Alg1231W4T2 - Linear dependence

Thursday, 16 August 2018

10:53 PM

Untitled picture.png Machine generated alternative text:
A set of vectors {VI, v2,. in a vector space V is called linearly dependent if we can find scalars , 0, ,rn, not all zero, such that 
rlV1 + 
where the right hand side is the zero vector. 
To demonstrate that a given set of vectors is linearly dependent, it is a good idea to use the definition and row-reduction to explicitly find scalars that 
demonstrate the linear relation. 
If our investigation leads us to conclude that the QLly way. to arrange a linear relation with right hand side being O is to use 
then the set cannot be linearly dependent (i.e. the set is a linearly independent set). 
Let's show that 
u 
1 
3 
6 
2 
3 
-6 
1 
-3 
—30 
is actually a linearly dependent set by finding scalars r, s and t (not all zero) such that 
For example 
Using this we can express one of the vectors as a linear combination of the others, for example 
Note: 
• There are three valid answers for this last part. 
• If your answer is v = 10u — 11 w then you should enter it using the syntax v=10*u-11*w (note: this isn't actually the answer). 
















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Untitled picture.png Machine generated alternative text:
45 
0 
-15 
- -16 
-10 
20 
15 
—5 
10 
15 
Using the app above and experimenting with the sliders, find integers r and s such that 
in each of the cases below: 
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Untitled picture.png Machine generated alternative text:
45 
0 
-15 
- -16 
-10 
20 
15 
—5 
10 
15 
Using the app above and experimenting with the sliders, find integers r and s such that 
in each of the cases below: 

Untitled picture.png Machine generated alternative text:
12 
ii) u 
—2 
and v 
3 
2 
and v 
-3 
1 
and w 
—3 
-5 
and w 
—1 
—4 
3 
11 
-8 
In this case we need 
O and s 
. In this case we need 
3 
O and s 
Now, find scalars r, s and t (not all zero) such that 
if u 
2 
and v 
—1 
4 
and w 
3 
32 
14 
In this case we could use 

Untitled picture.png Machine generated alternative text:
You have been asked to find out if the set B 
VI ¯ 
{VI, v2, v3} of vectors is linearly independent, where 
1 
—1 
2 
1 
1 
1 
—1 
—4 
and if not, find a linear relation between them. Recall that this really amounts to seeing if you can find solutions to the system 
1 
—1 
2 
1 
1 
—1 
—4 
which are not all zero. Of course y z 0 
is a solution, but it doesn't count 
Fortunately, you have mastered the delicate art of row reduction from MATH1131 , so you can readily deduce that a solution in fact does exist, namely 
Note: enter you answer using the syntax [1,2 , 3] . 
This means the set B {VI , v2, v3} is a linearly dependent set 
An example of a linear relation between these vectors is 
13 
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Untitled picture.png Machine generated alternative text:
You have been asked to find out if the set B 
VI ¯ 
{VI, v2, v3} of vectors is linearly independent, where 
1 
—1 
2 
1 
1 
1 
—1 
—4 
and if not, find a linear relation between them. Recall that this really amounts to seeing if you can find solutions to the system 
1 
—1 
2 
1 
1 
—1 
—4 
which are not all zero. Of course y z 0 
is a solution, but it doesn't count 
Fortunately, you have mastered the delicate art of row reduction from MATH1131 , so you can readily deduce that a solution in fact does exist, namely 
Note: enter you answer using the syntax [1,2 , 3] . 
This means the set B {VI , v2, v3} is a linearly dependent set 
An example of a linear relation between these vectors is 
13 

Untitled picture.png Machine generated alternative text:
The set C 
, where 
is a linearly dependent set. We can show this by using row reduction to find a non-zero solution to the system 
For example 
And, in this case, all non-zero solutions are just scalar multiples of this one. 
Interestingly [select all that apply]: 
we can write VI as a linear combination of v2, v3 and v 4 
we can write v2 as a linear combination of VI , and v 4 
O we can write v3 as a linear combination of VI , v2 and v 4 
we can write v 4 as a linear combination of VI, v2 and v3. 
















Untitled picture.png Machine generated alternative text:
When considering the span of a set of vectors, a vector in the set which can be expressed as a linear combination of the others in the set can be 
considered redundant, in the sense that removing it will not affect the span. 
From the our working above we can conclude that [select all that apply]: 
span{v2, v3, v 4} 
span{V1 v3' 
span{V1 v2' "'4 } 
span{V1 v2' v3}• 

Untitled picture.png Machine generated alternative text:
cosh (x) 
12 —10 
-2 
—8 
—6 
-4 
-6 
4 
-2 
—6 
2 
4 
6 
a = 1.52 
8 
10 
Forget about face recognition, James. The latest SPECTRE internet password requires you to combine functions in the infinite dimensional vector space of 
, coshx, and sinhæ together with 
functions to obtain a "0-key" to crack into their evil web-portal. The pass key app here shows the functions e ,e 
the linear key combination 
f@) = aec + be +ccoshx +dsinhx. 
Your mission, if you choose to accept it, is to find some combination of coefficients that will: 
i) ensure that the function 
aec + be¯x + ccoshx + dsinhx 
is the zero function (or at least very close to it), and 
ii) at least one of a, b, c, d is exactly equal to 1.
Untitled picture.png Machine generated alternative text:
cosh (x) 
12 —10 
-2 
—8 
—6 
-4 
-6 
4 
-2 
—6 
2 
4 
6 
a = 1.52 
8 
10 
Forget about face recognition, James. The latest SPECTRE internet password requires you to combine functions in the infinite dimensional vector space of 
, coshx, and sinhæ together with 
functions to obtain a "0-key" to crack into their evil web-portal. The pass key app here shows the functions e ,e 
the linear key combination 
f@) = aec + be +ccoshx +dsinhx. 
Your mission, if you choose to accept it, is to find some combination of coefficients that will: 
i) ensure that the function 
aec + be¯x + ccoshx + dsinhx 
is the zero function (or at least very close to it), and 
ii) at least one of a, b, c, d is exactly equal to 1. 

Untitled picture.png Machine generated alternative text:
Your solution James? Clearly the answer, or at least one answer, M, is given by the vector 
b 
<-1.5, 0.5, 1, 2> 
c 
1 
2 
Recall: the Maple notation for the vector 
3 
4

 

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