Welcome to an introduction to matrix exponentials,
as well as how to determine matrix exponentials
for two simple cases.
Matrix exponentials provide another way
of finding a fundamental matrix solution
for a system of differential equations
with constant coefficients in the form
of x prime equals P times x.
If this would be just one equation,
when P is a number or a one by one matrix,
then the solution would be x equals e to the power of Pt.
This doesn't make sense if P is a larger matrix,
but essentially the same computation that led
to the above works for matrices
when we define e to the power of Pt properly.
Let us first write down the Taylor series
for e to the power of at for some number a.
You may recall this series from calc II.
Recall that k factorial is equal
to one times two times three,
all the way out to times k,
and zero factorial is equal to one.
If we differentiate the series term by term
we get the result below.
Notice if we factor out a we do have
a times e to the power of at.
Similarly, for an n by n matrix
we define the matrix A exponential
as e to the power of A equals I plus A plus 1/2A squared
plus 1/6A cubed, and so on.
If we compare this definition
to the definition above for e to the power of at
notice one is replaced with the identity matrix I,
the number a is replaced with matrix A, and there is no t.
Let us not worry about convergence.
The series really does always converge.
We usually write P times t as t times P by convention
when P is a matrix.
With this small change and by the exact same calculation
as above we have that the derivative of e to the power of tP
equals P times e to the power of tP.
Now, P and hence e to the power of tP is an n by n matrix.
What we are looking for is a vector.
In a one by one case we would at this point multiply
by an arbitrary constant to get the general solution.
In this case, we multiply by a column vector c.
Looking at the theorem below,
if P is an n by n matrix then the general solution
to x prime equals P times x is x equals
e to the power of tP times c
where c is an arbitrary constant vector.
In fact, x of zero equals the constant vector c.
And again, here, e to the power of tP
is a matrix exponential.
And of course, we can check this by differentiating,
which is shown below.
So, again, the important part here is
e to the power of tP is a fundamental matrix solution
of the homogeneous system.
So, if we compute the matrix exponential
we have another method for solving
constant coefficient homogeneous systems.
It also makes it easy to solve for the initial conditions.
To solve x prime equals A times x
with the initial condition x of zero equals b,
the solution is x equals e to the power
of tA times vector B.
This follows because e to the power of zero A
is equal to the identity matrix,
and therefore x of zero is equal
to the identity matrix times vector b,
which does return vector b.
Next, we have an important property that we use
when determining matrix exponentials.
If A and B are matrices,
if A times B equals B times A, meaning the matrices commute,
then e to the power of A plus B equals
e to the power of A times e to the power of B.
Otherwise, e to the power of A plus B doesn't equal
e to the power of A times e to the power of B in general.
And again, this is important when determining
matrix exponentials, which we'll see shortly.
And now let's take a look at two simple cases
on determining matrix exponentials.
In some instances it may work
to just plug into the series definition.
Suppose the matrix is diagonal.
For example, let's consider matrix D,
which is the two by two matrix with entries a zero, zero b.
We know that D to the power of k has entries
a the power of k zero, zero and b to the power of k.
And now if we apply the definition for e to the power of D
we have e to the power of D equals I plus D
plus 1/2D squared plus 1/6D cubed, and so on.
Again, this is using the definition
here at the bottom of the screen.
And now we replace I with the two by two identity matrix,
and we replace D and its powers with the appropriate matrix.
From here if we take a look at the entries
in row one column one,
notice the sum is the Taylor series for e to the power of a.
Similarly, if we look at the entries in row two column two
we have the series for e to the power of b,
then their two entries remain zero.
So, now we have e to the power of D
is equal to the two by two identity matrix
with entries e to the a zero, zero and e to the b.
From here it follows that the matrix exponential,
e to the power of tD, has entries e to the power of at zero,
zero and e to the power of bt.
Notice how determining a matrix exponential
for a diagonal matrix is very straightforward.
Using this rationale for two by two matrices,
e to the power of I has entries e zero, zero and e.
