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Given the following matrices A, B, and C compute the eigenvalue and eigenvector for each matrix.
A =(-2, 6, 6, 3)
B= (4, 2, -1, 6)
C= (0, -1, 1, 2)

https://brainmass.com/math/linear-algebra/find-eigenvalues-vectors-421267

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To find eigenvalues:
Calculate the characteristic polynomial
(1.1)
Find the roots of the equation
(1.2)
These are the eigenvalues
To find the eigenvectors, for each eigenvalue solve the system of equations
(1.3)
Note that for any constant and eigenvector v, the vector is an eigenvector as well with the same eigenvalue:

Case 1:

Then:

(1.4)
Equating it to zero we find two roots:

(1.5)
To find the eigenvectors we define and we set the first equation:
(1.6)
Solving the system:

Note that we must have (otherwise we get the trivial solution)
(1.7)
This means that the two equations are linearly dependent and the relations between and are:
(1.8)
So if we set we get

The eigenvector is:
(1.9)
The second eigenvector:
(1.10)

It is easy to see that again we get a linear dependent system and we obtain:
(1.11)
The eigenvector is:
(1.12)
To summarize, the eigenvalues and their associated eigenvectors of the matrix are:
(1.13)

Case 2:

Then:

(2.1)
Equating it to zero we find two roots:

(2.2)
Where
Solving the system:

Note that we must have (otherwise we get the trivial solution)
(2.3)
This means that the two equations are linearly dependent and the relations between and are:
(2.4)
So if we set we get
The eigenvector is:
(2.5)
The second eigenvector:

It is easy to see that again we get a linear dependent system and we obtain:
(2.6)

The eigenvector is:
(2.7)
To summarize, the eigenvalues and their associated eigenvectors of the matrix are:
(2.8)

Case 3:

Then:

(3.1)
Equating it to zero we find two roots:
(3.2)
We have a single eigenvector with multiplicity of 2
Solving the system:

This is obviously a linearly dependent system with:
(3.3)
The eigenvector is:
(3.4)
Now, is there another eigenvector?
In our case:
(3.5)
It is easy to see that
(3.6)
And since the dimension of A is , the degeneracy is:
(3.7)
If we write the vector as a linear combination of the standard basis vectors:
(3.8)
We see that:
(3.9)
The geometric multiplicity (the number of basis vectors we need to use to express the eigenvector as a linear combination) is 2.
Since the geometric multiplicity equals the algebraic multiplicity of the eigenvalue, this is the only eigenvector of the matrix.

If we want to get the Jordan form of the matrix (it cannot be diagonalized) we find a second generalized eigenvector :
(3.10)
In our case this gives:

The generalized eigenvector is:
(3.11)
It is NOT an eigenvector of A, but if we set:
(3.12)
It is easy to see that
(3.13)
And we get the Jordan canonical form of A:
(3.14)

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