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Given 6 vectors in R4 and given that they span R4 find a basis.
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England schools
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England university
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Scotland schools
Taxonomy: mathcentre
Taxonomy: Kind of activity
Taxonomy: Context
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History
Mohamed Mehbali 7 years, 6 months ago
Created this as a copy of Find a basis given a spanning set of vectors.Name | Status | Author | Last Modified | |
---|---|---|---|---|
Find a basis given a spanning set of vectors | Doesn't work | Newcastle University Mathematics and Statistics | 20/11/2019 14:50 | |
Find a basis given a spanning set of vectors | draft | Mohamed Mehbali | 31/08/2017 12:59 | |
Michael's copy of Find a basis given a spanning set of vectors | draft | Michael Hoffmann | 18/07/2018 23:07 | |
John's copy of Find a basis given a spanning set of vectors | draft | John Steele | 13/05/2019 03:56 |
There are 9 other versions that do you not have access to.
Name | Type | Generated Value |
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mm5 | list |
[ 1, 0 ]
|
||||
mm4 | list |
[ 1, 0 ]
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||||
nt3 | string |
|
||||
mm6 | list |
[ 0, 1 ]
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||||
nt5 | string |
|
||||
al | integer |
-1
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||||
ga1 | integer |
0
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||||
mm2 | list |
[ 0, 0.5 ]
|
||||
ga | integer |
1
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||||
ep | integer |
0
|
||||
ont3 | string |
not
|
||||
ont2 | string |
|
||||
ont6 | string |
|
||||
ont5 | string |
not
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||||
ont4 | string |
not
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||||
message5 | string |
There are now 4 linearly indep
|
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message4 | string |
|
||||
be1 | integer |
-1
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||||
tmm | list |
Nested 2×6 list
|
||||
be | integer |
0
|
||||
nt2 | string |
not
|
||||
nt4 | string |
|
||||
v1 | list |
[ 1, -3, 3, 0 ]
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v2 | list |
[ -1, 3, -3, 0 ]
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||||
v3 | list |
[ 3, -1, -1, 4 ]
|
||||
v4 | list |
[ -1, 1, 1, 0 ]
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||||
mm3 | list |
[ 1, 0 ]
|
||||
v6 | list |
[ 0, -1, 0, 1 ]
|
||||
nt6 | string |
not
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||||
b | integer |
2
|
||||
c5 | integer |
-1
|
||||
al1 | integer |
-1
|
||||
a | integer |
2
|
||||
c | integer |
-1
|
||||
p1 | list |
[ -1, 3, -3, 0 ]
|
||||
mm | list |
Nested 6×2 list
|
||||
q | list |
[ -1, 3, -3, 0 ]
|
||||
p2 | list |
[ 0, -1, 0, 1 ]
|
||||
r | list |
[ -2, 4, -2, 0 ]
|
||||
u | integer |
3
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||||
t | list |
[ 0, 0, 0, 0 ]
|
||||
v | list |
[ -3, 3, 0, -1 ]
|
||||
y | list |
[ 3, -1, -1, 4 ]
|
||||
x | list |
[ 1, -3, 3, 0 ]
|
||||
z | list |
[ -1, 1, 1, 0 ]
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||||
v5 | list |
[ -3, 3, 0, -1 ]
|
Generated value: list
- u
- mm
- nt5
- ont5
Gap-fill
Ask the student a question, and give any hints about how they should answer this part.
Your task is to find a basis for R4 by finding a linearly independent subset of these vectors.
Start from v1 and work through each vector in turn.
Determine if a vector is a linear combination of the previous vectors in the list.
If it is not such a linear combination then include it in the basis by choosing Yes, otherwise choose No.
Note that if a vector vi for i=2,…5 is a linear combination of the previous vectors in the list then it will satisfy a simple relation of the form vi=avj+bvk where a can be 0,1 or −1 similarly for b.
When you get to v6 it will be obvious if it is in the spanning set or not. (why?)
Use this tab to check that this question works as expected.
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