The Cayley-Hamilton Theorem states that a square matrix  is a solution of its own characteristic equation. To find this, first find the polynomial formed from the determinant of . This matrix is 

Multiply  by each element in the identity:

Subtract elementwise:

Now find the determinant of this matrix. It is lower triangular - it has no nonzero elements above its main diagonal - so its determinant is the product of its diagonal elements. Therefore, the determinant of this matrix is

.

This makes the characteristic equation 

.

By the Cayley-Hamilton Theorem, the matrix  is a solution to this equation, so the correct choice is the above equation, modified for matrix arithmetic:

.

The Cayley-Hamilton Theorem states that a square matrix  is a solution of its own characteristic equation. To find this, first find the polynomial formed from the determinant of . This matrix is 

Multiply  by each element in the identity:

Subtract elementwise:

Now find the determinant of this matrix. It is upper triangular - it has no nonzero elements below its main diagonal - so its determinant is the product of its diagonal elements. Therefore, the determinant of this matrix is

This makes the characteristic equation 

.

By the Cayley-Hamilton Theorem, the matrix  is a solution to this equation, so the correct choice is the above equation, modified for matrix arithmetic:

.

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, .