0:06

Good, so, let's look at these.

Â We said they're judicious in both frames.

Â They're fixed in both frames.

Â That means the B frame components and N frame components have to be the same.

Â So the e hat vector, if you composed into both axes gives you the same thing.

Â So, there's really no distinction if this is, if this e hat is in the B frame or

Â the N frame.

Â If e hat is what takes you from N to B, they're the same.

Â 1:09

I have e hat in the N frame.

Â And I end up with e hat in the B frame, all right?

Â That's essentially what that other side was talking about.

Â That's where we're going to be.

Â But with this theorem, Euler says, on his right,

Â that really you can drop these letters and say, look, they're really the same.

Â 1:34

So Robert, if you take a vector the three by one basically

Â times a matrix and you get back the same vector what is that vector called?

Â How does it relate to the matrix?

Â >> [INAUDIBLE] Eigen.

Â >> Eigen.

Â Eigen what is Eigen it's a German word?

Â >> Self.

Â >> Self essentially.

Â It's a self if you hear eigenvector, eigenvalue it's due to a self mapping.

Â This thing maps back onto itself.

Â So that's why I had you guys solve this eigenvector,

Â eigenvalue problem by hand you know so much fun.

Â As much we don't have to solve here an eigenvector, eigenvalue problem.

Â The beauty here is with this stuff,

Â I can show an analytic answer of how to go from a DCN and get this vector.

Â 2:28

Do you just get back the same vector?

Â What was your name again?

Â >> David.

Â >> David, thank you.

Â >> No, if the eigenvalue's one, you would but yeah, here it constant.

Â >> Okay, so for every eigenvector, there is an associated eigenvalue.

Â Hopefully not news to even enough because you have done that homework one.

Â So, I'm not spending much time on eigenvector, eigenvalues.

Â So good, so that is always there.

Â So you have to know that is the eigenvalue.

Â The eigenvalue tells you by how much do you stretch or shrink that vector.

Â And this has to do now also with the question of whether it has to be

Â unit length and so forth.

Â When you doing rotations and if this is more to unit like one of the sadden

Â you end up to take that put it regular dc in math.

Â You definitely subtraction vectors you will not get something that auto normal in

Â that kind of habit.

Â You have to read normalize the rotation matrix.

Â Its a lot of extra work.

Â Now since you guys are all expert eigenvalues, eigenvectors.

Â If you have a three by three matrix Maurice how many eigenvalues

Â would you have?

Â >> Three.

Â 3:40

>> They're not necessarily none.

Â >> Exactly, trick question, could be either, [LAUGH] right?

Â So you have always eigenvalue, eigenvector problem.

Â There's all the stuff,

Â hey assuming they're unique, this is Math and how you do it?

Â If they're unique do you have unique eigenvectors?

Â 4:40

>> Non-unique.

Â >> Non-unique, actually.

Â Now the plane in which v one and two line is going to be unique.

Â But within that plane you can define it with an infinities of

Â combinations of vectors, right.

Â So does just remember that.

Â So now into eigen, if we talk about this and think it through and think, okay,

Â the zero rotation is identity matrix,

Â an identity matrix has a triple repeated root.

Â It has three plus one's basically the eigenvector is anything.

Â 5:09

Any access right and that's what we described earlier.

Â We're talking about those ambiguities that we're going to get into, right.

Â So it directly relates to eigenvectors.

Â So let's say we do have unique eigenvalues and

Â I'm going to go back to a plus one times v equal to a times v.

Â 5:45

>> Really not sure, actually.

Â I would want to say no.

Â >> It's 50/50.

Â Now my next question [LAUGH].

Â >> I don't know really like why it'd be no.

Â >> Let's just think it through, all right?

Â I'm not expecting it to be completely expert to linear algebra and

Â that stuff but this really becomes.

Â If this math is property.

Â If this math is property, if this property is mathematically correct and it does it.

Â I pick a vector of 1, 2, 3 times +1,

Â does three by three matrix give me 1, 2, 3?

Â If I see this math, is there a different

Â vector that you can envision that would also satisfy this equation?

Â 6:34

Yes, Cody. >> You can multiply by scalar.

Â >> Exactly, if one, two, three is an icon vector, so is two, four, six.

Â So you can multiply them to scalar and in particularly, you can flip the sign.

Â 7:04

>> If you have repeated eigenvalues you don't have unique.

Â There is thin to eigen space it's kind of a subspace of the manifold.

Â It is unique, that plane.

