{"id":4020,"date":"2026-07-06T12:26:21","date_gmt":"2026-07-06T12:26:21","guid":{"rendered":"https:\/\/www.mathros.net.ua\/en\/?p=4020"},"modified":"2026-07-06T14:26:05","modified_gmt":"2026-07-06T14:26:05","slug":"eigenvalues-of-a-matrix-lu-decomposition-method","status":"publish","type":"post","link":"https:\/\/www.mathros.net.ua\/en\/eigenvalues-of-a-matrix-lu-decomposition-method.html","title":{"rendered":"Eigenvalues of a Matrix: How to Find Them Using LU Decomposition"},"content":{"rendered":"

Eigenvalues of a matrix<\/a> can be found not only through the characteristic polynomial<\/a>. In numerical methods, another approach is often used: the original matrix is gradually replaced by matrices similar to it, which have the same eigenvalues. During the calculations, these matrices get closer to a triangular form. For a triangular matrix, eigenvalues are easy to determine from the main diagonal.<\/p>\n

Eigenvalues of a Matrix: Why Triangular Form Helps<\/h2>\n

Let us start with a simple question. Why do we so often try to obtain a triangular matrix<\/a> when solving a problem about eigenvalues?<\/p>\n

The reason is that the eigenvalues of a triangular matrix are equal to its diagonal elements. In other words, instead of doing complicated calculations, it is enough to find the elements on the main diagonal.<\/p>\n

For example, suppose we have a triangular matrix<\/p>\n

\\[
\nT=
\n\\begin{pmatrix}
\n\\lambda_1 & 0 & \\dots & 0\\\\
\nt_{21} & \\lambda_2 & \\dots & 0\\\\
\n\\vdots & \\vdots & \\ddots & \\vdots\\\\
\nt_{n1} & t_{n2} & \\dots & \\lambda_n
\n\\end{pmatrix}.
\n\\]<\/p>\n

Then the determinant in the characteristic equation has the form<\/p>\n

\\[
\n\\det(T-\\lambda I)=(\\lambda_1-\\lambda)\\cdot(\\lambda_2-\\lambda)\\cdot\\dots\\cdot(\\lambda_n-\\lambda).
\n\\]<\/p>\n

Therefore, the roots of the characteristic equation coincide with the diagonal elements of the matrix. So, the eigenvalues of the matrix \\( T \\) are<\/p>\n

\\[
\n\\lambda_1,\\lambda_2,\\dots,\\lambda_n.
\n\\]<\/p>\n

This is where the main idea of the method comes from. If the original matrix \\( A \\) is gradually brought closer to a triangular matrix, then its eigenvalues can be found much more easily.<\/p>\n

But here an important detail appears right away. We cannot change the matrix in any way we want. Its eigenvalues must be preserved during the transformations. That is why similar matrices are used. If matrices are similar, they have the same eigenvalues.<\/p>\n

So, the task comes down to this: to construct a sequence of similar matrices that gradually moves toward a more convenient, triangular form.<\/p>\n

LU Decomposition of a Matrix: How to Find L and U<\/h2>\n

Now let us find out what is used to construct such a sequence. This method uses LU decomposition.<\/p>\n

Suppose we are given a square matrix \\( A \\) of size \\( n\\times n \\):<\/p>\n

\\[
\nA=
\n\\begin{pmatrix}
\na_{11} & a_{12} & \\dots & a_{1n}\\\\
\na_{21} & a_{22} & \\dots & a_{2n}\\\\
\n\\vdots & \\vdots & \\ddots & \\vdots\\\\
\na_{n1} & a_{n2} & \\dots & a_{nn}
\n\\end{pmatrix}.
\n\\]<\/p>\n

It is represented as the product of two matrices<\/p>\n

\\[
\nA=L\\cdot U,
\n\\]<\/p>\n

where \\( L \\) is a lower triangular matrix, and \\( U \\) is an upper triangular matrix.<\/p>\n

To be specific, we will use the version in which the diagonal elements of the matrix \\( U \\) are equal to one: \\( u_{ii}=1 \\).<\/p>\n

