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Groepsopdrachten en antwoorden (incompleet)
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Antwoorden van groepsopdrachten van Matrix Algebra. Let op! Deze verzameling is incompleet; alleen week 1-5.
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1 Groepsopdrachten1.1 Week 1
Exercise 1 (Exercise 1). Los de volgende uitdrukking op naar de vector v in
termen van de vectoren a en b:
x + 2a − b = 3(x + a) − 2(2a − b)
.
x + 2a − b = 3(x + a) − 2(2a − b)
x = 3(x + a) − 2(2a − b) − 2a + b
x = 3x + 3a − 2(2a − b) − 2a + b
−2x = 3a − 2(2a − b) − 2a + b
−2x = −3a + 3b
3 3
x= a− b
2 2
Exercise 2 (Exercise 2). Bewijs dat
1 1
u·v= ||u + v||2 − ||u − v||2
4 4
voor alle vectoren u en v in Rn .
1 1 1 1
||u + v||2 − ||u − v||2 = ((u + v) · (u + v)) − ((u − v) · (u − v))
4 4 4 4
1 1
= ((u · u) + 2(u · v) + (v · v)) − ((u · u) − 2(u · v) + (v · v))
4 4
1 1 1 1 1 1
= (u · u) + (u · v) + (v · v) − (u · u) + (u · v) − (v · v)
4 2 4 4 2 4
1 1
= (u · v) + (u · v)
2 2
=u·v
Exercise 3 (Exercise 3). Beschouw de vectoren u en v in Rn , waarbij u 6= 0.
(a) Bewijs dat proju (v) loodrecht staat op v − proju (v).
(b) Gebruik het voorgaande en de stelling van Pythagoras om te bewijzen dat
||proju (v)|| ≤ ||v||.
(c) Bewijs dat de ongelijkheid ||proju (v)|| ≤ ||v|| equivalent is aan de
Cauchy-Schwarz Inequality.
1
, (a)
u · v
u · v
proju (v) · (v − proju (v)) = u· v− u
u · u u · u
u·v u·v u · v
= u·v− u· u
u·u u·u u·u
(u · v)2 u · v 2
= − u·u
u·u u·u
(u · v)2 (u · v)2
= − u·u
u·u (u · u)2
(u · v)2 (u · v)2
= − =0
u·u u·u
And thus are proju (v) and v − proju (v) orthogonal.
(b)
||v||2 = ||proju (v)||2 + ||v − proju (v)||2
||v||2 − ||v − proju (v)||2 = ||proju (v)||2
≥0
Thus there holds that ||v||2 ≥ ||proju (v)||2 and therefore, because lengths
are always nonnegative ||v|| ≥ ||proju (v)||
(c) Cauchy-Schwarz: |u · v| ≤ ||u|| ||v||
||proju (v)|| ≤ ||v|| ⇔
u · v
u ≤ ||v|| ⇔
u·u
u·v
||u|| ≤ ||v|| ⇔
u·u
|u · v|
||u|| ≤ ||v|| ⇒
||u||2
|u · v| ≤ ||u|| ||v||
Exercise 4 (Exercise 4). Los het volgende stelsel van vergelijkingen op:
x1 + x2 + x4 = 0
x2 + x3 + x4 = 0
x2 + x3 − x4 = 0
1 1 0 1 0 1 0 −1 0 0 1 0 −1 0 0
0 1 1 1 0 → 0 1 1 1 0 → 0 1 1 0 0
0 1 1 −1 0 0 0 0 −2 0 0 0 0 1 0
x1 =t
x2 = −t
x3 =t
x4 =0
2
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