Method of Eigenvalues and Eigenvectors

Solved Problems

Solution.

We calculate the eigenvalues \({\lambda_i}\) of the matrix \(A\) composed of the coefficients of the given equations:

In this example, the auxiliary equation has two distinct real roots.

Find an eigenvector \({\mathbf{V}_1}\) corresponding to the eigenvalue \({\lambda _1} = 3.\) Substituting \({\lambda _1} = 3,\) we get the vector-matrix equation for \({\mathbf{V}_1}:\)

Let the eigenvector \({\mathbf{V}_1}\) have components \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T}\) (where the superscript \(T\) denotes transposition). Then the previous equation can be written as

After matrix multiplication we get a system of two equations:

Both equations are linearly dependent. From the second equation we find the relation between the coordinates of the eigenvector: \({V_{11}} = {V_{21}}.\) Suppose that \({V_{21}} = 1.\) Then \({V_{11}} = 1.\) Thus, the eigenvector \({\mathbf{V}_1}\) has coordinates \({\mathbf{V}_1} = {\left( {1,1} \right)^T}.\)

Similarly, we find the second eigenvector \({\mathbf{V}_2}\) corresponding to \({\lambda _2} = -3.\) Let \({\mathbf{V}_2} =\) \( {\left( {{V_{21}},{V_{22}}} \right)^T}.\) Then

We get the system of two identical equations:

From this we find the coordinates of the eigenvector \({\mathbf{V}_2}:\)

Hence, \({\mathbf{V}_2} = {\left( {-5,1} \right)^T}.\)

Thus, the system has two different eigenvalues and two eigenvectors. The general solution is given by

where \({C_1},\) \({C_2}\) are arbitrary numbers.

Solution.

We seek a solution of the form

where \(\lambda\) is the eigenvalue of the matrix \(A\) composed of the coefficients of these equations, and \(\mathbf{V}\) is the corresponding eigenvector. Solve the auxiliary equation:

We obtain two eigenvalues as a pair of complex conjugate numbers. Find the eigenvector \({\mathbf{V}_1}\) for the eigenvalue \({\lambda _1} = 1 + 4i\) from the following equations:

Both equations are linearly dependent. From the second equation we have:

Thus, the eigenvector \({\mathbf{V}_1}\) is

Consequently, the complex number \({\lambda _1} = 1 + 4i\) produces a solution of the form

Transform the exponential function by Euler's formula:

The solution \({\mathbf{X}_1}\left( t \right)\) takes the form:

or after multiplication

In the complex solution, the real and imaginary parts are linearly independent. Separating them, we find the general solution:

Thus, the general solution has the form

where \({C_1},\) \({C_2}\) are arbitrary numbers.

Solution.

The matrix of the system is diagonal:

Therefore, we can just say that the eigenvectors are

However, we will construct a solution, following the general algorithm. Calculate the eigenvalues of the matrix \(A:\)

The matrix has a single eigenvalue with algebraic multiplicity \(2.\) If we substitute the number \({\lambda _1} = 3\) into the system of equations for the eigenvector \(\mathbf{V},\) we obtain a singular case:

It is clear that any nonzero vector \(\mathbf{V}\) will be the eigenvector for the given matrix \(A.\) Therefore, it is convenient to take the following two linearly independent vectors as a basis of eigenvectors:

Note that we have the case where the eigenvalue \({\lambda _1} = 3\) has the same algebraic and geometric multiplicity: \({k_1} =\) \( {s_1} = 2.\) This corresponds to the Case \(2\) in our classification.

The general solution of the system of equations can be written as

Solution.

We calculate the eigenvalues of \(A:\)

Expand the determinant along the first column:

You may notice that one of the roots of the cubic equation is the number \(\lambda = 1.\) Then we get

The quadratic equation, in turn, has roots \(\lambda = 2,3.\) Therefore, the matrix \(A\) has three distinct real eigenvalues:

Now we define an eigenvector for each of the eigenvalues.

Determine the vector \({\mathbf{V}_1}\) for the number \({\lambda _1} = 1\) by solving the vector-matrix equation

Denoting \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}},{V_{31}}} \right)^T},\) we can write this equation in the form

As a result, we have a system of linear algebraic equations:

In this system, the first and third equations are the same, i.e., the rank of the system is \(2.\) We leave two independent equations and take \({V_{31}}\) as a free variable. This yields:

Thus, the eigenvector \({\mathbf{V}_1}\) has coordinates

where for simplicity we set \(t = 1.\)

Similarly, we find the coordinates of the second eigenvector \({\mathbf{V}_2}\) corresponding to the number \({\lambda _2} = 2.\) We put \({\mathbf{V}_2} =\) \( {\left( {{V_{12}},{V_{22}},{V_{32}}} \right)^T}.\) Then we have the following equations:

Let \({V_{32}} = t.\) From the third equation we find: \({V_{22}} = {V_{32}} = t.\) Substituting in the first equation, we get:

Hence, the eigenvector \({\mathbf{V}_2}\) is

At \(t = 1\) one can write: \({\mathbf{V}_2} =\) \( {\left( { - 1,1,1} \right)^T}.\)