Then the matrices \\( L \\) and \\( U \\) have the form<\/p>\n

\\[
\nL=
\n\\begin{pmatrix}
\nl_{11} & 0 & \\dots & 0\\\\
\nl_{21} & l_{22} & \\dots & 0\\\\
\n\\vdots & \\vdots & \\ddots & \\vdots\\\\
\nl_{n1} & l_{n2} & \\dots & l_{nn}
\n\\end{pmatrix},\\qquad
\nU=
\n\\begin{pmatrix}
\n1 & u_{12} & \\dots & u_{1n}\\\\
\n0 & 1 & \\dots & u_{2n}\\\\
\n\\vdots & \\vdots & \\ddots & \\vdots\\\\
\n0 & 0 & \\dots & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n

The elements of these matrices are calculated step by step. First, we find the elements of the matrix \\( L \\):<\/p>\n

\\[
\nl_{ij}=a_{ij}-\\sum_{k=1}^{j-1} l_{ik}\\cdot u_{kj}, \\qquad i\\ge j.
\n\\]<\/p>\n

After that, we calculate the elements of the matrix \\( U \\):<\/p>\n

\\[
\nu_{ij}=\\frac{1}{l_{ii}}\\cdot \\left(a_{ij}-\\sum_{k=1}^{i-1} l_{ik}\\cdot u_{kj}\\right), \\qquad i<j.
\n\\]<\/p>\n

Here it is important to remember the condition \\( l_{ii}\\ne 0 \\). Why is it needed? Because when calculating the elements of the matrix \\( U \\), we divide by \\( l_{ii} \\). If this element is equal to zero, then the usual LU decomposition without row or column permutations cannot be applied.<\/p>\n

So, in this method, we usually assume that the required LU decomposition exists. This applies not only to the original matrix \\( A \\), but also to all matrices that appear later in the iterative process.<\/p>\n

Thus, LU decomposition gives us a clear way to split the current matrix into two triangular matrices. Next, we will show how these two matrices are used to obtain a new matrix with the same eigenvalues.<\/p>\n

Iterative Process: Constructing Similar Matrices<\/h2>\n

Let the original matrix be<\/p>\n

\\[
\nA_0=A.
\n\\]<\/p>\n

At the first step, we perform the LU decomposition of the matrix \\( A_0 \\):<\/p>\n

\\[
\nA_0=L_0\\cdot U_0.
\n\\]<\/p>\n

After that, we change the order of the factors and construct a new matrix:<\/p>\n

\\[
\nA_1=U_0\\cdot L_0.
\n\\]<\/p>\n

At first glance, this is just a rearrangement of the factors. But this rearrangement is exactly what gives us a similar matrix.<\/p>\n

Indeed, from the equality<\/p>\n

\\[
\nA_0=L_0\\cdot U_0
\n\\]<\/p>\n

we get<\/p>\n

\\[
\nU_0=L_0^{-1}\\cdot A_0.
\n\\]<\/p>\n

Then<\/p>\n

\\[
\nA_1=U_0\\cdot L_0=L_0^{-1}\\cdot A_0\\cdot L_0.
\n\\]<\/p>\n

So, the matrix \\( A_1 \\) is similar to the matrix \\( A_0 \\). Therefore, the matrices \\( A_1 \\) and \\( A_0 \\) have the same eigenvalues.<\/p>\n

Next, the same process is repeated for the matrix \\( A_1 \\):<\/p>\n

\\[
\nA_1=L_1\\cdot U_1,\\qquad
\nA_2=U_1\\cdot L_1.
\n\\]<\/p>\n

In general form, the algorithm is written as<\/p>\n

\\[
\nA_k=L_k\\cdot U_k,\\qquad
\nA_{k+1}=U_k\\cdot L_k,
\n\\]<\/p>\n

where \\( k=0,1,2,\\dots \\).<\/p>\n

Each next matrix \\( A_{k+1} \\) is similar to the previous matrix \\( A_k \\), because<\/p>\n

