A  matrix will have three (not necessarily distinct) 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 of this equation, so is its complex conjugate, . 

The characteristic equation of a  equation sets a degree-3 polynomial equal to 0. Since 2, , and  are its zeroes, this polynomial is

,

The characteristic equation is 

.

The two eigenvectors given,  and , are linearly independent, since they are not scalar multiples of each other; therefore, they form a basis of the 2-dimensional eigenspace of .  is an eigenvector corresponding to  if and only if it is a linear combination of the two given basis vectors, or, equivalently, if there exist  so that 

Rewrite this as follows:

This is equivalent to a system of three linear equations in two variables:

Solve the system by first, rewriting the second equation as 

The first equation becomes

Substitute in the first equation:

Now substitute for  and  in the third equation:

, the correct response.

The two eigenvectors given,  and , are linearly independent, since they are not scalar multiples; since the eigenspace corresponding to  has dimension 2, they form a basis of the eigenspace.  is an eigenvector corresponding to  if and only if it is a linear combination of the two given basis vectors, so we seek to find  so that 

This is 

This is the system of four linear equations in two variables:

If we look at the first and fourth equations, we can immediately see that this system is inconsistent. There is no  that solves this system. It follows that  is not an eigenvector of  corresponding to .

One property of eigenvalues is that if  is nonsingular, the set of eigenvalues of  is exactly the set of reciprocals of eigenvalues of . The eigenvalues of  are , so the eigenvalues of these numbers are the reciprocals of these - in order, . This is not the same set.

 is an eigenvalue of  if it is a solution of the characteristic polynomial equation

Set this equation:

The determinant can be found most easily by adding the products of each entry in one of the rows or columns to its corresponding cofactor. Since the second row has only one nonzero entry, we use this one:

, 

the minor formed by striking out row 2 and column 2:

Take the upper-left to lower-right product, and subtract the upper-right to lower-left product:

Thus, 

,

and 

This makes  and 1 the solutions of the characteristic equation, the latter with multiplicity 2. It follows that  has as its eigenvalues the set .

Let  for some complex values of the variables.

 is an eigenvector of  corresponding to eigenvalue  if

.

 is an eigenvector corresponding to eigenvalue 2, so

and 

 is an eigenvector corresponding to eigenvalue 2, so, similarly, 

Two systems of two linear equations in two variables are created. Each can be solved separately.

Multiplying the bottom equation by 2, then adding:

The second system is

Solve similarly:

This expression is equivalent to saying "find the characteristic polynomial of , and set it equal to ." This is the most frequently used method of finding eigenvalues of a matrix.

For example,, has  as one of its eigenvalues, but is clearly not invertible since its determinant is .

The eigenvalues of a matrix  are the solutions of the characteristic polynomial equation 

.

 is a matrix with only real entries, so the coefficients of the polynomial must be real as well; it follows that any imaginary solutions are in conjugate pairs. , an eigenvalue of , is a solution of the characteristic equation; consequently, its complex conjugate  is a solution, and thus, an eigenvalue of .

One property of eigenvalues is that if  is nonsingular, the set of eigenvalues of  is exactly the set of reciprocals of eigenvalues of . Since  is an eigenvalue of , it follows that  is an eigenvalue of .

While the question can be answered by direct definition, it is arguably easier to answer it by noting that the columns of  each have entries adding up to 1 - the characteristics of a stochastic matrix; each column has the same four entries, and

.

A matrix with this property has eigenvalue 1 - in fact, it is the dominant eigenvalue.

An eigenvector of  corresponding to an eigenvalue  is that vector  such that 

.

Set ; this becomes

Since  is the dominant eigenvalue, as is raised to a large enough power,  approaches ,  the desired eigenvector. Finding, say, , this vector approaches

.

Multiplying this by 4, it can easily been seen that   is an eigenvector of  corresponding to eigenvalue 1.