Hi everyone my name is Claire Tomlin I'm
a professor of electrical engineering
and Computer Sciences at Berkeley and
this is the twenty-third module in a
series that we're recording to support
the course
EE CS 221 a linear system theory so in
previous modules we've been talking
about the solution to the linear
time-invariant system which involves
computing the matrix exponential e to
the a T where a is an N by n matrix and
so we're used to exponential functions
but we're not this this concept of a
matrix exponential where this is an N by
n matrix and how we compute it is
through the inverse Laplace transform of
si minus a inverse again for this n by n
matrix a so this can be a hard
computation as we talked about and so
we're going to spend a little bit of
time talking about how to simplify this
computation and one of the ways in which
we try to do this is to simplify the
matrix a to get it into a simple
canonical form where it's much easier to
do this calculation and the canonical
form that we typically simplify it to is
called the jordan form or in the case in
which there are n linearly independent
eigenvectors of the matrix a that
simplifies to a diagonal form so let's
start by talking about I gain values and
eigenvectors which is going to be our
basis for this transformation to either
the diagonal form or the jordan form so
eigenvalues and eigenvectors of the
matrix a let's assume that a is a matrix
in R and by N and we define an
eigenvector eigenvalue pair of the
matrix a as follows so e i which in
general is a complex vector in CN is
said to be an eigenvector is an
eigenvector
of a with associated eigen value eigen
value lambda I which can be a complex
number so it's a complex scalar if the
following condition holds a e I is equal
to lambda I e I ok and and we assume in
this the AI is not equal to zero so
let's just write that here so that's not
equal to the zero vector of course the
eigen value could be zero so this is
interesting if we just look at this
equation this is an N by n matrix this
is an N by 1 vector and now we have a
scalar and an N by 1 vector so if we
look at this sort of amazing fact that
all the information about that n by n
matrix has collapsed when we post
multiply it by this vector e aí into
this eigenvalue and eigenvector pair so
I mean the here we have an N by n matrix
and this is only a 1 by 1 scalar ok so
that's interesting how do we solve for
the eigenvalues and eigenvectors well if
we just look at that formulation we can
see that these lambda i's so the lambda
i's are solutions to and we'll just
bring the lambda ie I pair on to the
other side and write it out as
determinant of Si minus a is equal to 0
okay so I've done a couple of steps
there I've brought this over I've
assumed that if there's going to be a
non-trivial eigen vector in the null
space of a minus lambda I then I need to
have that the determinant of a minus
lambda I is equal to 0 or rewritten as
determinant of Si minus a is equal to 0
okay so solving this for s gives us the
eigenvalues of the matrix a okay so now
we can say a few things about that so if
we have a matrix a which is an N by n
matrix then we can solve for these
eigenvalues and we find that an N by n
matrix could have n possibly complex so
a real matrix could have n possibly
complex eigenvalues so in general we
could rewrite this formulation here
let's just keep that on the board this
is actually it's got a name it's called
the characteristic polynomial of the
matrix a so we use the terminology chi
hat a of s to be the characteristic
polynomial it's just defined as the
determinant of Si minus a and in general
we can write that as in terms of its
expansion s minus lambda 1 s minus
lambda 2 up to ask minus lambda n ok
good so we have a formulation for the
eigenvalues how to solve them and the
eigenvectors so the eigenvectors once we
have the eigenvalues we can solve for
the eigenvectors just by looking at the
null space what's in the null space of a
minus lambda i i for each index i okay
so we have the following interesting
result now given this n by n matrix a
and given its eigenvalues and
eigenvectors we can say the following
that so this is a theorem if a so it's a
real n by n matrix has n distinct
meaning different from each other
i ghen values and we'll call them lambda
1 lambda 2 up to lambda n so they can be
possibly complex as we know but they're
different from each other then
there exists so the theorem states that
there exist n linearly independent
eigenvectors
so we can write that as follows then the
set e 1 e 2 so this set of vectors which
are the eigenvectors associated to those
eigenvalues forms a basis for RN ok ok
so basically we're saying that those
there's n of them we're saying that
they're linearly independent when we say
they form a basis for RN and and the
proof of that is pretty easy we just
write it out we write out what it what
what the result of having n distinct
eigen values of the matrix a is okay so
we can prove it by saying suppose we
have n distinct eigen values but now
suppose these are not linearly
independent and we can derive a
contradiction okay so if if the set e
aí let's just write it as that for i
equals 1 up to n is not linearly
independent then we can write out some
linear combination of them such that the
coefficients are not all 0 yet the the
overall linear combination equates to 0
so that we can say there exists some
scalars which are possibly complex such
that alpha 1 e 1 plus alpha 2 e 2 plus
alpha and e n is equal to 0
yet the alpha eyes don't necessarily
have to be 0 so at least one of those
alpha eyes is not 0 such that at least
one of the alpha eyes is not 0 ok and
then we can go on so let's just assume
without loss of generality that it's
alpha
that's not zero so without loss of
generality we'll assume that alpha 1 is
not equal to zero okay so what does that
mean if we take this equation here let's
call that equation star and let's pre
multiply that equation by the following
by the following matrix a so pre
multiply the equation star by a -
lambda' - I a minus lambda 3i all the
way up to a minus lambda n I and so
we're going to multiple pre multiply
