Solutions Exam 2021
15 juni 2022
1.
(A) False. For any B a basis of V , we have that |B| = dim(V ) = N > m.
(B) True. Example: V = R3 and S = {~e1 , ~e2 }.
(C) True. Example: V = R3 and S = {e~1 , 2e~1 }.
(D) True. By definition in the case of a finite S, we have that
(m
)
X
span(S) =
ai si |ai ∈ R, si ∈ S, 1 ≤ i ≤ m .
i=1
To prove:
(i) span(S) ⊂ V, and
(ii) span(S) is closed under vector addition and scalar multiplication.
Try to prove this yourself!
(E) True. If S is linearly dependent, there exists a 1 ≤ k ≤ m such that
m
X
ai si = sk .
i=1,i6=k
Then S \ {sk } is either linearly independent, or there exists a 1 ≤ k 0 ≤ m
such that
m
X
bi si = sk0 .
0}
i=1,i∈{k,k
/
Then S \ {sk , sk0 } is either linearly independent or there exists ... etc. You
can repeat this procedure until you are left with a linearly independent set.
In the special case where N > 1 and S = {~0}, you can remove ~0. The
empty set is also linearly independent.
(F) False. If S is linearly dependent, adding vectors will never make it linearly
independent.
1
2.
The input of h are the coördinates with respect to E2 = {~e1 , ~e2 } and the output
is the coördinate with respect to E1 = {1}. So
x
x
2
h:R →R:
7→ 1 a
= (x + ay) · 1 = x + ay.
y
y
3.
⇐⇒
~v · w
~ = |~v | · |w|
~ cos θ
√
√
√
3 · 0 + 3 · x = 32 + 3 · 02 + x2 ·
√
√
3 · x = 2 3 · |x| · 21
⇐⇒
x = |x| with x 6= 0.
⇐⇒
1
2
So the angle between ~v and w
~ is 60◦ for x > 0.
4.
Define the basis B = h1, X − 1, X 2 , X 3 i and rewrite 2X 3 + bX 2 + cX + 1 as a
linear combination of basis elements:
2X 3 + bX 2 + cX + 1 = (1 + c) · 1 + c · (X − 1) + b · X 2 + 2 · X 3 .
Then
f (2X 3 + bX 2 + cX + 1)
=
(1 + c) · f (1) + c · f (X − 1) + b · f (X 2 ) + 2 · f (X 3 )
=
(1 + c) · 1 + c · (X 2 + 1) + b · X 10 + 2 · 0
=
1 + 2c + cX 2 + bX 10 .
5.
(a)
Ker(h) =
>
>
4
(x, y, z, w) ∈ R |h((x, y, z, w) ) =
−2x
y+w
0
=
,
0
implying that x = 0, y = −w and z = z, so that
Ker(h) = w(0, −1, 0, 1)> + z(0, 0, 1, 0)> with w, z ∈ R .
As we can write Ker(h) as the span of two linearly independent vectors,
the nullity of h is two.
(b) A possible basis is h(0, −1, 0, 1)> , (0, 0, 1, 0)> i, see (a).
(c)
Range(h) =
−2
0
0
x
+y
+w
with x, y, w ∈ R ,
0
1
1
which is the span of three vectors. However, only two of these are linearly
independent, so the rank of h is 2.
2
.
So indeed the dimension of the domain of h equals the sum of its nullity and its rank.
(d) The matrix H must be compatible with
H~v = ~u,
with ~v ∈ R4 and ~u ∈ R2 , so H must have 4 columns and 2 rows.
6.
See 1.D, only this time S can have an infinite amount of elements, so that the
span of S is
( k
)
X
span(S) =
ai si |k ∈ N, ai ∈ R, si ∈ S, 1 ≤ i ≤ k .
i=1
7.
Assume the green dot is on f (x0 , y0 ) and g(x1 , y1 ).
(a) For any infinitesimally small number > 0, we have that
f (x0 − , y0 ) < 0.8 and f (x0 + , y0 ) ≥ 0.8,
implying that f (x, y) increases in f (x0 , y0 ) in the x-direction ⇒
(b) Idem for f (x, y) in f (x0 , y0 ) in the y-direction ⇒
∂f
∂y
∂f
∂x
> 0.
∂g
∂x
< 0.
> 0.
(c) For any infinitesimally small number > 0, we have that
g(x1 − , y1 ) > 1.7 and g(x1 + , y1 ) ≤ 1.7,
implying that g(x, y) decreases in g(x1 , y1 ) in the x-direction ⇒
(d) Idem for g(x, y) in g(x1 , y1 ) in the y-direction ⇒
∂g
∂y
< 0.
9.
(a) On the one hand
x1 + x2
x1 + x2 + y1 + y2 + 1
h1
=
.
y1 + y2
x1 + x2 + y1 + y2 − 1
On the other hand
x1
x2
x1 + y1 + 1
x2 + y2 + 1
x1 + y1 + x2 + y2 + 2
h1
+h1
=
+
=
.
y1
y2
x1 + y1 − 1
x2 + y2 − 1
x1 + y1 + x2 + y2 − 2
This implies that h1 (~v1 + ~v2 ) 6= h1 (~v1 ) + h1 (~v2 ), so h is not a homomorphism.
3
(b) Remove the constants:
x
x+y
h2 : R → R :
7→
.
y
x+y
2
2
1
1
(d) h2 (~e1 ) =
and h2 (~e1 ) =
, so that
1
1
1 1
H2 =
.
1 1
(e) |H2 | = 1 − 1 = 0. This means that h2 squeezes every gridsquare of R2 into
a line.
(f) A possible change we could make is
x
x+y
2
2
h3 : R → R :
7→
,
y
x−y
1
1
with h3 (~e1 ) =
and h3 (~e2 ) =
, so that
1
−1
1 1
H3 =
.
1 −1
(g) Solution:
(H3 )−1 =
1
2
1
1
1
.
−1
10.
(a) Strategy: take the matrix with the basisvectors as columns and check
whether its determinant is non-zero.
(b) The change-of-base matrix from D to E3 is the easiest, because every
element in D is already written with respect to E3 ! So


