Matrix Algebra and Linear Models: Problem Sheet 4
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Matrix Algebra and Linear Models: Problem Sheet 4
Present all your answers in complete sentences . For R exercises, include screenshots and/or code . Questions
or sub- questions with an asterisk ∗ comprise your homework.
∗ 1. Do parts (i)-(iv) and (vii) by hand; you may use a calculator for parts (v) and (vi).
(i) Find x, y ∈ R such that
(a) ei = x + iy (b) 2ei = x + iy
(ii) Find r ∈ R+ , ϕ ∈ [0, 2π) such that
(a) 1 + i = reiϕ (b) − 5i = reiϕ
(iii) Let z1 = 1 + i, z2 = 2 − i. Compute Re(z1 ), Im(z2 ), z1 + 2 , z1 /z2 , and |z1 − z2 |.
(v) (You may use a calculator in this part.) Find z5 such that cos(z5 ) = ^2.
(vi) (You may use a calculator in this part.) Compute the principal value of !e(1)"1+i . (vii) Compute the determinant and the inverse of the matrix #11(+) i 1 i$
2. Use De Moivre’s formula
cos(nϕ) + isin(nϕ) = (cosϕ + isinϕ)n
to derive the following relations
cos(3ϕ) = 4 cos3 ϕ − 3cosϕ sin(3ϕ) = −4sin3 ϕ + 3sinϕ .
3. Use Euler’s identity eiϕ = cosϕ+isinϕ to show the following representations for trigonometric functions:
eiϕ − e −iϕ eiϕ + e−iϕ
4. The following extension of the rational numbers is analogous to the construction of the complex numbers from the real numbers.
Consider numbers of the form z = x +^2y where x and y are rational numbers. We call the set of all these numbers Q(^2), i.e., Q(^2) = {x +^2y ; x, y ∈ Q}. Show that if z1 , z2 ∈ Q(^2) then
(i) z1 + z2 ∈ Q(^2)
(ii) z1 z2 ∈ Q(^2)
(iii) If z1 0 then 1/z1 ∈ Q(^2) (hint: use the fact that ^2 is irrational.)
(iv) If z1 0 then z2 /z1 ∈ Q(^2)
5. In this problem we use complex numbers as a tool to prove a geometric statement. Let v = (x1 , y1 ) and w = (x2 , y2 ) be two vectors in R2 and let A(v, w) be the oriented area of the parallelogram spanned by v and w, i.e.,
|A(v, w)| is the area and A(v, w) > 0 if w is to the left of v and A(v, w) < 0 if w is to the right of v.
w
v
(i) Compute A(e1 , e2 ) and A(e2 , e1 ) where e1 = (1, 0) and e2 = (0, 1).
(ii) Show that A(v, w) = &v&&w&sin(θ), where θ ∈ [ −π , π] is the angle between v and w.
Hint: You can use without proof the fact that
the area of a parallelogram is the length of the
base times the height, A = ah.
a
(iii) Show that
A(v1 , v2 ) = x1 y2 − x2 y1 .
Hint: Consider ℑ(z1 z2 ) for z1 = x1 + iy1 = r1 eiϕ1 and z2 = x2 + iy2 = r2 eiϕ2 .
6. Let n be a positive integer, a complex number z is called an n’th root of unity if zn = 1 .
(i) Show that if z is an n’th root of unity, then |z| = 1.
(ii) Find all roots of unity for n = 2 and n = 3 and plot their location in the complex plane.
(iiii) For an arbitrary n ∈ N, show that there are exactly n different roots of unity and describe their
location on the unit circle.
∗ 7. Inverse Trigonometric and Hyperbolic Functions: By definition, the inverse sine w = arcsin z is the
relation such that sin w = z . The inverse cosine w = arccos z is the relation such that cos w = z . The inverse tangent, inverse cotangent, inverse hyperbolic sine, etc., are defined and denoted in a similar fashion. (Note that all these relations are multivalued.) Using cos w = (eiw + e−iw)/2 show that
arccosz = w = −iln(z +^z2 − 1)
∗ 8. Let
T1 (x, y) = (x + y , 2x − y , x + 2y) , T2 (x, y , z) = (2z , x + y + z),
T5 (x, y) = (x + y , x − y) , T6 (x + y , y + z , x + y + z) = (x + z , y + z , x + 2y + z)
(a) For each i ∈ {1, . . . , 6}, find positive integers m and n such that Ti : Rm → Rn .
(b) For each i ∈ {1, . . . , 6}, state whether Ti is linear or not. If it is, find the corresponding matrix MTi that represent the linear transformation Ti . (Hint: T6 IS linear. How would you find MT6 ?)
(c) For the Ti which are linear, state whether they’re surjective, injective, both surjective and injective (bijective), or neither.
(d) Which of T1 ◦ T2 , T2 ◦ T1 , T1 ◦ T1 and T2 ◦ T2 is/are well-defined? For whichever is/are well-defined write the matrix associated with the corresponding linear transformation.
(e) If it exists, what is the inverse of the linear transformation T5 , i.e, what is T1 (x, y)?
∗ 9. Let S : Rn → Rm and T : Rp → Rn be two linear transformations. Take as given the rank-nullity
theorem, and the following fact discussed in class and proven in the notes:
nullity(S ◦ T) = nullity(T) + dim(ker(S) ∩ Im(T))
Using these prove the following:
(a)
rank(S ◦ T) = rank(T) − dim(ker(S) ∩ Im(T))
(Hint: Rank-nullity theorem.)
(b)
rank(S ◦ T) ≤ rank(T), rank(S ◦ T) ≤ rank(S)
(Hint: The previous part, and also the fact that for any matrix M , rank(M) = rank(MT ).)
(c)
nullity(S ◦ T) ≥ nullity(T), nullity(S ◦ T) ≥ nullity(S) + p − n
(d) S ◦ T = 0 if and only if image(T) ⊆ kernel(S).
(e) If S is invertible, then rank(S ◦ T) = rank(T) and nullity(S ◦ T) = nullity(T).
2022-11-30