this equation here so the summation from
I equals 1 to N of alpha ie I so that's
equal to zero so this is equal to zero
and and so each time what we're going to
do is bring so if you think about this
pre multiplication we're going to bring
the eigenvectors back into this term of
matrices and we can use the fact that
these are eigenvectors to reduce each of
these terms to scalars okay so if we
think about that successive
multiplication then we see that we
quickly get that alpha 1 times lambda 1
minus lambda 2 times lambda 1 minus
lambda 3 all the way down to lambda 1
minus lambda n e1 is equal to zero okay
so that just resulted from pre
multiplying this equation by this factor
here and now we can derive the
contradiction because we know that e 1
is not equal to zero
it's an eigen vector of the matrix a and
we've assumed that these eigenvalues are
all distinct so each of these terms is
nonzero and so we derive the
contradiction that alpha 1 must be equal
to 0 but we assumed that it was not zero
so here we have that this implies that
alpha 1 is equal to zero which leads to
a contradiction okay so that tells us
that if we have distinct eigen values
all of these eigenvalues are different
from each other then a has n linearly
independent eigenvectors okay so that's
really important when we start thinking
about the diagonalization of a matrix
but we'll do that in the next module but
at the end of this module I just like to
give you a little bit of an intuitive or
geometric interpretation of what
eigenvalues and eigenvectors really mean
and so I said that it's it's sort of
amazing when you think about an
eigenvalue that all the information
about the matrix a is contained when you
post multiply that matrix a by the
eigenvector ii i you you get that that
that information about a or the
properties of a are just collapsed into
that scalar value okay so we can think
about relating this back to our linear
systems interpretation so we have a
matrix a it's got its eigenvalues and
eigenvectors now let's think about let's
think about the just the linear
time-invariant system without input X
dot is equal to ax we know what the
solution to this looks like we've spent
a lot of time on that now so we know the
solution just looks like e to the a T
let's assume that T 0 is equal to 0
without loss of generality times the
initial state x0 ok so suppose you had
an initial state X 0 which was some
linear combination or let's just write
it this way it's actually some scalar so
I'll call this some scalar but it's um
could be a function of time times the
eigenvector e I okay so that's one of
the eigenvectors of the matrix a well if
you have that then you can see the
following that if you so now this is a
scalar so if we plug this into this
equation here we'll get e to the a T
times the eigenvector AI ok so what's
that you remember AI is an eigen vector
of the matrix a so e to the a T times e
I let's just think about the expansion
of e to the a T so that's I plus a
t plus a squared T squared over 2
factorial times e I okay so now let's
bring e I in and multiply everything
here so we get that's equal to e i+ a te
I which is just lambda i e
I T plus and then a squared so a squared
e I if we think about that that's a
times AEI which is lambda I a e I which
is just lambda I squared e I etc so we
can simplify this by basically bringing
the ein and collapsing that matrix a so
here we get lambda I squared over 2
factorial t squared etc which is equal
to e to the lambda I T just our standard
exponential function where this is a
scalar post multiplied by e I okay so
that tells us that if we use this as an
initial condition then X of T is equal
to e to the lambda I T times e I and
then we've post multiplied by this
scalar function so that's just a scalar
which affects the magnitude of this so
in general if you have an initial
condition which is a linear combination
of the eigen vectors of the matrix a
then you can see that you can write out
the solution in terms of a linear
combination of these terms here these
are called the modes of the system
related to the individual eigen values
of the matrix a ok so generalize this
initial condition now to a linear
combination of those eigen vectors okay
so so geometric interpretation well if
lambda I is real so let's just go back
to the case where our initial condition
is is one of the eigen vectors of the
matrix a if lambda I is real then it
says that if you start on that eigen
vector E I you're always going to stay
on that eigen vector so you start out at
you know at some
that eigenvector then the system will
stay on that eigenvector evolving that
way or that way depending on the you
know the nature of the eigenvalue
whether it's positive or negative okay
so if you start out on an eigenvector
you're always going to stay on an
eigenvector you start out on a linear
combination of eigenvectors which if
we've just shown that the eigenvectors
form a basis for RN if the eigenvalues
are distinct then you'll stay in a
linear combination of those eigenvectors
and you can actually compute the terms
in terms of the eigenvectors
if and so you can generalize now to
suppose that you start out so suppose
you consider this formulation where
lambda is now complex so you're starting
out on one of the complex eigenvalues
and and the associated eigenvectors are
going to be complex so then you can show
that your solution is going to stay
within a linear in the space spanned by
a linear combination of that eigenvector
and it's complex conjugate so that's a
that's a neat extension of what we've
just written here for the case in which
lambda I is complex okay so what we've
talked about in this module we've
introduced so motivated by the fact that
we'd like to simplify the matrix a in
some way and by that I mean we're going
to perform a similarity transform on the
matrix a to make the computation of e to
the 8e easier we've started to think
about how to simplify this by or perform
the similarity transform by defining the
eigenvalues and the eigenvectors of the
matrix a and in the next module we're
going to use that for constructing this
similarity transform thanks very much
