okay these are review problems for the
test number three in Math 220 the
topics are pretty much I can values
eigen vectors and eigen spaces and the
trajectories of given matrices so you
take a look at that first problem says
describe the trajectories corresponding
to the matrix a that's a point three
point five negative 0.5 and point three
so let's start on that first you may
notice this is of the form a B the
opposite of B a and these have complex
eigenvalues and eigenvectors those
eigenvalues are supposed to be of the
form lambda is equal to a plus or minus
bi you can show that pretty quickly in
this one but saying if you take point 3
minus lambda and 0.5 and then negative
0.5 and point 3 minus lambda you set
that determinant equal to 0 you'll get
point 3 minus lambda times point 3 minus
lambda so that's point 3 minus lambda
squared and then minus the product of
negative point 5 and point 5 which is
positive will be a negative 0.25 so
it'll be plus point 2 5 or 0.5 squared
if you will and then we subtract point 2
5 from both sides way 3 minus lambda
squared is equal to negative 0.25 now we
have to take the square root of both
sides so they take the square root and
the square root of 0.25 is 0.5 of course
and since it's negative its 0.5 I we get
the absolute value of 0.3 minus lambda
it's equal to 0.5 I so 0.3 minus lambda
should equal positive or negative 0.5 I
and adding lambda to both sides and then
subtracting and adding point 5i from
both sides we get 0.3 minus or plus or
plus or minus 0.5 I and so that's our
those are our eigen values now based on
those eigen values we can describe the
trajectories because the trajectories if
you recall on when you have complex
eigen values like this the and they're
the same conjugates of one another
then the the magnitude of lambda tells
you how that those vectors will change
as they are as we apply this matrix
multiplication to them so we get lambda
the magnitude of lambda is equal to the
square root of a squared plus B squared
and a is 0.3 and B is 0.5 and so that's
going to give us the square root of
that's point 0 9 plus 0.25 which is
point 3 for the square root of that's
about point 6 I think we do that on the
calculator I think we'll get point 5 8
to something so I'll say approximately
point 5 8 so that is we usually call
that are the magnitude of lambda we call
R that's the distance of this the
magnitude of that back those vectors so
that means that every vector in this
trajectory will be 58 percent or 0.5 8
times the magnitude of the previous
vector if you remember this ends up
being a either a contraction or an
expansion of R 0 0 R and then it's a
rotation well first it's a rotation that
doesn't really matter here cosine fee
and either plus or minus sign fee here
in our case we have a minus sign fee and
so this minus in this corner means it's
going to be a clockwise rotation
remember if it's up here it's a
counterclockwise in the positive
direction so this is cosine fee that's
the standard matrix of rotation for a
clockwise rotation and so we know that
that
the cosine of fee from this we also know
that the cosine of fee which we use to
find the magnitude of the rotation is
equal to a that coefficient over R so in
this case it's going to be a in this
case is 0.3 and R is the square root of
0.3 for so that means fee is going to
equal the inverse cosine of 0.3 over the
square root of 0.3 for put that in your
calculator in degree mode that turns out
to be about 59 degrees so that gives us
pretty simple way subscribing that
rotation then the the vector any vector
in that trajectory will be contracted
it'll be shortened be about 58% of what
it was previously and it will undergo a
clockwise rotation of 59 degrees so that
means a vector
would undergo a 59 degree clockwise
rotation and a magnitude change with a
factor or by a factor of R in this case
which is about 0.5 eight-game so that is
the conclusion of that by the way that
would be sorry since that's that we
would say that is spiraling towards or
toward the origin spiraling inward
toward the origin so these vectors if we
just a rough sketch if you took a vector
out here and applied this to it it would
be a little more than half its length
and then would take about a 60 degree
rotation or you could rotate it 60 and
then shrink it and that's it that's like
at about a 45 so here's a vector and
then it goes down to here it's about
half as long then oh it would go down to
about another 60 degrees
right about here and be even smaller
that length and be cut in half then
another 60 degrees would be cut in half
then another 60 degrees so it's spiral
in 2 right here it kind of look kind of
like this if you follow the tips of
those vectors and it would get closer
and closer to the origin who never
really equals 0 it's always gonna be
about 58% of the previous but it'll
spiral in pretty quick ok
problem number two is to find the
eigenvalues eigenvectors and then a
basis for the eigenspace spaces and the
actual eigenspaces for the matrix a and
then plot the trajectory of the vector 3
4 under that transformation so here's
the matrix 3 1 0 5 so how do we find the
eigenvalues well this is a triangular
matrix so the eigenvalues are just the
main diagonal elements 3 & 5 so have the
four the eigenvalue 3 the eigen space
