Construction of the General Solution of a System of Equations Using the Jordan Form

Solved Problems

Solution.

We write the auxiliary equation for the matrix and find the eigenvalues:

Compute the eigenvectors for each eigenvalue.

Substituting \({\lambda _1} = 1,\) we find the vector \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T}:\)

We see that the rank of this matrix is \(1.\) Consequently, the geometric multiplicity of the eigenvalue \({\lambda _1} = 1\) is

Accordingly, there is one eigenvector. Its coordinates are

Similarly, we calculate the eigenvector \({\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}}} \right)^T}\) for the eigenvalue \({\lambda _2} = 5:\)

Let \({V_{22}} = t.\) Then

Here you can see that we have the case of simple eigenvalues (Case \(1\)). The general solution is expressed as

Solution.

As usual, first we determine the eigenvalues by solving the characteristic equation:

Hence, the matrix of the system has one eigenvalue \({\lambda _1} = - 1\) of multiplicity \({k_1} = 2.\)

Find the eigenvectors for the value \({\lambda _1}.\)

As it can be seen, in this case any vector is an eigenvector. Therefore, we can choose the unit vectors as a pair of linearly independent vectors:

Here we meet with the Case \(2:\) a system of two differential equations has one eigenvalue, the algebraic and geometric multiplicity of which is equal to \(2.\) The general solution is written as

Solution.

We write the auxiliary equation and find its roots.

The matrix of the system has one eigenvalue \({\lambda _1} = 3\) with algebraic multiplicity \({k_1} = 2.\)

Determine the eigenvectors corresponding to the number \({\lambda _1} = 3.\) Let \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T}.\) We obtain:

Let \({V_{21}} = t.\) Then the coordinates of \({\mathbf{V}_1}\) are

Check whether we have calculated the eigenvector \({\mathbf{V}_1}\) correctly. By definition, the eigenvector should satisfy the relationship

Substituting the known values, we get

This combination of parameters \(\left( {{k_1} = 2,{s_1} = 1} \right)\) corresponds to the Case \(3,\) in which the solution is described by a single Jordan block. To construct the general solution, we need to determine the generalized eigenvector \({\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}}} \right)^T}.\) We find it from the matrix equation:

Let us make another check to make sure that the generalized eigenvector \({\mathbf{V}_2}\) is calculated correctly. Use the formula for the reduction of the original matrix \(A\) to the Jordan normal form \(J:\)

Here the matrix \(H\) is composed of the found vectors:

The inverse matrix \({H^{ - 1}}\) will be equal to:

where \({{A_{ij}}}\) are cofactors of the elements of the matrix \(H,\) and \(\Delta\) is its determinant.

After substitution we see that the result of the transformations is the Jordan form:

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

Solution.

We compose the auxiliary equation and calculate its roots:

Expand the determinant along the third row:

Hence, the matrix has three distinct eigenvalues: \({\lambda _1} = 0,\) \({\lambda _2} = 1,\) \({\lambda _3} = 2.\)

Compute the eigenvectors \({\mathbf{V}_1},{\mathbf{V}_2},{\mathbf{V}_3}\) for these eigenvalues. For the eigenvalue \({\lambda _1} = 0\) we find \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}},{V_{31}}} \right)^T}:\)

Determine the rank of the system of equations:

In this case, the rank of the matrix is \(2,\) so the geometric multiplicity of \({\lambda _1}\) is equal to \(1.\) To find the vector \({\mathbf{V}_1}\) associated with the value \({\lambda _1},\) we set \({V_{31}} = t.\) As a result, we have

Consequently,

The correctness of the calculation of the eigenvector \({\mathbf{V}_1}\) can be verified using the definition of the eigenvector. Substituting the coordinates of \({\mathbf{V}_1},\) we obtain for \({\lambda _1} = 0:\)

Similarly, we find the vector \({\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}},{V_{32}}} \right)^T}\) for the eigenvalue \({\lambda _2} = 1:\)

Thus, we see that \(\text{rank}\left( {A - {\lambda _2}I} \right) = 2.\) The geometric multiplicity of \({\lambda _2} = 1\) is \({s_2} = 1.\) Let \({V_{32}} = t\) and express the other coordinates \({V_{12}},\) \({V_{22}}\) in \(t:\)

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

Verification:

Now we find the vector \({\mathbf{V}_3} = {\left( {{V_{13}},{V_{23}},{V_{33}}} \right)^T},\) associated with the eigenvalue \({\lambda _3} = 2:\)

It is seen that \(\text{rank}\left( {A - {\lambda _3}I} \right) = 2.\) Putting \({V_{33}} = t,\) calculate the coordinates \({V_{13}}, {V_{23}}:\)

Hence,

Check again:

Thus, we have found all the eigenvectors. Now we can write the general solution of the system, which in this case is given by

Solution.

