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 V in whose matrix A can be expressed as diagonal form, that we will call D.
This change of basis is operated as follows:

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

This is interesant because, rather than geometric implications and consequences of eigenvectors and eigenvalues for many applications, but D is much simpler and easier to operate than the original A.
For instance if we want to calculate A5 we can do it as follows:

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

As we will see, not all matrices are diagonalizable that means: not for all matrix A it is possible to find a basis in wich A
can be expressed as a diagonal form.

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

In both cases, Jordan and diagonal forms we need to calculate eigenvalues and eigenvectors of A.

# 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 keen in invariant sub-spaces by this application (the eigenspaces), this is 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 it determinant value is different than zero, that means:

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

When solving this determinant, it shows up up a polynomial in the λ variable called

Multiplicity of each eigenvalue as root of characeristic polynomial is called

With each eigenvalue j there are some

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

The dimension of eigenspace \( E_{j} \) is called

Therefore, the calculation of the eigenvalues of a matrix A is as easy or as difficult as calculate the roots of a polynomial. You can see the following example

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

We are keen in invariant sub-spaces by this application (the eigenspaces), this is 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 it determinant value is different than zero, that means:

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

When solving this determinant, it shows up up 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 by λj. and they are called**eigenvalues**of A.Multiplicity of each eigenvalue as root of characeristic polynomial is called

**algebraic multiplicity**of eigenvalue λj.With each eigenvalue j there are some

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

The dimension of eigenspace \( E_{j} \) is called

**geometric multiplicity**of eigenvalue λj.Therefore, the calculation of the eigenvalues of a matrix A is as easy or as difficult as calculate the roots of a polynomial. You can see the following example

### Eigenvalues and eigenvectors example

Let be the matrix:

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

Then, we begin to calculate the characteristic polynomial, as we said

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

The eigenvalues are:

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

Then calculate the eigenvectors foreach 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, made up by the egenvectors, in wich matrix A is expressed as diagonal matrix

The basis change matrix is made up 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, both 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), if 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 begin to calculate the characteristic polynomial, as we said

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

The eigenvalues are:

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

Then calculate the eigenvectors foreach 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, made up by the egenvectors, in wich matrix A is expressed as diagonal matrix

The basis change matrix is made up 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, both 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), if 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 A is a diagonalizable matrix with repeated eigenvalue value on the diagonal.**2)**The eigenspace corresponding to the eigenvalue has dimension 1: In this case A matrix is not diagonalizable.3x3 Jordan form example here

Other one here