Given a square matrix A nxn in the field K (real, complex, ..., any other field ...).

\( \begin{pmatrix}
a_{11} & a_{12} ... & a_{1n}\\
a_{21} & a_{22} & a_{2n} \\
... & & ... \\
a_{n1} & a_{21}... & a_{nn}
\end{pmatrix} \)

In many cases, it is possible to find a basis of vectors of V in which the matrix A is expressed as diagonal form, that we will call D, the change of basis is operated as follows

\( A=PDP^{-1} \)

This is interesant because (apart from geometric knowledge and know the consequences of eigenvectors and
eigen values in many applications) D is much simpler and easy to operate, for example if we want calculate A5 it is valid to make the operation

\( A^{5}=PD^{5}P^{-1} \)

As we will see, not all matrix are diagonalizable (that is, not for all matrix A it is possible to find a basis in wich A has a diagonal form).

However it is possible to obtain a change of basis in which matrix A takes a
simpler form called Jordan form.

In both cases, to calculate the Jordan form or to calculate diagonal form we need to calculate eigenvalues and eigenvectors.

# Eigenvectors and Eigenvalues

### Basic Concepts and principles

### Extended Therory

Consider the linear map (also called endomorphism) given by

\( \begin{matrix} F:V\rightarrow V & \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \in V \rightarrow F(x) = Ax\in V \end{matrix} \)

We are interested in invariant sub-spaces by this application (eigenspaces), ie the subspace spanned by those vectors v such that

\( Av = \lambda v \Leftrightarrow (A - \lambda I)v = 0 \)

So, the eigenspace is the kernel of \((A - \lambda I)\)

This last equation is in fact a linear equations system wich have solution if and only if the determinant is zero, ie

\( det (A - \lambda I) \neq 0 \)

when solving this determinant, appears a polynomial in the λ variable called

λj

and they are called

To each eigenvalue j will correspond some

\( E(\lambda_{j}) = {x \in V : Ax= \lambda_{j}v} \)

Therefore, the calculation of the eigenvalues of a matrix A is as easy (or difficult) as calculate the roots of a polynomial,

see the following example

\( \begin{matrix} F:V\rightarrow V & \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \in V \rightarrow F(x) = Ax\in V \end{matrix} \)

We are interested in invariant sub-spaces by this application (eigenspaces), ie the subspace spanned by those vectors v such that

\( Av = \lambda v \Leftrightarrow (A - \lambda I)v = 0 \)

So, the eigenspace is the kernel of \((A - \lambda I)\)

This last equation is in fact a linear equations system wich have solution if and only if the determinant is zero, ie

\( det (A - \lambda I) \neq 0 \)

when solving this determinant, appears a polynomial in the λ variable called

**characteristic polynomial of application A**or simply**characteristic polynomial**whose roots are in the field K we denote byλj

and they are called

**eigenvalues**of A.To each eigenvalue j will correspond some

**eigenvectors**vi. The space spanned by the eigenvectors is called**eigenspace**associated to each eigenvalue λj and we denote it by E(λj). Ie the eigenspace associated to eigenvalue λj is\( E(\lambda_{j}) = {x \in V : Ax= \lambda_{j}v} \)

Therefore, the calculation of the eigenvalues of a matrix A is as easy (or difficult) as calculate the roots of a polynomial,

see the following example

### Eigenvalues and eigenvectors worked example

Lets the matrix

\(A =\begin{pmatrix} 1 & -4\\ -1 & 1 \end{pmatrix}\)

Then, we start calculating the characteristic polynomial

\( det (A - \lambda I) = \begin{vmatrix} 1-\lambda & -4\\ -1 & 1- \lambda \end{vmatrix} = (1-\lambda)^2 -4 = (\lambda-3)(\lambda+1) \)

The eigenvalues are

\( \lambda_{1}=3, \lambda_{2}=-1 \)

We calculate the eigenvectors foreach one eigenvalue, we begin with corresponding to \( \lambda_{2} = 3 \) eigenvalue

\( \left.\begin{matrix} -2x-4y=0 \\ -x-2y=0 \end{matrix}\right\} \)

whose solution is the space spanned by the eigenvector

\( v_{1}=\begin{pmatrix} -2\\ 1 \end{pmatrix} \)

Now we operate with the eigenvalue \( \lambda_{2} = -1 \)

2x-4y=0, whose solution is

\( v_{2}=\begin{pmatrix} 2\\ 1 \end{pmatrix} \)

So, there exists a basis, those formed by the egenvectors, in wich matrix A is expressed as diagonal matrix

The change of basis matrix is those formed by eigenvalues in its colummns, as follows

\( P=\begin{pmatrix} -2 & 1\\ 1& 1 \end{pmatrix} \)

A diagonal is

\( P=\begin{pmatrix} 3 & 0\\ 0 & -1 \end{pmatrix} \)

\( A=PDP^{-1} \)

In this simple example, the eigenvalues ?of characteristic polynomial are distinct and subspace for each one of them has dimension 1. For this reason the matrix is ??diagonalizable

We will see the details in the next section of Linear Algebra, but we anticipate that if the characteristic polynomial would have a single double root (with multiplicity two) is, we have a single eigenvalue, then the situation might have been very different and would there are two possibilities:

3x3 Jordan form example here

Other one here

\(A =\begin{pmatrix} 1 & -4\\ -1 & 1 \end{pmatrix}\)

Then, we start calculating the characteristic polynomial

\( det (A - \lambda I) = \begin{vmatrix} 1-\lambda & -4\\ -1 & 1- \lambda \end{vmatrix} = (1-\lambda)^2 -4 = (\lambda-3)(\lambda+1) \)

The eigenvalues are

\( \lambda_{1}=3, \lambda_{2}=-1 \)

We calculate the eigenvectors foreach one eigenvalue, we begin with corresponding to \( \lambda_{2} = 3 \) eigenvalue

\( \left.\begin{matrix} -2x-4y=0 \\ -x-2y=0 \end{matrix}\right\} \)

whose solution is the space spanned by the eigenvector

\( v_{1}=\begin{pmatrix} -2\\ 1 \end{pmatrix} \)

Now we operate with the eigenvalue \( \lambda_{2} = -1 \)

2x-4y=0, whose solution is

\( v_{2}=\begin{pmatrix} 2\\ 1 \end{pmatrix} \)

So, there exists a basis, those formed by the egenvectors, in wich matrix A is expressed as diagonal matrix

The change of basis matrix is those formed by eigenvalues in its colummns, as follows

\( P=\begin{pmatrix} -2 & 1\\ 1& 1 \end{pmatrix} \)

A diagonal is

\( P=\begin{pmatrix} 3 & 0\\ 0 & -1 \end{pmatrix} \)

\( A=PDP^{-1} \)

In this simple example, the eigenvalues ?of characteristic polynomial are distinct and subspace for each one of them has dimension 1. For this reason the matrix is ??diagonalizable

We will see the details in the next section of Linear Algebra, but we anticipate that if the characteristic polynomial would have a single double root (with multiplicity two) is, we have a single eigenvalue, then the situation might have been very different and would there are two possibilities:

**1)**The eigenspace corresponding to the eigenvalue has dimension 2: In this case it would be diagonalizable matrix with repeated eigenvector value on the diagonal.**2)**The eigenspace corresponding to the eigenvalue has dimension 1: In this case the matrix is not diagonalizable.3x3 Jordan form example here

Other one here

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