Calculate the roots of the characteristic equation:

You may notice that one of the roots of the cubic equation is the number \({\lambda _1} = - 1.\) Then, taking out the factor \(\left( {\lambda + 1} \right),\) we obtain:

Thus, the system has two eigenvalues: \({\lambda _1} = - 1\) of multiplicity \({k_1} = 1\) and \({\lambda _2} = 2\) of multiplicity \({k_2} = 2.\)

Determine the eigenvectors. For the number \({\lambda _1} = - 1,\) the matrix rank is

We see that \(\text{rank}\left( {A - {\lambda _1}I} \right) = 2.\) This means that only one eigenvector \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}},{V_{31}}} \right)^T}\) corresponds to the given eigenvalue:

Find its coordinates from the system

Let \({V_{31}} = t.\) Therefore,

Then, the eigenvector \({\mathbf{V}_1}\) will be equal to

Now we consider the second eigenvalue \({\lambda _2} = 2\) with algebraic multiplicity \({k_2} = 2.\) Determine the rank of \(A - {\lambda _2}I\) and geometric multiplicity \({s_2}:\)

Hence,

In this case, the matrix has two eigenvectors (i.e., we have Case \(5\)). If we denote \({\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}},{V_{32}}} \right)^T},\) \({\mathbf{V}_3} = {\left( {{V_{13}},{V_{23}},{V_{33}}} \right)^T},\) then the coordinates of these two vectors will satisfy the equations

Choosing coordinates \(y, z\) as free variables and setting them equal to \(\left( {0,1} \right)\) for \({\mathbf{V}_2},\) and \(\left( {1,0} \right)\) for \({\mathbf{V}_3},\) we get the following linearly independent vectors:

Putting all the components together, we can write the general solution in the form

Solution.

We calculate the eigenvalues:

It can be seen that there are two eigenvalues: \({\lambda _1} = 0\) of multiplicity \({k_1} = 2\) and \({\lambda _2} = 1\) of multiplicity \({k_2} = 1.\)

Compute the rank of the matrix \(A - {\lambda _1}I:\)

Consequently, \(\text{rank}\left( {A - {\lambda _1}I} \right) = 2\) and, respectively, the geometric multiplicity \({s_1}\) of the eigenvalue \({\lambda _1} = 0\) is equal to:

It is clear that we are dealing with the Case \(6,\) where the Jordan form contains two blocks, one of which is associated with one independent eigenvector and one generalized eigenvector. First we find the ordinary eigenvector \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}},{V_{31}}} \right)^T}\) by solving the matrix equation

which is equivalent to the system

Let \({V_{31}} = t.\) Then

We get

Now we calculate the generalized eigenvector \({\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}},{V_{32}}} \right)^T}:\)

We can choose any vector satisfying these equations. Suppose \({V_{32}} = 0.\) Then the remaining coordinates are

Thus, the coordinates of the generalized eigenvector \({\mathbf{V}_2}\) are

Consider now the eigenvalue \({\lambda _2} = 1.\) The corresponding eigenvector \({\mathbf{V}_3} = {\left( {{V_{13}},{V_{23}},{V_{33}}} \right)^T}\) is given by

Let \({V_{33}} = t.\) Then

Hence,

Verify that the eigenvectors are calculated correctly by the formula

The determinant of the matrix \(H\) is

Form the matrix \(B\) of cofactors:

Hence,

Transposing the matrix \(B,\) we write the inverse matrix \({H^{ - 1}}:\)

Calculate the product of three matrices:

We obtained the Jordan form \(J\) containing the eigenvalues \({\lambda _1} = 0\) on the diagonal of the first block and the eigenvalue \({\lambda _2} = 1\) in the second block.

The general solution is given by

Solution.

We write the characteristic equation and find the eigenvalues:

The equation has one root \({\lambda _1} = 1\) of multiplicity \({k_1} = 3.\) Determine the rank of the matrix \(A - {\lambda _1}I:\)

The rank is \(1.\) Therefore, the geometric multiplicity of the eigenvalue \({\lambda _1} = 1\) is

It follows that the Jordan form consists of two blocks and corresponds to Case \(7.\)

We find the eigenvectors\({\mathbf{V}_1}\) and \({\mathbf{V}_2}\) associated with the eigenvalue \({\lambda _1} = 1.\) Let the coordinates of the vector \({\mathbf{V}_1}\) be \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}},{V_{31}}} \right)^T}.\) Solve the equation

We can choose two coordinates arbitrarily. A simple pair of linearly independent vectors is obtained by setting \(y = 1, z = 0\) for the vector \({\mathbf{V}_1}\) and \(y = 0, z = 1\) for the vector \({\mathbf{V}_2}.\) Substituting these values into the last equation, we find the coordinates \(x\) of the eigenvectors \({\mathbf{V}_1}\) and \({\mathbf{V}_2}:\)

Here we must bear in mind that in the system of equations of rank \(1,\) there is an infinite number of eigenvectors (lying in the plane \(x + 2y - 5z = 0\)). At the same time, in this step the vectors \({\mathbf{V}_1}\) and \({\mathbf{V}_2}\) are not necessary to be a part of a Jordan basis.

