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# Matrices Eigenvalues, eigenspace, and eigenbasis.

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Matrices Eigenvalues, eigenspace, and eigenbasis.

See attached file.

For each of the matrices, find all (real) eigenvalues. Then find a basis of each eigenspace, and find an eigenbasis, if you can.
[■(7&[email protected]&9)]
[■(1&[email protected]&1)]
[■(6&[email protected]&7)]
[■(0&[email protected]&2)]
[■(1&1&[email protected]&2&[email protected]&0&3)]
[■(1&1&[email protected]&1&[email protected]&0&1)]
[■(■(0&[email protected]&1)&■(0&[email protected]&1)@■(0&[email protected]&0)&■(0&[email protected]&1))]

https://brainmass.com/math/linear-algebra/matrices-eigenvalues-eigenspace-eigenbasis-446911

## SOLUTION This solution is FREE courtesy of BrainMass!

1.
The matrix is:
(1.1)
This is an upper triangular matrix, hence the eigenvalues are the elements of the diagonal:
(1.2)
The first eigenvector:

(1.3)
The second eigenvector:

(1.4)
We have two distinct eigenvalues and two distinct eigenvectors. The algebraic multiplicity (m=1) of each of the eigenvalues equal their respective geometric multiplicity, hence these two eigenvectors form an eigenbasis.

2.
The matrix is:
(2.1)
The eigenvalues are:

(2.2)
The first eigenvector:

Therefore:
(2.3)
The second eigenvector:

(2.4)
As before we have two distinct eigenvalues and two distinct eigenvectors. The algebraic multiplicity (m=1) of each of the eigenvalues equal their respective geometric multiplicity, hence these two eigenvectors form an eigenbasis.

3.
The matrix is:
(3.1)
The eigenvalues are:

(3.2)
The first eigenvector:

(3.3)

The second eigenvector:

(3.4)

As before we have two distinct eigenvalues and two distinct eigenvectors. The algebraic multiplicity (m=1) of each of the eigenvalues equal their respective geometric multiplicity, hence these two eigenvectors form an eigenbasis.

4.

The matrix is:
(4.1)
The eigenvalues:

(4.2)
For the eigenvectors we have a single eigenvalue, hence:

We have a single eigenvector
(4.3)
We have a single eigenvalue with algebraic multiplicity , but with a single associated eigenvector (geometric multiplicity ) therefore this single vector cannot be an eigenbasis.
5.
The matrix is:
(5.1)
This is an upper triangular matrix, hence the diagonal elements are the eigenvalues:
(5.2)
The first eigenvector:

(5.3)

The second eigenvector:

(5.4)
the third eigenvector:

(5.5)
We have three distinct eigenvalues and three distinct eigenvectors. The algebraic multiplicity (m=1) of each of the eigenvalues equal their respective geometric multiplicity, hence these three eigenvectors form an eigenbasis.

6.
The matrix is:
(6.1)
This is an upper triangular matrix, hence the diagonal elements are the eigenvalues:
(6.2)
Since the matrix is in a canonical Jordan form we know it has no eigenbasis.
The single eigenvector is:

(6.3)
Obviously, we cannot span with a single vector. therefore, this eigenvector does not form an eigenbasis.

7.
The matrix is:
(7.1)
It is an upper triangular matrix hence the eigenvalues are the diagonal elements are the eigenvalues:
(7.2)
This is an obviously a defective matrix, hence its eigenvectors do not form an eigenbasis.
The first two eigenvectors:

(7.3)

The third eigenvector:

(7.4)
We have two eigenvalues (each with algebraic multiplicity ) and for the geometric multiplicity of the eigenvalue is 2 as well.
However the geometric multiplicity of is only 1, hence the matrix is defective and has no eigenbasis.

a.
The matrix is:
(8.1)
This is a lower triangular matrix hence the eigenvalues are the diagonal elements:
(8.2)

b.
The matrix is:

This is a block-diagonal matrix. Its diagonal is composed of two matrices, hence its determinant is the product of the matrices' determinants.

The eigenvalues are:

(8.3)

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