Preface

This is a tutorial made solely for the purpose of education and it was designed for students taking Applied Math 0340. It is primarily for students who have some experience using Mathematica. If you have never used Mathematica before and would like to learn more of the basics for this computer algebra system, it is strongly recommended looking at the APMA 0330 tutorial. As a friendly reminder, don'r forget to clear variables in use and/or the kernel.

Finally, the commands in this tutorial are all written in bold black font, while Mathematica output is in regular fonts. This means that you can copy and paste all comamnds into Mathematica, change the parameters and run them. You, as the user, are free to use the scripts to your needs for learning how to use the Mathematica program, and have the right to distribute this tutorial and refer to this tutorial as long as this tutorial is accredited appropriately.

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Eigenvalues and Eigenvectors

If A is a square \( n \times n \) matrix and v is an \( n \times 1 \) column vector, then the product \( {\bf A}\,{\bf v} \) is defined and is another \( n \times 1 \) column vector. It is important in many applications to determine whether there exist nonzero column vectors v such that the product vector \( {\bf A}\,{\bf v} \) is a constant multiple (which we denote as λ) of v.

If a homogeneous equation

\[ {\bf A} \, {\bf v} = \lambda\,{\bf v} \]
has a nontrivial solution v (meaning it is not identically zero), then the vector v is called the eigenvector, corresponding to eigenvalue \( \lambda .\) This may happen only when the determinant of the system \( \lambda \,{\bf v} - {\bf A}\, {\bf v} = {\bf 0} \) , namely, \( \det \left( \lambda\, {\bf I} - {\bf A} \right) =0 ,\) where I is the identity matrix. This determinant is called the characteristic polynomial and we denote it by \( \chi (\lambda ) = \det \left( \lambda\, {\bf I} - {\bf A} \right) . \) Therefore, eigenvalues are the nulls of the characteristic polynomial and they are the roots of the equation \( \chi (\lambda ) = 0. \) The characteristic polynomial is always a polynomial of degree n, where n is the dimension of the square matrix A. It can be expressed through eigenvalues:
\[ \chi (\lambda ) = \det \left( \lambda\, {\bf I} - {\bf A} \right) = \lambda^n - \left( \mbox{tr} {\bf A} \right) \lambda^{n-1} + \cdots + (-1)^n \,\det {\bf A} , \]
where \( \mbox{tr} {\bf A} = a_{11} + a_{22} + \cdots + a_{nn} = \lambda_1 + \lambda_2 + \cdots + \lambda_n \) is the trace of the matrix A, that is, the sum of its diagonal elements, which is equal to the sum of all eigenvalues (including their multiplicities).

The set of all eigenvalues is called the spectrum of the matrix A.

 

 

Eigensystem[B] gives a list {values, vectors} of the eigenvalues and eigenvectors of the square matrix B.

B := {{0, 1}, {-1, 0}}
Out[1]= {{0, 1}, {-1, 0}}
Eigenvalues[B]
Out[2]= {I,-I}
Eigenvectors[B]
Out[3]= {{-I,1},{I,1}}
Eigensystem[B]
Out[4]= {{I, -I}, {{-I, 1}, {I, 1}}}

Consider a defective matrix:
A = {{1, 1, 0}, {0, 0, 1}, {0, 0, 1}}
Eigenvalues[A]
Out[2]= {1, 1, 0}

Therefore, we know that the matrix A has one double eigenvalue \( \lambda = 1, \) and one simple eigenvalue \( \lambda = 0 \) (which indicates that matrix A is singular). Next, we find eigenvectors

Eigenvectors[A]
Out[3]= {{1, 0, 0}, {0, 0, 0}, {-1, 1, 0}}

So Mathematica provides us only one eigenvector \( \xi = \left[ 1,0,0 \right] \) corresponding to the eigenvalue \( \lambda =1 \) (therefore, it is defective) and one eigenvector v = <-1,1,0> corresponding eigenvalue \( \lambda = 0. \) To check this, we introduce the matrix B1:

B1 = IdentityMatrix[3] - A
Eigenvalues[B1]
Out[5]= {1, 0, 0}

which means that B1 has one simple eigenvalue \( \lambda = 1 \) and one double eigenvalue \( \lambda =0. \) Then we check that \( \xi \) is the eigenvector of the matrix A:

B1.{1, 0, 0}
Out[6]= {0, 0, 0}

Then we check that v is the eigenvector corresponding to \( \lambda = 0: \)

A.{-1, 1, 0}
Out[7]= {0, 0, 0}

To find the generalized eigenvector corresponding to \( \lambda = 1, \) we use the following Mathematica command

LinearSolve[B1, {1, 0, 0}]
Out[8]= {0, -1, -1}

This gives us another generalized eigenvector \( \xi_2 = \left[ 0,-1,-1 \right] \) corresponding to the eigenvalue \( \lambda = 1 \) (which you can multiply by any constant). To check it, we calculate:

B1.B1.{0, 1, 1}
Out[9]= {0, 0, 0}

but the first power of B1 does not annihilate it:

B1.{0, 1, 1}
out[10]= {-1, 0, 0}

 

The characteristic polynomial can be found either with Mathematica's command
CharacteristicPolynomial or directly.

A := {{0, 1}, {-1, 0}}
CharacteristicPolynomial[A, lambda]
Out[2]= 1 + lambda^2
sys[lambda_] = lambda*IdentityMatrix[2] - A
Out[3]= {{lambda, -1}, {1, lambda}}
p[lambda_] = Det[sys[lambda]]                (* characteristic polynomial *)
Out[4]= 1 + lambda^2
Solve[p[lambda] == 0] // Flatten
Out[5]= {lambda -> -I, lambda -> I}       (* I is the imaginary unit *)

This can be obtained manually as follows:

A = {{1, 2}, {2, 4}}
Out[1]= {{1, 2}, {2, 4}}
sys[lambda_] = lambda*IdentityMatrix[2]-A
Out[2]= {{-1 + lambda, -2}, {-2, -4 + lambda}}
p[lambda_] =Det[sys[lambda]]
Out[3]= -5 lambda + lambda^2
Find the roots of the characteristic equation (eigenvalues of the matrix A):

Solve[p[lambda]==0]
Out[4]= {{lambda -> 0}, {lambda -> 5}}
Capture the eigenvalues:
{lambda1,lambda2} = x/.Solve[p[x]==0]
Out[5]= {0, 5}
To show the basis of the null space of the matrix A:
v1 = NullSpace[sys[lambda1]][[1]]
Out[6]= {-2, 1}

Two matrices A and B are called similar if there exists a nonsingular matrix S such that \( {\bf A} = {\bf S}\,{\bf B}\,{\bf S}^{-1} .\) Similar matrices always has the same eigenvalues, but their eigenvectors could be different. Let us consider an example of two matrices, one of them is a diagonal one, and another is similar to it:

A = {{1, 0, 0}, {0, 2, 0}, {0, 0, 0.5}}
S = {{2, -1, 3}, {1, 3, -3}, {-5, -4, 1}}
B = Inverse[S].A.S
Out[3]= {{-25., -45., 36.}, {39.5, 70., -55.5}, {30.5, 53., -41.5}}
Therefore, the matrix B is similar to the diagonal matrix A. We call such matrix B diagonalizable.
Eigenvalues[B]
Out[4]= {2., 1., 0.5}
Eigenvectors[B]
Out= {{0.457144, -0.706496, -0.540262}, {-0.451129, 0.701757, 0.55138}, {-0.46569, 0.698535, 0.543305}}
Eigenvectors[A]
Out= {{0., 1., 0.}, {1., 0., 0.}, {0., 0., 1.}}

Therefore, these two similar matrices share the same eigenvalues, but they have distinct eigenvectors.

 

 

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