# Matrices Eigenvalues, eigenspace, and eigenbasis.

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.

[â– (7&[email protected]&9)]

[â– (1&[email protected]&1)]

[â– (6&[email protected]&7)]

[â– (0&[email protected]&2)]

[â– (1&1&[email protected]&2&[email protected]&0&3)]

[â– (1&1&[email protected]&1&[email protected]&0&1)]

[â– (â– (0&[email protected]&1)&â– (0&[email protected]&1)@â– (0&[email protected]&0)&â– (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.

â€ƒ

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.

â€ƒ

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)

â€ƒ

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.

â€ƒ

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)

â€ƒ

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.

â€ƒ

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)

Â© BrainMass Inc. brainmass.com December 24, 2021, 10:08 pm ad1c9bdddf>https://brainmass.com/math/linear-algebra/matrices-eigenvalues-eigenspace-eigenbasis-446911