Welcome to a lesson
on how to determine matrix exponentials
for the general case where we have an n by n matrix
with n linearly independent eigenvectors.
In the previous lesson, we found matrix exponentials
for simple cases such as for a diagonal matrix
and a two by two matrix A,
such that A equals lambda I plus B,
where B squared equals zero.
Now we look at the general case
to determine a matrix exponential.
To begin, we need the following interesting result
about matrix exponentials.
For two square matrices, A and B with B invertible,
we have e to the power of BAB inverse
equals B, e to the A, B inverse.
This can be seen by writing down the Taylor series first.
Let's begin with the square of BAB inverse.
This is equal to two factors of BAB inverse.
Notice in the middle here we have B inverse times B,
which is equal to the identity matrix.
And then A times I times A is equal to A squared
giving us BA squared B inverse.
And by the same reasoning, the K power of BAB inverse
equals B, A to the K, B inverse.
Now we write down the Taylor series for
e to the power of BAB inverse using the definition below.
E to the power of BAB inverse equals
I plus one factor of BAB inverse
plus 1/2 times a square of BAB inverse
plus 1/6 times the cube of BAB inverse, and so on.
From here we replace I with B times B inverse.
Notice all the terms have a factor of B
as well as a factor of B inverse.
For the next step we factor out B on the left
and B inverse on the right.
Notice in the middle here,
we're left with e to the power of A.
So now we have e to the power of BAB inverse
equals B, e to the A, B inverse.
Now that we've proven this, we'll come back to this later.
Now, given a square matrix A,
we can usually write A equals EDE inverse
where D is a diagonal matrix and E is an invertible matrix.
This procedure is called diagonalization.
If we can do that, the computation of the exponential
becomes straightforward as E to the D
is just taking the exponential of the entries
on the diagonal.
And if we include a factor of T,
we can compute the exponential.
We have E to the power of TA
equals E times E to the TD times E inverse.
Now it's important to remember to diagonalize matrix A
we need n linearly independent eigenvectors of A.
Otherwise this method of computing,
the exponential does not work.
From here, we let lambda one through lambda n
be the eigenvalues of matrix A,
and the vectors V1 through VN
be the corresponding eigenvectors.
Then we form matrix E using the eigenvectors as the columns,
and we also make a diagonal matrix D with the eigenvalues
on the diagonal as shown here.
And now I compute A times E.
So on the right we have A times E.
Multiplying by A we have the next line.
And then we replace A times V1 with lambda one times V1.
We replace A times V2 with lambda two times V2
all the way out to replacing A times the eigenvector VN
with lambda N times VN.
Next we factor out the eigenvalues on the right,
which gives us matrix D.
And notice on the left we have matrix E.
So now we have A times E equals E times D.
Since the columns of E are linearly independent,
as they're linearly independent eigenvectors of A,
we know A is invertible.
And since A times E equals E times D,
we multiply on the right by E inverse.
Simplifying, which gives us A equals EDE inverse.
From here if we exponentiate both sides of the equation
with the base of E,
we have E to the A equals E to the EDE inverse.
Here's where we use the property
that we proved at the beginning.
E to the power of ED inverse
is equal to E times E to the D times E inverse,
which gives us E of the A equals E times E to the D
times the inverse.
Multiplying the matrix by T,
we have the formula we use to determine
E to the power of TA.
Again, E to the power of TA is equal to
E times E to the power of TD,
which is the matrix shown here on the right,
times E inverse.
This formula gives the formula for computing
a fundamental matrix solution, E to the TA,
for the system X prime equals A times X
in the case where we have n linearly
independent eigenvectors.
This computation still works when the eigenvalues
and eigenvectors are complex,
but then we have to compute with complex numbers.
We won't deal with that case now.
Let's look at an example
We're asked to compute fundamental matrix solution
using the matrix exponential for the system.
Then compute the particular solution
for the initial conditions,
X of 0 equals 4, and Y of 0 equals 2.
Let's set this up on the next slide.
We let a be the coefficient matrix with entries 1, 2, 2, 1,
and we first compute the eigenvalues,
which are 4 and -1,
and the corresponding eigenvectors are 1, 1, and 1, -1.
You may want to pause the video and verify this.
The next step is to setup A equals EDE inverse.
So again, we have the given matrix A equals
matrix E is the matrix formed using the
eigenvectors as the columns.
So we have the eigenvector V1 and the eigenvector V2,
and then we have times matrix D
where matrix D is the diagonal matrix
where we have the eigenvalues along the main diagonal,
and then we have times E inverse.
Now that we have this, we know the matrix exponential,
E to the power of TA is equal to E times
E to the power of TD times E inverse.
So again, the first matrix is matrix E,
and then E to the power of TD is the matrix with entries
E to the power of 3T, 0, 0 and E to the negative T,
and then we have times E inverse.
The next step is to determine E inverse,
which is shown below.
E inverse is equal to negative 1/2
times the two by two matrix with entries, -1, -1, -1, 1.
So the next step we pull the negative 1/2 to the front
and then multiply the first two, two by two matrices,
which results in the two by two matrix with entries
E to the 3T, E to the negative T, E to the 3T,
and negative E to the negative T.
Now you multiply by the next matrix
and then perform together multiplication
to get the matrix exponential shown here on the far right.
From here, if we were looking for just the general solution
we would have the vector function XY equals
E to the power of TA times the constant vector C,
which is shown here on the right.
But again, we're looking for a particular solution
because we are given the initial conditions,
X of 0 equals 4, and Y of 0 equals 2.
So if we did use a general solution
and substituted zero for T,
notice how we would get C sub 1 equals 4
and C sub 2 equals 2,
which actually is the original vector B
given by the initial conditions.
And therefore it's much easier just to use the property
that E to the power of 0A equals the identity matrix.
And when you find the particular solution
is just E the power of TA times the vector B,
which is the vector 4, 2 given by the initial conditions.
So we have our particular solution is
the vector function X, Y equals again,
the matrix exponential times the constant vector 4, 2,
which if we multiply out and simplify,
is shown here on the right where X equals 3E to the 3T
plus E to the negative T,
and y equals 3E to the 3T minus E to the negative T.
I hope you found this helpful.