0 3 12
ID,E3 = 0 1 1  .
1 0 0
For the change-of-base matrix in the other direction, we simply take the
inverse of ID,E3 , which gives


5
0
0
2
2
IE3 ,D = (ID,E3 )−1 =  1 − 12 0  .
5
−1 3
0
(c)

GD,D = IE3 ,D · G · ID,E3
3
= 0
0
(d) The vectors in D are the eigenvectors of G.
4
0
1
0

0
0 .
−4
11.
We can find the critical points by setting
  
  
∂g
2(x
+
y
+
2)
0
 =  ,
∇g(x, y) =  ∂x  = 
∂g
2(x − 4y − 3)
0
∂y
which is equivalent to solving the system
1 1
x
−2
=
1 −4
y
3
with the unique solution (x, y) = (−1, −1).
To determine the type of critical point, we have to look at the second derivative
 

 2
∂2g
∂ g
2
2
2
∂x∂y 
.
=
H2 (x, y) =  ∂x2
∂ g
∂2g
2
−8
2
∂y∂x
∂y
Because the determinant |H2 | < 0, g(x, y) has a saddlepoint in (−1, −1).
12.
Use the transformation
x
= r cos φ
y
= r sin φ
dxdy
= rdrdφ.
Assuming b is even, we can now rewrite the integral as
Z π3 Z 1
sin(r2 )rdrdφ.
0
0
We will first solve the inner-integral
Z 1
sin(r2 )rdr
0
via a change of variables
u
= r2
du
=
2rdr.
If r = 0, then u = 0 and if r = 1, then u = 1, so the boundaries don’t change:
Z
Z 1
1
1
1 1
sin(u)du = [− cos(u)]10 = (1 − cos(1)).
sin(r2 )rdr =
2
2
2
0
0
We substitute this into the original integral:
Z π3 Z 1
Z π3
π
1
1
π
2
sin(r )rdrdφ = (1−cos(1))
dφ = (1−cos(1))[φ]03 = (1−cos(1)).
2
2
6
0
0
0
5
13.
(a) We can rewrite this differential equation a bit:
⇐⇒
⇐⇒
dP
= −M + rP
dt
dP
= dt
−M
+ rP
Z
Z
dP
= dt = t + k with k ∈ R.
−M + rP
We can solve the integral with respect to P via a change of variables
u
= −M + rP
du = rdP.
Now we can rewrite the integral
Z
Z
1
1
1
du
dP
=
= ln(u) = ln(−M + rP ).
−M + rP
r
u
r
r
Subsituting this into the equation, gives us
1
r
ln(−M + rP ) = t + k
⇐⇒
ln(−M + rP ) = r(t + k)
⇐⇒
−M + rP = Kert with K ∈ R
⇐⇒
P (t) = 1r (Kert + M ).
(b) We determine the value K such that P (t = T ) = 0:
1
rT
r (Ke
⇐⇒
+ M) = 0
M
K = − rT .
e
6