# Linear Algebra : Eigenvalues and Eigenvectors

## Example Questions

### Example Question #41 : Eigenvalues And Eigenvectors

Which of the following is an eigenvalue of  ?

Explanation:

Find the characteristic equation of  by obtaining the determinant of .

The determinant of this  matrix can be found by adding the upper-left to lower-right products:

Adding the upper-right to lower-left products:

And subtracting the latter from the former:

The characteristic equation is

Factor the polynomial completely

The solution set of this equation is comprised of the zeroes of both polynomial factors:

Only  is a choice.

### Example Question #42 : Eigenvalues And Eigenvectors

.

Is  an eigenvalue of , and if so, what is the dimension of its eigenspace?

No.

Yes; the dimension is 3.

Yes; the dimension is 2.

Yes; the dimension is 1.

No.

Explanation:

Assume that  is an eigenvalue of . Then, if  is one of its eigenvectors, it follows that

, or, equivalently,

,

where  are the  identity and zero matrices, respectively.

, so

Changing to reduced row echelon form:

We do not need to go further to see that this matrix will not have a row of zeroes. This means the rank of the matrix is 3, and the nullity is 0. If this happens, the tested value, in this case , is not an eigenvalue.

### Example Question #43 : Eigenvalues And Eigenvectors

matrix has exactly two distinct eigenvalues: 1 and . Which of the following cannot be its characteristic equation?

Any of the four choices are possible characteristic equations.

Explanation:

matrix has four not necessarily distinct eigenvalues, , which are the solutions of its characteristic polynomial equation

If there are only two distinct eigenvalues, 1 and , then one of the following three situations happens:

1 has multiplicity 3 and  has multiplicity 1, in which case ,

and the characteristic equation is

or

1 has multiplicity 2 and  has multiplicity 2, in which case ,

and the characteristic equation is

or

1 has multiplicity 1 and  has multiplicity 3, in which case

and the characteristic equation is

or

Of the four equations given as choices, only  cannot be a characteristic equation of the matrix.

### Example Question #41 : Eigenvalues And Eigenvectors

Which of the following statements follows from the Cayley-Hamilton Theorem?

None of the other choices gives a correct response.

and  have the same eigenvalues.

has one eigenvalue of multiplicity 2.

has two distinct real eigenvalues.

Every vector in  is an eigenvector of .

None of the other choices gives a correct response.

Explanation:

By the Cayley-Hamilton Theorem, a matrix is a solution of its own characteristic polynomial equation. None of the choices addresses the characteristic equation of .

### Example Question #45 : Eigenvalues And Eigenvectors

.

Is 2 an eigenvalue of , and if so, what is the dimension of its eigenspace?

Yes; the dimension is 3.

Yes; the dimension is 1.

Yes; the dimension is 2.

No.

Yes; the dimension is 1.

Explanation:

Assume that 2 is an eigenvalue of . Then, if  is one of its eigenvectors, it follows that

, or, equivalently,

,

where  are the  identity and zero matrices, respectively.

, so

Changing to reduced row echelon form:

The matrix is in reduced row echelon form. There is one column which does not contain leading 1s, so 2 is indeed an eigenvalue. The eigenspace is the set of all vectors  such that  - that is, the set of vectors . The eigenspace has dimension 1.

### Example Question #41 : Eigenvalues And Eigenvectors

Evaluate  so that the sum of the eigenvalues of  is 10.

The sum of the eigenvalues of  is 10 regardless of the value of .

Explanation:

The sum of the eigenvalues of a square matrix is equal to its trace, the sum of its diagonal elements. Examine these elements, which are in red below:

Set the trace equal to 10 and solve for :

### Example Question #41 : Eigenvalues And Eigenvectors

True or false: 1 is an eigenvalue of .

True

False

True

Explanation:

Examine the columns of :

The entries of each column add up to 1:

Column 1:

Column 2:

Column 3:

It follows that  could be a stochastic matrix for a state system; one of the properties of such a matrix is that one of its eigenvalues must be 1.

### Example Question #41 : Eigenvalues And Eigenvectors

real matrix has as two of its eigenvalues  and . Give its characteristic equation.

Insufficient information is provided to answer the question.

Insufficient information is provided to answer the question.

Explanation:

matrix will have four eigenvalues, which are the zeroes of its characteristic polynomial equation. Since all of the entries of the matrix are real, all of the coefficients will be real as well. It follows that any imaginary zeroes must occur in conjugate pairs, so, since  is a zero, so it

We now know three of the zeroes, and since only one eigenvalue is unknown, it must be a real value. However, there is no restriction on this zero. Therefore, we have no way of determining the fourth zero - and, consequently, no way of figuring out the characteristic equation with any certainty.

### Example Question #49 : Eigenvalues And Eigenvectors

for some .

has as its eigenvalues  and . Which of the following is equal to ?

Explanation:

The sum of the eigenvalues of a matrix is equal to the trace of a matrix - the sum of its diagonal elements; the product of a matrix is equal to its determinant.

Therefore, to find , it is necessary to first find . The trace of  is equal to diagonal sum , so set that equal to the sum of the given eigenvalues and solve:

The matrix is therefore

can be found by setting the determinant - the upper-left to lower-right product minus the upper-right to lower-left product - equal to the product of the eigenvalues:

,

the correct choice.

### Example Question #50 : Eigenvalues And Eigenvectors

Which of the following holds for the eigenvectors of ?

All eigenvectors of  are scalar multiples of .

All eigenvectors of  are scalar multiples of .

Every vector in  is an eigenvector of

All eigenvectors of  are scalar multiples of .

All eigenvectors of  are scalar multiples of .

All eigenvectors of  are scalar multiples of .

Explanation:

is a lower triangular matrix - all elements above its main diagonal are equal to 0. Such a matrix has as its eigenvalues its diagonal elements. All three diagonal elements in  are 1, so 1 is the only eigenvalue of .

To find the eigenvector(s) of  corresponding to , first, find the matrix

Perform the Gauss-Jordan elimination process to get this matrix in reduced row-echelon form:

This matrix is interpreted as

arbitrary.

The eigenvectors must all take the form

for some scalar  - equivalently, all eigenvectors of  are scalar multiples of

.