Good day Prof.,
We will like to confirm to you the agreed date for the scheduled class to be Saturday, 20 Feb., 2016 by 8:00AM.
We will like to confirm your availability so that we can pass the information around.

So far, I have not received any enquiries on the quiz 1.1, 1.2! Does it mean that you have all been able to work them out successfully? Does it mean you are all ignoring them and refusing to think? If the latter, that is what the path to failure is made of!

We assure that we will give it focus sir.
Personally, I’m trying to catch up with the other part of the last class which is entirely new to me, but very soon, I will attempt the problem and give my feedback.
Have a pleasant day sir.

attempt to quiz 1;a.v=b.v for all vectors v if and only if a=b. from a.v=b.v,rearranging we have a.v-b.v=0,therefore
v.(a-b)=o.for this to be zero,it implies that,v=(a-b),substituting v we have (a-b).(a-b)=0.dot product of this=/a-b/=0,which gives,sqrt((a-b)^2)=0.expanding we have sqrt(a.a-a.b-b.a+b.b)=0…eqn1,but a.a=1,b.b=1,so for eqn1 to hold,then a=b.
attempt to quiz2,using the above argument,ie putting v=(a-b) into v*(a-b)=0,we have,(a-b)*(a-b)=o.cross product of this gives /a-b/=0,which gives,sqrt((a-b)^2)=0.expanding we have sqrt(a*a-a*b-b*a+b*b)=0…eqn2,but a*a=0,b*b=0,so for eqn2 to hold,then a=b

You made a good effort. There are errors though:
1. You cannot assume that a or b are unit vectors. It is therefore wrong to assume that a.a=1 or b.b=1.
2. The fact that a*b=0 does not prove that a=b. It only shows that a and b are colinear. In which case, a=(alpha)b where alpha is a scalar.
One correct answer to the quizzes can be found in slide 78

If two members of the same set are subject to an operation whose product is always a member of the set, it means you cannot generate a nonmember of the set by performing such operations on set members. That operation has closure. For example, the vector space is closed under addition because whenever you add two vectors, the answer you get is another vector! Conversely, the dot operation does not have closure. If you perform the dot between two vectors, what you get is a scalar – not a member of the vector space! The vector space is not closed under the dot operation. To laugh, it means the door was open for the dot operation to go out and produce “bastard” babies that are non-members!

Yes. There are three products defined between two vectors: 1. Dot product which you know very well. 2. Cross product, which you are talking about now, and 3. Tensor product which acts on two vectors to create a tensor. Some call the dot product an “inner” product, while the tensor product is called the “outer” or Kronecker product.
Of the three products, only the cross product produces vectors. The others are not closed because inner product produces scalars while tensor product produces second-order tensors.

What you call a function is, mathematically speaking, a map. Whenever you can create a correspondence between one set and onother, or a subset of itself, you have a map. A map can be represented by a table of values, a drawing, or by a formula. The latter is the one you are most familiar with. But strictly, a map can be represented in several ways. A geographical map is a function because it creates a correspondence between a spatial space and a sheet of paper! These two regions are collection of points – or, sets!

Sir, does it really matter if the base vector is either in superscript or subscript when dealing with contravariant and covariant components? Or can I use them interchangeably?

When you use the contravariant components, you are referring to the subscript base vectors. When you are using the covariant components, you are using the supperscript base vectors. In one case you are using the original basis while in the other, you are referring to the dual basis. Either of these is valid; you must use the correct components and you can see that there is always a contravariance between the index of the components and that of the base vectors. Choose whichever is more convenient or just use ant representation you like!

Good day all,
Regarding our Saturday class, we realized that the movement restriction due to environmental sanitation might pose a threat, and as a counter measure, we should all consider getting to the school earlier before the commencement of the sanitation program to enable us have a smooth class.
Otherwise, we may consider starting our class by 11am prompt.
If you have any better suggestion, present it and lets have an agreement to propose to Prof. Fakinlede.
I wish everyone a pleasant day!

Good day,sir. what do we mean when we say ” two linearly independent vectors” and how is the product of 2vectors equal to a scalar? My warmest regards ,sir

Any two non collinear vectors are linearly independent. Three vectors are linearly independent if no two are collinear and the three are not contained on the same plane. It is impossible for four or more vectors to be linearly independent in the 3-D Euclidean point space of our physical experience.

I think vectors which are linearly independent are vectors whose angle between them should not be zero. The angle between them perhaps 45 degrees, 90 degrees( orthogonal) e.t.c. That is x>0<180..
Whenever you have the product of two vectors as a scalar it only means that it is an "inner product or dot product"- which could represent the magnitude of both vectors, angles and therefore measure areas and volumes.
I hope you understand me?

When the product of two vectors equal a scalar:It means that you can find out the actual value of the resultant of both vectors( magnitude), you can also find out the volumes and areas which are enclosed by both vectors and you can also find out the angle between both vectors.
Hope it is much clearer?

The answer you have given is necessary but not sufficient. If four vectors have non-zero angles between them, they will still be linearly dependent because the 3-D space of our physical intuition cannot admit a set of free vectors that are more than three in number!