would be determined by taking setting a
minus lambda I times X sum all the eigen
space would be all the vectors X that
would make that equal to 0 remember an
eigen space is a null space really
special null space and so let's let's go
ahead and do that we'll subtract 3 from
the main diagonals right that's what it
means so we'll get 3 minus 3 is 0 and we
get 1 there and then 0 and 5 minus 3
would be 2 so we basically get this the
eigen space is the solutions of this
homogeneous equation and so here we see
that X sub 1 is free and X sub 2 is a
basic variable this is 1 X sub 2 is 0
and 2 X sub 2 is 0 so we know that exit
2 from this is just equal to 0 as simple
as that
so the eigen the basis for the
eigenspace is going to be eigen space is
going to be 1 for the free variable and
0 for that and the whole eigen space is
gonna be X sub 1 times 1 0 all possible
vectors that would solve this so here's
the basis vector and this is the eigen
space all corresponding to lambda
equaling
three now for lamb be equal to five we
get three three minus five on that hang
on five minus five so that should be
negative two one and zero zero now of
course this is a homogeneous system
we're looking for a solution for and so
we can clearly see here that X sub one
is a fixed variable and X sub two is our
free variable so this tells us that
negative two X sub one plus one X sub
two is equal to zero so negative two X
sub 1 is equal to negative one x sub two
so X sub one is equal to one-half X sub
two so that gives us a basis vector of X
sub 1 is 1/2 of X sub 2 that's a half
there and 1/4 X sub 2 because it's the
free variable and the whole space would
be this again is our basis factor and
this eigen space is X of 2 times 1/2 1
now by the way the basis could be any
multiple of this would be a basis vector
since it's just one vector and any
multiple of this would be the basis a
bait would be a basis for this eigen
space as well so we could have we could
have said this is one too for that
matter
and this would be X of 2 times 1 2 if
you'd like this is all for lambda equal
in 5 and now it says plot the trajectory
so this is our this is our basis this is
our eigen space and there's our eigen
value for those corresponding eigen
values so if we plotted that I guess
I'll get a piece of graph paper
he's a graph paper these I want to do
this I have to plot the trajectory a
little more closely it says to plot the
trajectory of this eigen the vector 3/4
which is clearly not an eigenvector okay
so let's zoom back out there so it's not
an eigenvector so it won't be on the
eigen space but it the eigenspaces will
influence it so if we plot those eigen
spaces for lambda equal 3 we get 1 0
that's just along that horizontal axis
horizontal axis there this is since
these are all positives I think we're
gonna go start this below and that's the
eigenspace for lambda equal in 3 because
that was 1 0 so that would be 1 0 would
be that vector it's all multiples of it
and then the eigenspace for lambda equal
5 is more all multiples of 1 2 so if we
go over 1 and up 2 we get that point do
you go over 1 and up 2 all these all
these would be eigen vectors on the
eigen space and between so this is the
eigenspace for lambda equal 5 and that
is a repeller that's heading away from
the origin this is also a repeller
heading away from the origin I didn't
draw that in class today and I should
have been I was treating this as an
attractor coming towards the origin when
I talked about that in class so and that
was an error the vector is 3/4 so the
vector three four one two three one two
three four is close to that eigen space
but this is a
repeller also so3 this this eigenspace
does have a pretty good influence on
this it's not just influenced by that
eigenspace so if we do a trajectory we
would take this matrix three one zero
five and multiply it by the vector three
four if we do that calculation we get
nine and four which is 13 and we get
zero and 20 which of course is 20 and so
that's the first point trajectory so 13
20 would be over there's 1 2 3 4 5 8 9
10 11 12 13 and then 20 2 4 6 8 10 12 14
16 18 1e that's gonna be way up here so
let me do that again okay 2 4 6 12 14 20
now look at that it was from here to
here if this was an attractor this would
be drawing in and when I drew it
sketched it quickly this morning I
thought oh that's going to be going into
that because that was a eigenvalue of 3
which also draws it to the right so this
trajectory actually won't go into that
eigen space like the ones we did
yesterday in class because this is a
repeller they're both repellers if we
did this trajectory again we would take
3 1 0 5 and multiply that by 13 20 and
that would give us 39 and 1 is 59 and
then 0 that would be 0 and a hundred so
59 and 100 which is off our chart but if
you did a vector in the direction of 59
100 would be about 30 and 50
or 15 and 25 that's just a little
further this way 15 and 25 would be it
would be in this direction way you know
4 times the distance quite a ways out so
anyway that that trajectory doesn't go
actually towards that directly so anyway
that's just 1/2 of me you can go like
this say okay goes along there that's
the trajectory at first of that
particular vector for at least one
iteration I'm going to continue the next
two problems on another video