Â And if it's a double repeated eigenvalue, that plane is unique at that instant but

Â you can use any sets, combinations.

Â >> Isn't that what the geometric multiplicity is once so

Â you could have repeated eigenvalues with unique vectors like the identity matrix.

Â 7:41

It is, you could have a- >> It's a two dimensional space.

Â If I only have a double repeated eigenvalue,

Â I have a two dimensional subspace within which those two eigenvectors must lie.

Â >> But there are some examples whether the deficiency there,

Â that are actually one dimensional.

Â 7:56

>> Not that I'm aware of these real numbers maybe with complex or

Â some other stuff, but with regular real numbers, no.

Â You end up with subspace, that is then defined through them.

Â We'll take this off line to see what we are really talking about there, but

Â typically no.

Â If you have a double repeated eigenvalue, then when you do the reduction to get

Â the eigenvectors, you come up with an ambiguity.

Â And it's always that if it is a double repeated,

Â it is an two dimensional sub-manifold of the fold mapping.

Â And the fold mapping is a three dimensional mapping.

Â So it is a plane within that three dimensional sub-space.

Â So the key is to remember eigenvalues, eigenvectors are not even what

Â math lab gives you, people ask me, can I use math to find my answer?

Â I go, go for it.

Â At least to double check your answer, you'd have to do this process.

Â But it's going to give you a vector.

Â And it uses some algorithm that says, I'm going to pick one, two, three.

Â Why not minus one, two, three?

Â It's its choice.

Â But the choice of the vector impacts your attitude description.

Â And when we do estimation we'll find similar eigenvalue,

Â eigenvector kind of ambiguities.

Â So this discussion again.

Â It's going to pay off multiple times that we do this so

Â there's some issues there that we really have to consider when we do this.

Â The other thing is MathLab, mathematic and

Â most algorithms tend to give you the unit vector.

Â Instead of giving you one, two, three,

Â it's going to give you that normalized back to a unit of length.

Â Which for us is perfect, because in our application,

Â we're actually looking for a unit vector.

Â But eigenvectors by themselves do not have to be unit vectors you could have 500

Â being an eigenvector.

Â So if you think of identity, really it's the identity operator

Â 9:59

This must be an eigenvector and

Â actually it must be the eigenvector corresponding to a plus one eigenvalue.

Â If you have a general rotation, you will tend to get one that is plus one and

Â the other eigenvalue will tend to be a complex conjugate set of number.

Â So if you have to solve that

Â cubic polynomial to get the characteristic stuff.

Â One is plus one, the other two are complex conjugates, okay, cool.

Â So you're looking for the plus one in math lab, you put in a DCM, get some numbers.

Â There's two complex numbers.

Â One of the real one, plus one hopefully.

Â They're all unitary, actually, because it's a orthonomal metrics and

Â now you know which eigenvector to get.

Â It won't always be the first,

Â it won't always be the second it won't always be the third.

Â Math lab has no idea what you're trying to solve.

Â Just look for which one is plus one and

Â that's the corresponding eigenvector you have to pick.

Â So that's one way you can find this.

Â Now, let's talk about uniqueness.

Â With these sets, e hat and fee.

Â 10:56

If I have one rotation.

Â I'll make it very simple.

Â This is my zero rotation facing you guys.

Â And my new rotation b is a 45 degree rotation to my left, your right.

Â Right so I'm doing for me a plus 45 degree rotation.

Â So I've got 45 degrees with my big feet and my e hat is up.

Â 11:25

Lewis?

Â >> If you do the negative of the axis or the long way around?

Â >> Right, and that's basically a different combination.

Â So let's use the up axis.

Â I went to the left.

Â Okay, but we know attitudes reside in the SO3 group which means there's a close set.

Â My attitude can't go off to infinity.

Â The worse I can be off is 180 degrees so instead of 45 this way.

Â You could've gone basically all the way around, and then, what is it?

Â 360 minus 45, is that 315?

Â So you go 315 minus 315, about the up axis.

Â That gets you there as well, right?

Â That's one.

Â And then you mentioned well, you could go the negative axis.

Â And again, being an eigenvector, I can flip the sign.

Â There's nothing you need that you have to go about this one.

Â I could've gone about the down, I would now have to go minus 45,

Â because this would be a positive.

Â Or I go a plus 350 to get there.

Â 12:22

Then we have four combinations that you can get.

Â That's important because then when you extract these parameters from the DCM,

Â I should expect four possible answers.