You are not correct here. Linear independence is NOT about angles. If you have only two vectors, linear independence is guaranteed once they are not co-linear. If there are three vectors, pairwise non co-linearity will not guarantee linear independence. In addition, you MUST have that the three are NOT co-planar. Lastly, if the vectors are four or more, they will be linearly dependent, no matter what because the space of our physical experience DOES NOT allow for more than three linearly independent vectors.

Good day prof, can i also refer to the base vector as the covariant base vector just as the contravariant base is been used in place of the dual base vector?

To find a contravariant component, you take the dot product with the contravariant base. Similarly with a covariant component. Remember that the covariant component is actually the component of the contravariant base. You always need the base and the dual when you are dealing with non-orthogonal bases.

To find a contravariant component, you take the dot product with the contravariant base. Similarly with a covariant component. Remember that the covariant component is actually the component of the contravariant base. You always need the base and the dual when you are dealing with non-orthogonal bases.

A set of basis vectors is a way to simplify our reference to the vector. The vector itself is independent of the basis you are using to refer to it. If you choose a basis, there is immediately another set that also forms a different basis. This related set is called the dual. their relationship is called the reciprocity relationship.

There is a small untidiness on the slide you refer to. I have corrected it and will effect it on the web by Friday. If it remains unclear by Saturday, I will go through it in class.

Dear all,
Trust we had a great week?
Since the movement restriction during tomorrow’s environmental sanitation programme is suspended due to the JAMB UTME Exam, we can now hold our class by 8:00am as scheduled.

Sir, In the effort to secure classroom for Saturday’s class during the Easter Celebration of Christ Work. Is it possible that the request the Potters to release the keys to classroom for Saturday?

The LG Design Lab will be opened on Saturday. I have already talked to Emmanuel who will see to it that we have access. AS long as the members of the course want a class, they will get it.

Good day Prof.,

We will like to confirm to you the agreed date for the scheduled class to be Saturday, 20 Feb., 2016 by 8:00AM.

We will like to confirm your availability so that we can pass the information around.

Best regards,

Iretioluwa

Yes I will teach the class on Saturday morning at 8-11am. Three hours with a 15 mins break midway.

So far, I have not received any enquiries on the quiz 1.1, 1.2! Does it mean that you have all been able to work them out successfully? Does it mean you are all ignoring them and refusing to think? If the latter, that is what the path to failure is made of!

We assure that we will give it focus sir.

Personally, I’m trying to catch up with the other part of the last class which is entirely new to me, but very soon, I will attempt the problem and give my feedback.

Have a pleasant day sir.

Good day Sir.

We are still working on them Sir and will get across accordingly Sir.

attempt to quiz 1;a.v=b.v for all vectors v if and only if a=b. from a.v=b.v,rearranging we have a.v-b.v=0,therefore

v.(a-b)=o.for this to be zero,it implies that,v=(a-b),substituting v we have (a-b).(a-b)=0.dot product of this=/a-b/=0,which gives,sqrt((a-b)^2)=0.expanding we have sqrt(a.a-a.b-b.a+b.b)=0…eqn1,but a.a=1,b.b=1,so for eqn1 to hold,then a=b.

attempt to quiz2,using the above argument,ie putting v=(a-b) into v*(a-b)=0,we have,(a-b)*(a-b)=o.cross product of this gives /a-b/=0,which gives,sqrt((a-b)^2)=0.expanding we have sqrt(a*a-a*b-b*a+b*b)=0…eqn2,but a*a=0,b*b=0,so for eqn2 to hold,then a=b

You made a good effort. There are errors though:

1. You cannot assume that a or b are unit vectors. It is therefore wrong to assume that a.a=1 or b.b=1.

2. The fact that a*b=0 does not prove that a=b. It only shows that a and b are colinear. In which case, a=(alpha)b where alpha is a scalar.

One correct answer to the quizzes can be found in slide 78

In slide 30.what do you mean closure exits for the operation(Addition)

If two members of the same set are subject to an operation whose product is always a member of the set, it means you cannot generate a nonmember of the set by performing such operations on set members. That operation has closure. For example, the vector space is closed under addition because whenever you add two vectors, the answer you get is another vector! Conversely, the dot operation does not have closure. If you perform the dot between two vectors, what you get is a scalar – not a member of the vector space! The vector space is not closed under the dot operation. To laugh, it means the door was open for the dot operation to go out and produce “bastard” babies that are non-members!

In other words, only a cross product operation has closure?

Yes. There are three products defined between two vectors: 1. Dot product which you know very well. 2. Cross product, which you are talking about now, and 3. Tensor product which acts on two vectors to create a tensor. Some call the dot product an “inner” product, while the tensor product is called the “outer” or Kronecker product.

Of the three products, only the cross product produces vectors. The others are not closed because inner product produces scalars while tensor product produces second-order tensors.

Thank you sir

Sir, I would like to know the idea behind mapping.

Thank you.

What you call a function is, mathematically speaking, a map. Whenever you can create a correspondence between one set and onother, or a subset of itself, you have a map. A map can be represented by a table of values, a drawing, or by a formula. The latter is the one you are most familiar with. But strictly, a map can be represented in several ways. A geographical map is a function because it creates a correspondence between a spatial space and a sheet of paper! These two regions are collection of points – or, sets!

