip(v1,v2)