By definition, if  is an eigenvector of a matrix , there is a scalar  such that

Let , the given eigenvector of . Then 

 for some real . From this equation, it follows that any scalar multiple of eigenvector  of a matrix is also an eigenvector of the matrix, since:

,

making  an eigenvector as well.

Let . Each element in  is four times the corresponding element in , so  - that is, it is a scalar multiple of a known eigenvector of . That makes  itself an eigenvector of .

To find the eigenvectors, we first need to find the eigenvalues. We can do by calculating

.

Setting this equal to , we have .

Hence , are our eigenvalues.

Now we find the eigenvectors associated with each eigenvalue. Let's start with .

First plug the eigenvalue into the equation , and solve the resulting system of equations.

Solving this system gives  once it is parametrized .

Hence  is one of our eignvectors.

Repeating the above process with  gives the equations

, with the solution . Hence  is the other eigenvector.

To find the eigenvalues, we first need to characteristic polynomial of our matrix

This determinant is easy to evaluate since it is of an upper triangular matrix; we simply multiply the diagonal entries together.

.

The roots of this polynomial are , but since the root  has multiplicity , we only have  distinct eigenvalues, not .

Suppose  is an invertible matrix with nonzero eigenvalue . Suppose  is the associated eigenvector. It can be shown that  is an eigenvalue of . We have

We can multiply both sides by  to get

Divide both sides by  to get

So we get that  is an eigenvalue of 

It it is certainly the case that with a square matrix  with eigenvector ,  has an infinite number of eigenvectors. Take a nonzero real number . Then  is also an eigenvector because

So the infinite list of vectors  are all eigenvectors - therefore,  will have an infinite number of eigenvectors. They are all from the same eigenspace, though, so nothing impressive was done. 

This is false. Suppose  is an eigenvector of .  is also an eigenvector:

But  and  are obviously not linearly independent: 

An eigenvalue of  is a zero of the characteristic equation formed from the determinant of , so find this determinant as follows:

Subtracting elementwise:

The determinant can be found by taking the upper-left-to-lower-right product and subtracting the upper-right-to-lower-left product; set this to 0 to get the characteristic equation.

For 4 to be an eigenvalue,  must be a solution of this equation, so substitute and solve for :

Applying the quadratic formula:

 is an upper triangular matrix - that is, the only element above its main (upper left to lower right) diagonal is a zero. As a consequence, its eigenvalue(s) are the elements in its main diagonal, both of which are . It therefore follows that for the matrix to have 4 as an eigenvalue, .

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.

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.