The eigenvalues of a matrix are the zeroes of its characteristic equation.

We can try to extract one or more zeroes using the Rational Zeroes Theorem, which states that any rational zero must be the positive or negative quotient of a factor of the constant and a factor of the leading coefficient. Since the constant is 57, and the leading coefficient of the polynomial is 1, any rational zeroes must be one of the set divided by 1, with positive or native numbers taken into account. Thus, any rational zeroes must be in the set

By trial and error, it can be determined that 3 is a solution of the equation:

True.

It follows that is a factor. Divide by ; the quotient can be found to be , which is prime.

The characteristic equation is factorable as

We already know that  is an eigenvalue; the other two eigenvalues are the zeroes of , which can be found by way of the quadratic formula:

An eigenvalue is a dominant eigenvalue if its absolute value is strictly greater than those of all other eigenvalues. Calculate the absolute values of all three eigenvalues:

Therefore,

.

Since no single eigenvalue has absolute value strictly greater than that of the other two, the matrix has no dominant eigenvalue.

The eigenvalues of a matrix are the zeroes of its characteristic equation.

can be solved as follows:

First, factor the polynomial by using the substitution . The equation can be changed to

The polynomial can be factored, using the "reverse FOIL" method, to

Substituting back:

Factoring further:

The zeroes of the polynomial are the zeroes of the individual factors, which can be calculated to be

An eigenvalue is a dominant eigenvalue if its absolute value is strictly greater than those of all other eigenvalues. The absolute values of all three eigenvalues are:

Since no single eigenvalue has absolute value strictly greater than that of the other three, the matrix has no dominant eigenvalue.

An eigenvalue is a dominant eigenvalue if its absolute value is strictly greater than those of all other eigenvalues. The absolute values of all four eigenvalues are as follows:

;

therefore, the dominant eigenvalue is .

The eigenvalues of a matrix are the zeroes of its characteristic equation.

can be solved by factoring out the polynomial. First, factor by grouping:

The zeroes of the polynomial are the zeroes of the individual factors, which can be calculated to be

An eigenvalue is a dominant eigenvalue if its absolute value is strictly greater than those of all other eigenvalues. The absolute values of all three eigenvalues are:

,

 so 9 is the dominant eigenvalue of .

A necessary and sufficient condition for a matrix to have 0 as an eigenvalue is for the matrix to have determinant 0. has an all-zero column:

This is a sufficient condition for a matrix to have determinant 0. It follows that 0 is an eigenvalue of .

One way to identify the matrix with these eigenvalues is to note that the sum of the eigenvalues of a matrix is equal to the trace of the matrix, and the product of the eigenvalues is equal to its determinant. Therefore, for a matrix to have 4 and 7 as eigenvalues, its trace must be 11 and its determinant must be 28.

The trace of a matrix is the sum of the elements in its main diagonal; the determinant is the product of these two elements minus the product of the other two:

Trace:

Determinant:

Trace:

Determinant:

Trace:

Determinant:

Trace:

Determinant:

All four matrices have the requisite trace and determinant, so all four have the eigenvalues 4 and 7.

An eigenvalue of a matrix is the dominant eigenvalue if its absolute value is strictly greater than that of all of its other eigenvalues. The absolute values of the three eigenvalues are:

.

, so is the dominant eigenvalue.

The first step into solving for eigenvalues, is adding in a  along the main diagonal. 

Now the next step to take the determinant.

Now lets FOIL, and solve for .

Now lets use the quadratic equation to solve for .

So our eigen values are

In this problem, we will get three eigen values and eigen vectors since it's a symmetric matrix.

To find the eigenvalues, we need to minus lambda along the main diagonal and then take the determinant, then solve for lambda.

This can be factored to

Thus our eigenvalues are at 

Now we need to substitute  into or matrix in order to find the eigenvectors.

For .

Now we need to get the matrix into reduced echelon form.

This can be reduced to 

This is in equation form is , which can be rewritten as . In vector form it looks like, . 

We need to take the dot product and set it equal to zero, and pick a value for , and .

Let , and .

Now we pick another value for , and  so that the result is zero. The easiest ones to pick are , and .

So the orthogonal vectors for  are , and .

Now we need to get the last eigenvector for .

After row reducing, the matrix looks like

So our equations are then

, and , which can be rewritten as , .

Then eigenvectors take this form, . This will be orthogonal to our other vectors, no matter what value of , we pick. For convenience, let's pick , then our eigenvector is.