Section 5.3

Two Applications

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```Warning, the protected names norm and trace have been redefined and unprotected
```

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An Application: Exponential of a Matrix

Exponential Form of a Matrix

The differential equation = y'(t) = a*y(t) is a prototype equation that models

problems of growth and decay. For example, if y(t) is the count of bacteria at time t and

if the rate of change of bacteria is proportional to the count at time t, then the differential

equation models this phenomenon with a being the constant of proportionality.

The solution is and c is an arbitrary constant that can be determined

using the initial condition y(0) .

The exponential function is given by the expansion

> exp(a*t):=Sum((a*t)^i/i!,i=0..infinity);

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Since this function is a solution of many important problems including the problem of

exponential growth or decay, a natural question to ask in this setting:

is there an equivalent expansion where the parameter a is replaced by a matrix A?

If so, how do we compute this expression?

This result is needed if we were to replace the scalar equation by the matrix equation

Y'(t) = A* Y(t),

where Y(t) = [ , ...., ] and solve the resulting system. The solution of this system is Y(t) = where Y(0) is the initial condition.

To have an analogous exponential function for matrices

> exp(A*t):=Sum( (A*t)^i/i!,i=0..infinity);

we need to compute the powers of the matrix A for different type of matrices.

Case of a Diagonal Matrix

Let A be the 2x2 identity matrix

> A:=diag(1,1);

Since all powers of A are equal to A

> exp(A*t):=evalm(A)*Sum( (t)^i/i!, i =0..infinity);

This is equivalent to

> exp(A*t):=evalm((A)*sum( t^i/i!, i =0..infinity));

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Let A be the 2x2 diagonal matrix

> A:=diag(2,3);

Since A is a diagonal matrix, its powers are easy to compute. In particular, for the matrix A

> A^2=evalm(A^2);
A^3=evalm(A^3);
A^4=evalm(A^4);

In general,

> A^i = matrix([[2^i,0],[0,3^i]]);

Therefore

> exp(A*t):=Sum(matrix([[2^i,0],[0,3^i]])*t^i/i!,
i=0..infinity);

That is

> exp(A*t):=Sum( matrix([[(2*t)^i/i!,0],[0,(3*t)^i/i!]]),
i =0..infinity);

or

> exp(A*t):= matrix([[Sum((2*t)^i/i!,
i=0..infinity),0],[0,Sum((3*t)^i/i!,
i=0..infinity)]]);

When the sums are evaluated , we get the matrix

> exp(A*t):= matrix([[sum((2*t)^i/i!,
i=0..infinity),0],
[0,sum((3*t)^i/i!,
i=0..infinity)]]);

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Case of a Diagonalizable Matrix

Now consider the 2x2 matrix

> A:=matrix([[3,4],[3,2]]);

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The eigenvalues of the matrix A are: 6 and -1 and the corresponding eigenvectors are

> v1:=vector([4,3]); v2:=vector([1,-1]);

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Thus the matrix A is similar to a diagonal matrix A = with

> P:=matrix([[4,1],[3,-1]]):D1:=diag(6,-1): evalm(A)=evalm(P)*evalm(D1)*evalm(inverse(P));

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Let us use the fact that A is diagonalizable to compute various powers of A:

> A^2=evalm(A^2);P*D1^2*P^(`-1`)=
multiply(P,multiply(D1^2,inverse(P)));

> A^3=evalm(A^3);P*D1^3*P^(`-1`)=
multiply(P,multiply(D1^3,inverse(P)));

> A^4=evalm(A^4);P*D1^4*P^(`-1`)=
multiply(P,multiply(D1^4,inverse(P)));

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Then

> A^n=P*D1^n*P^(`-1`);

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Furthermore, the exponential of the matrix is:

> exp(D1*t):=diag(exp(6*t),exp(-t));

> exp(A*t):=evalm(P&*exp(D1*t)&*inverse(P));

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To get the exponential of a diagonalizable matrix A proceed as follows :

1. compute the eigenvalues and eigenvectors of the given matrix

2. construct the matrix P whose columns are the eigenvectors and form the product

A = where is the diagonal matrix whose diagonal entries are the eigenvalues of A.

3. construct the exponential matrix

4. the exponential of the given matrix is .

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Initial value problem

Solve the initial value problem

> A:=matrix([[3,2],[0,4]]): Y:=matrix([[y1],[y2]]):

> evalm((d/dt)*Y) = evalm(A)*evalm(Y);

with initial data Y(0)

> Y0:=matrix([[1],[2]]);

1. Compute the eigenvalues of A (use the nostep mode)

> eigenvals(A);

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2. Compute the eigenvectors of matrix A (use the nostep mode)

> eigenvects(A);

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An eigenvector corresponding to the eigenvalue 3 is the vector [1,0] and an eigenvector corresponding to the eigenvalue 4 is the vector [2,1].

3.Construct the matrix P and the diagonal matrix .

> P:=matrix([[1,2],[0,1]]); D1:=diag(3,4);

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4. construct the exponential of :

> exp(D1*t):=diag(exp(3*t),exp(4*t));

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The exponential matrix of A is

> exp(A*t):=evalm(P&*exp(D1*t)&*inverse(P));

Therefore the solution to the initial value problem is

> Y:=evalm(exp(A*t)&*Y0);

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Exercises

6, 7, 8.