And e to the power of aI, where a is some constant,
is equal to the two by two matrix with entries
e to the power of a zero, zero and e to the power of a.
And of course, this property does apply
for larger square diagonal matrices.
And now let's look at the second simple case.
Now that we know how to easily determine
a matrix exponential for a diagonal matrix,
this makes exponentials of certain
other matrices easy to compute.
For example, when we have a two by two matrix
that has a repeated eigenvalue with one defect
we can easily determine the matrix exponential.
For this case let's look at an example.
Let's consider matrix A
that has entries five four, negative one one.
This matrix has an eigenvalue of three
with an algebraic multiplicity of two with one defect.
Because of this we can easily determine
the matrix exponential.
First, the given matrix A is equal to lambda I plus B.
What we really want here is matrix B.
Notice matrix B is equal to A minus lambda I,
which in our case means we have matrix A
minus three times the two by two identity matrix,
which gives us matrix B which is the two by two matrix
with entries two four, negative one negative two.
What's special about this matrix B is that B squared
is equal to the zero matrix,
and therefore B to the power of k is equal to zero
for all k greater than zero, which means if we apply
the matrix exponential definition below
for e to the power of B,
e to the power of B is equal to I plus B
because all the terms past I plus B are equal to zero
because we have B squared, B cubed, and so on,
all of which are zero.
So, from here if we want to calculate e to the power of tA
this would be equal to e to the power of t times A
where A can be written as three I plus B,
which indicates e to the power of tA
is equal to e to the power of three tI plus tB.
And since the matrices three tI and tB commute
we can write e to the power of three tI plus tB
as e to the power of three tI times e to the power of tB.
And e to the power of three tI is a diagonal matrix,
which gives us the two by two matrix
with entries e to the power of three t zero,
zero and e to the power of three t.
And because e to the power of B is equal to I plus B,
e to the power of tB is equal to the sum of I and tB.
Next, we determine I plus tB,
which gives us a two by two matrix shown here
with entries one plus two t four t,
negative t and one minus two t.
And then finally, we multiply the two matrices,
which gives us the two by two matrix with entries
the quantity one plus two t times e to the three t
four te to the three t, negative te to the three t,
and finally, the quantity one minus two t
times e to the three t.
This is the matrix exponential for matrix A,
which means this is a fundamental matrix solution
to x prime equals Ax.
And therefore, if we want,
the general solution would be x equals e to the power of tA
times the constant vector c,
which is shown here on the right.
Matrices B such that B to the power of k equals zero
for some k are called nilpotent.
Computation of the matrix exponential for nilpotent matrices
is easy by just writing down the first k terms
of the Taylor series.
Before we go let's summarize our findings.
First, we care about matrix exponentials
because it provides another way to solve the system,
the system of differential equations,
represented by the equation x prime equals Ax
where the general solution is x equals e to the power of tA
times the constant vector c.
It also allows us to easily determine
a particular solution given our initial condition
x of zero equals the constant vector b.
That particular solution is just x equals
e to the power of tA times the constant vector b.
And then we also found an easy way to determine
matrix exponentials for diagonal matrices.
If we're given matrix A
where we have a, b, c along the main diagonal
the matrix exponential e to the power of tA
has the entries of e to the power of at,
e to the power of bt, and e to the power of ct
along the main diagonal, all other entries remain zero.
And then secondly, if we have a two by two matrix
that has an eigenvalue of algebraic multiplicity of two
with defect one we determine matrix B
where B is equal to A minus lambda I.
So, for the example below
matrix A has a repeated eigenvalue of five with one defect,
and matrix B is the matrix two one,
negative four negative two.
And therefore, e to the power of tA
is equal to e to the power of t times
the sum of five A and B,
which because the two matrices commute
we can write it as e to the power of five tI
times e to the power of tB.
E to the power of tB is equal to I plus tB,
which I've calculated below,
and therefore the matrix exponential
is equal to the product of these two two by two matrices,
which is shown here on the right.
I hope you found this helpful.