Â There can't just be one answer.

Â So now the attitudes sets we go to if they're not unique and

Â we'll find ways to really take advantage of non-uniqueness in elegant ways.

Â 12:43

But I have to have Math to give you multiple answers, so that's it.

Â So I'm going to use here, this is my pi prime,

Â is basically my long way and pi is my short way.

Â It's a shorthand to differentiate.

Â There's two short descriptions, one with a positive angle, one a negative,

Â and then two long.

Â One with a positive, one with a negative, that get you there.

Â Which ones do you prefer, Trevor, short angles or long angles?

Â >> Bring old.

Â >> Yeah, most people prefer short, especially if you doing a control

Â application, which you'd rather control the space for app and say,

Â look your arrow is 45 degrees, or say no the arrow is -335, and you're doing this.

Â 13:21

So in the leadership this is called the unwinding problem.

Â And you want to have attitude sets and cease where you will prefer to go look,

Â it's just your white, just glance that way and off you go, all right?

Â So we tend to pick the shorter one but it's not required.

Â If I'm looking at these angles, let's say estimation.

Â I want to know how different are these two frames, I don't want to say, well track

Â men, guess what, your estimation men, your difference was 359 degrees.

Â That wasn't very good, all right.

Â But no, it was off by one degree, that's brilliant.

Â 13:58

So here is a four possible combinations, we had, 45 degrees,

Â about up, minus 45 about down, or you do.

Â Up about the long axis or minus down about the minus long.

Â Different combinations how you can do these two things and you can get there.

Â But here in the book I derive this, I'm not doing using class but

Â there's some basic algebra staring at DCM finding how if you add this,

Â this and this if you get the trace of a matrix.

Â The trace of a matrix is the sum of the diagonal elements here the c123.

Â If you do this math you will have a bunch of stuff that is going to be cosined.

Â 14:42

You will add up these diagonals.

Â You'll get e1 squared, e2 squared, e3 squared times sigma

Â while the e123 squares have to be one right because this is a unit vector.

Â That simplifies it, you bring in the definition.

Â And in the end you can solve for the cosine of that angle, which is nice.

Â But now if you look at the cosine curve

Â 15:23

That's it.

Â It doesn't know the problem you're trying to solve.

Â Any calculator inverse cosine just gives you an answer between zero and 180.

Â Which was perfect for inclination angles and extracting them, but

Â here there's actually other answers.

Â So if you look at the math and go where's my second answer coming from,

Â the long answer is simply if this is your cosine value that is the short answer.

Â And then basically, 360 minus that angle that you got from the calculator,

Â that's the long way around, right.

Â Then so you have to do your own math to get the other angle.

Â The calculators won't give this to you or math level only always give you one angle.

Â It doesn't give you multiple ones.

Â So good.

Â So basically that's the, then you could subtract 360 or here I'm taking

Â the angle minus 360, that's how I'm defining the long way around.

Â And that's what makes this math work.

Â Once you have that angle, then if you stare at this, you can see if you take 2,

Â 3 minus 3, 2, so 2, 3 is this element minus this element This part and

Â this part cancels it leaves you with these terms that sum up and we know the angle.

Â That's a known quantity so I can do that math and extract out e1.

Â So this is nice because this will now implicitly given me the correct

Â eigenvector and yeah,

Â so we have the short, the long, and then you have right eigenvector.

Â 17:04

Well no.

Â This is going to give you, if you pick the right,

Â the short of the long, this will all work out.

Â [COUGH] And then you use the other combinations that we showed.

Â Here instead of 45 here, you can do minus the other one

Â comes from the complement sets that you could generate.

Â But they are all equivalent and so you get them getting four.

Â Practically we always pick to positive angle.

Â The difference between this frame and this frame is one degree, not minus one.

Â It's true minus one, but we would practically speak to positive to

Â show the angle and then pick the axis that gets you there.

Â 17:38

So any questions?

Â This is just showing you how do we go from the DCM.

Â With Euler angles we have to go inverse sines and cosines and inverse tangents.

Â Here you take the trace to get the cosine of the angle and

Â then take differences of the matrix components to find e hat vector.

Â We don't have to solve an eigenvalue, eigenvector problem.

Â We can find it explicitly from this derivation.

Â And how to get from these principles stuff to here.

Â The book actually gives you a detailed derivation of that.

Â I'm not going through that in class.

Â If you're curious, you can follow along there, okay.

Â Now, so these angles are very handy.

Â