Now we calculate the coordinates of the third eigenvector \({\mathbf{V}_3}\) corresponding to the number \({\lambda _3} = 3.\) Denoting \({\mathbf{V}_3} = {\left( {{V_{13}},{V_{23}},{V_{33}}} \right)^T},\) we obtain the following equations:

We choose \({V_{33}} = t\) as the free variable. From the last equation we express \({V_{23}}:\)

Substituting \({V_{23}},\) \({V_{33}}\) in the first equation, we get:

Thus, the eigenvector ({\mathbf{V}_3}) has coordinates

The general solution can be written as

where \({C_1},\) \({C_2},\) \({C_3}\) are arbitrary constants.

Solution.

Let's start with finding the eigenvalues of the matrix \(A:\)

Factoring the left side, we get

It can be seen that the auxiliary equation has one real and two complex roots (as a pair of complex conjugate numbers):

The eigenvector \({\mathbf{V}_1}\) for the eigenvalue \({\lambda_1} = 1\) can be found in the same way as in the previous example. The coordinates of \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}},{V_{31}}} \right)^T}\) are defined by the system of linear equations:

After multiplying we get:

We see that the rank of the system of equations is \(2.\) Therefore, we can choose one independent variable, which we take as \({V_{31}} = t.\) Express the other variables in \(t:\)

So the first eigenvector has coordinates

Consider now the pair of complex conjugate roots \({\lambda _{2,3}} = \pm i.\) To find the part of the general solution associated with this pair of roots it is sufficient to take only one number, for example, \({\lambda _2} = + i\) and find for it the eigenvector \({\mathbf{V}_2},\) which can have complex coordinates. Next, we construct the particular solution \({\mathbf{X}_2}\) of the form

and identify the real and imaginary parts, which will represent two linearly independent solutions. By implementing this plan, we write the matrix-vector equation for the vector \({\mathbf{V}_2}:\)

We get the following system of equations:

Transform the first equation into a more convenient form, multiplying it by \( - {\frac{1}{{1 + i}}} :\)

Get rid of the complex numbers in the denominators of the coefficients:

Then the first equation becomes:

Let's go back to the system of equations and reduce it to a triangular form to determine its rank:

Transform the second equation:

Here the coefficient of the variable \({V_{32}}\) is

Hence, the second equation is

Similarly, we transform the third equation:

Calculate the coefficient of \({V_{32}}:\)

Then the third equation can be written as

that is, it coincides with the second equation.

So, the rank of the system is \(2,\) and it can be written in the following equivalent form:

We take as an independent variable \({V_{32}} = t.\) The other variables \({V_{22}}\) and \({V_{12}}\) can be successively expressed in terms of \(t:\)

Thus, we determined the eigenvector \({\mathbf{V}_2}\) with complex coordinates:

Now we construct the solution \({\mathbf{X}_2}\) on the basis of eigenvalue \({\lambda_2}\) and eigenvector \({\mathbf{V}_2}\) and expand it into real and imaginary parts.

Represent \({e^{it}}\) by Euler's formula:

Hence,

The calculated real and imaginary parts of the complex vector solution \({\mathbf{X}_2}\) are linearly independent. Taking into account the first component \({\mathbf{X}_1}\) corresponding to the eigenvalue \({\lambda_1},\) we can write the general real solution of the system as

where \({C_1},\) \({C_2},\) \({C_3}\) are arbitrary numbers.

Solution.

Determine the eigenvalues of the given matrix:

It can be noted that the cubic equation has a root \(\lambda = 1.\) Factoring out the term \(\left( {\lambda - 1} \right),\) we obtain:

The roots of the quadratic equation are: \(\lambda = 1, 4.\) Thus, the auxiliary equation is represented as

The initial matrix of the system is symmetric. So it will have three eigenvectors. This means that the algebraic and geometric multiplicity of the root \(\lambda = 1\) are the same (and equal to \(2\)).

Find the eigenvectors corresponding to the number \({\lambda _{1,2}} = 1.\) They can be found from the equations

We see that all three equations are identical. Leaving one equation, and choosing \({V_{21}} = u,\) \({V_{31}} = v\) as independent variables, we get:

It follows that the coordinates of the first eigenvector (with \(u = 1, v = 0\)) are \({\mathbf{V}_1} = {\left( { - 1,1,0} \right)^T}.\)

Accordingly, the coordinates of the second linearly independent eigenvector (when \(u = 0, v = 1\)) are \({\mathbf{V}_2} = {\left( { - 1,0,1} \right)^T}.\)

Now we define the third eigenvector \({\mathbf{V}_3}\) corresponding to the number \({\lambda _3} = 4:\)

Here we choose as a free variable \({V_{33}} = t.\) The other two coordinates are

Hence, the eigenvector \({\mathbf{V}_3}\) has the following coordinates:

The general solution of the system of differential equations is given by

where \({C_1},\) \({C_2},\) \({C_3}\) are arbitrary constants.