Sir, does it really matter if the base vector is either in superscript or subscript when dealing with contravariant and covariant components? Or can I use them interchangeably?

When you use the contravariant components, you are referring to the subscript base vectors. When you are using the covariant components, you are using the supperscript base vectors. In one case you are using the original basis while in the other, you are referring to the dual basis. Either of these is valid; you must use the correct components and you can see that there is always a contravariance between the index of the components and that of the base vectors. Choose whichever is more convenient or just use ant representation you like!

Thank you sir.

Good day all,

Regarding our Saturday class, we realized that the movement restriction due to environmental sanitation might pose a threat, and as a counter measure, we should all consider getting to the school earlier before the commencement of the sanitation program to enable us have a smooth class.

Otherwise, we may consider starting our class by 11am prompt.

If you have any better suggestion, present it and lets have an agreement to propose to Prof. Fakinlede.

I wish everyone a pleasant day!

Good day,sir. what do we mean when we say ” two linearly independent vectors” and how is the product of 2vectors equal to a scalar? My warmest regards ,sir

Any two non collinear vectors are linearly independent. Three vectors are linearly independent if no two are collinear and the three are not contained on the same plane. It is impossible for four or more vectors to be linearly independent in the 3-D Euclidean point space of our physical experience.

I think vectors which are linearly independent are vectors whose angle between them should not be zero. The angle between them perhaps 45 degrees, 90 degrees( orthogonal) e.t.c. That is x>0<180..

Whenever you have the product of two vectors as a scalar it only means that it is an "inner product or dot product"- which could represent the magnitude of both vectors, angles and therefore measure areas and volumes.

I hope you understand me?

When the product of two vectors equal a scalar:It means that you can find out the actual value of the resultant of both vectors( magnitude), you can also find out the volumes and areas which are enclosed by both vectors and you can also find out the angle between both vectors.

Hope it is much clearer?

The answer you have given is necessary but not sufficient. If four vectors have non-zero angles between them, they will still be linearly dependent because the 3-D space of our physical intuition cannot admit a set of free vectors that are more than three in number!

You are not correct here. Linear independence is NOT about angles. If you have only two vectors, linear independence is guaranteed once they are not co-linear. If there are three vectors, pairwise non co-linearity will not guarantee linear independence. In addition, you MUST have that the three are NOT co-planar. Lastly, if the vectors are four or more, they will be linearly dependent, no matter what because the space of our physical experience DOES NOT allow for more than three linearly independent vectors.

good day, every one. getting to school an hr after sanitation exercise will be difficult for some of us. why not 12pm?

Agree among yourselves. We can have the class at 6:30-9:30am 0r 12noon -3:00pm as you choose.

Good day prof, can i also refer to the base vector as the covariant base vector just as the contravariant base is been used in place of the dual base vector?

Yes. You can do that because it uses a lower index. Remember that its component must be conravariant

Sir, to evaluate a contravariant component, Do we take d inner product of d Vector& it contravariant base vector (V.g-superscript) ?

To find a contravariant component, you take the dot product with the contravariant base. Similarly with a covariant component. Remember that the covariant component is actually the component of the contravariant base. You always need the base and the dual when you are dealing with non-orthogonal bases.

To find a contravariant component, you take the dot product with the contravariant base. Similarly with a covariant component. Remember that the covariant component is actually the component of the contravariant base. You always need the base and the dual when you are dealing with non-orthogonal bases.

Sir is it true that In any euclidean space for every vector that exist is accompanied by a basis vector g(subscript i)?

A set of basis vectors is a way to simplify our reference to the vector. The vector itself is independent of the basis you are using to refer to it. If you choose a basis, there is immediately another set that also forms a different basis. This related set is called the dual. their relationship is called the reciprocity relationship.

Sir Slide 64: cross product of basis vectors.

it’s a little fuzzy now. Sir I need more lights.

Thank you

updated slide 61.cross product of basis vectors.

Thank you

The last equation in particular

There is a small untidiness on the slide you refer to. I have corrected it and will effect it on the web by Friday. If it remains unclear by Saturday, I will go through it in class.

The course is now beginning to make sense..

Thank you sir.

Dear all,

Trust we had a great week?

Since the movement restriction during tomorrow’s environmental sanitation programme is suspended due to the JAMB UTME Exam, we can now hold our class by 8:00am as scheduled.

Regards!

I want a confirmation that we can agree for 8:00 am Saturday for the SSG805 lecture.

Sir good day. I would like to request for the video animation on base vector and its dual to aid visualization.

Thank you.

Sir, In the effort to secure classroom for Saturday’s class during the Easter Celebration of Christ Work. Is it possible that the request the Potters to release the keys to classroom for Saturday?

So far they (potters) have declined the request.

The LG Design Lab will be opened on Saturday. I have already talked to Emmanuel who will see to it that we have access. AS long as the members of the course want a class, they will get it.

Saturday 26, 2016

Yes. 8-11am Saturday, March 26, 2016 at the LG Lab, University of Lagos!