Consider the \(2 \times 2\) Jordan block. Obviously, the Jordan chain consists of one independent eigenvector and one generalized eigenvector. We denote these vectors as \({\mathbf{U}_1}\) and \({\mathbf{U}_2}.\) They must satisfy the following matrix equations:

Verify that \({\left( {A - {\lambda _1}I} \right)^2} = \mathbf{0}:\)

Thus, any non-zero vector belongs to the kernel of the operator \({\left( {A - {\lambda _1}I} \right)^2}.\) Since the first column of the matrix \({A - {\lambda _1}I}\) is not equal to zero, one can take the unit vector of the axis \(Ox\) as the generalized eigenvector \({\mathbf{U}_2}:\) \({\mathbf{U}_2} = {\left( {1,0,0} \right)^T}.\)

Compute the vector \({\mathbf{U}_1}:\)

Let us show that the found vector \({\mathbf{U}_1}\) belongs to the kernel of the operator \({A - {\lambda _1}I},\) i.e. is an eigenvector of \(A:\)

So, we have two basis vectors \({\mathbf{U}_1}\) and \({\mathbf{U}_2}\) associated to the \(2 \times 2\) Jordan block. The other \(1 \times 1\) block contains one more eigenvector, which can be taken as any eigenvector of \(A\) that is not collinear with \({\mathbf{U}_1} = {\left( {3,1,1} \right)^T}.\) Take, for example, the vector \({\mathbf{V}_2} = {\left( {5,0,1} \right)^T}\) determined in the beginning of the solution.

Three linearly independent vectors \({\mathbf{U}_1},\) \({\mathbf{U}_2}\) and \({\mathbf{V}_2}\) form a Jordan basis. The general solution of the system is expressed as

Solution.

We form the characteristic equation and find its roots:

Hence, the matrix \(A\) has one eigenvalue \({\lambda _1} = - 3\) with algebraic multiplicity \({k_1} = 3.\)

Compute the rank of the matrix \(A - {\lambda _1}I:\)

It is seen that \(\text{rank}\left( {A - {\lambda _1}I} \right) = 2,\) so the geometric multiplicity of \({\lambda _1} = - 3\) is

Here we have the Case \(8,\) where there is a Jordan block of size \(3 \times 3.\) The corresponding Jordan chain will consist of one regular eigenvector \({\mathbf{V}_1}\) and two generalized eigenvectors \({\mathbf{V}_2},\) \({\mathbf{V}_3}.\) For these vectors, which form a Jordan basis, we will have the following relations:

We show that \({\left( {A - {\lambda _1}I} \right)^3} = \mathbf{0}:\)

Thus,

that is the kernel of the operator \({\left( {A - {\lambda _1}I} \right)^3}\) coincides with the whole space. Therefore, we can choose any arbitrary non-zero vector \({\mathbf{V}_3}\) to form the Jordan chain. Take, for example, the vector \({\mathbf{V}_3} = {\left( {1,0,0} \right)^T}\) and make sure that it does not belong to the kernel of the operator \({\left( {A - {\lambda _1}I} \right)^2}:\)

Now we calculate the vectors \({\mathbf{V}_2}\) and \({\mathbf{V}_1}\) from the chain of relationships

Our goal is to get a nonzero vector \({\mathbf{V}_1},\) i.e. construct a Jordan basis. If the vector \({\mathbf{V}_1} = \mathbf{0},\) one can take the vector \({\left( {0,1,0} \right)^T}\) or \({\left( {0,0,1} \right)^T}\) as the initial vector \({\mathbf{V}_3}.\) In one of the three options we will definitely obtain a non-zero vector \({\mathbf{V}_1}.\) This follows from the fact that the kernel of the operator \({\left( {A - {\lambda _1}I} \right)^2}\) is not the whole space, and, therefore, can not have three linearly independent vectors.

Continuing the calculation, we find \({\mathbf{V}_2}\) and \({\mathbf{V}_1}:\)

Thus, we have determined a Jordan basis consisting of the vectors

Now we check the result using the formula

Here the matrix \(H\) is composed of the basis vectors \({\mathbf{V}_1},\) \({\mathbf{V}_2},\) \({\mathbf{V}_3}:\)

The inverse matrix \({H^{ - 1}}\) is

Multiplying the matrices, we get

The result is a Jordan form with one block of size \(3 \times 3.\)

The general solution of the system is given by