\\[
\nA_{k+1}=L_k^{-1}\\cdot A_k\\cdot L_k.
\n\\]<\/p>\n

Therefore, all matrices<\/p>\n

\\[
\nA_0,A_1,A_2,\\dots,A_k
\n\\]<\/p>\n

have the same eigenvalues as the original matrix \\( A \\).<\/p>\n

Under certain conditions, this sequence gradually approaches a triangular matrix. After a sufficiently large number of iterations, the diagonal elements of the matrix \\( A_k \\) become approximations to the eigenvalues of the original matrix \\( A \\).<\/p>\n

Then we can write<\/p>\n

\\[
\nA_k=
\n\\begin{pmatrix}
\n\\lambda_1 & 0 & \\dots & 0\\\\
\na_{21}^{(k)} & \\lambda_2 & \\dots & 0\\\\
\n\\vdots & \\vdots & \\ddots & \\vdots\\\\
\na_{n1}^{(k)} & a_{n2}^{(k)} & \\dots & \\lambda_n
\n\\end{pmatrix}.
\n\\]<\/p>\n

So, the numbers<\/p>\n

\\[
\n\\lambda_1,\\lambda_2,\\dots,\\lambda_n
\n\\]<\/p>\n

are approximate eigenvalues of the original matrix \\( A \\).<\/p>\n

Stopping Criterion for Iterations: When to Finish the Calculations<\/h2>\n

In practice, the iterative process is not carried out forever. We need a rule that shows when the calculations can be stopped.<\/p>\n

Since the eigenvalues are read from the main diagonal of the matrix \\( A_k \\), it is natural to control the diagonal elements. If they almost do not change between two neighboring steps, then the process can be stopped.<\/p>\n

For example, the following condition is used:<\/p>\n

\\[
\n\\max_{1\\le i\\le n}\\left|a_{ii}^{(k+1)}-a_{ii}^{(k)}\\right|\\le \\varepsilon,
\n\\]<\/p>\n

where \\( \\varepsilon \\) is the given tolerance.<\/p>\n

This means that all diagonal elements have changed by no more than \\( \\varepsilon \\). Therefore, these elements can be considered approximate eigenvalues of the matrix \\( A \\).<\/p>\n

At the same time, it is important to remember that the convergence of the method depends on the properties of the matrix. Important factors include the existence of LU decomposition at each step, how much the absolute values of the eigenvalues differ, and the numerical stability of the calculations.<\/p>\n

For example, if the eigenvalues have different absolute values, that is,<\/p>\n

\\[
\n|\\lambda_1|>|\\lambda_2|>\\dots>|\\lambda_n|,
\n\\]<\/p>\n

then the iterative process usually behaves better. But if some eigenvalues have equal or very close absolute values, convergence may be slow or less stable.<\/p>\n

So, the LU decomposition method is convenient for explaining the iterative approach. However, in practice, its use always requires error control, checking convergence conditions, and careful attention to numerical stability.<\/p>\n

Eigenvalues of a Matrix: Practice with LU Decomposition<\/h2>\n

Now let us look at how the LU decomposition method works for specific matrices. In each example, we will construct a sequence of similar matrices and observe how their diagonal elements change. These elements gradually give approximate eigenvalues of the original matrix.<\/p>\n

Example 1. Find the eigenvalues of the matrix using the LU decomposition method with tolerance \\( \\varepsilon=0.1 \\):
\n\\[
\nA=
\n\\begin{pmatrix}
\n4 & 1\\\\
\n2 & 3
\n\\end{pmatrix}.
\n\\]<\/h3>\n

Denote the initial matrix by<\/p>\n

\\[
\nA_0=A=
\n\\begin{pmatrix}
\n4 & 1\\\\
\n2 & 3
\n\\end{pmatrix}.
\n\\]<\/p>\n

Perform the LU decomposition:<\/p>\n

\\[
\nA_0=L_0\\cdot U_0.
\n\\]<\/p>\n

Calculate the elements of the matrices \\( L_0 \\) and \\( U_0 \\):<\/p>\n

\\[
\n\\begin{gathered}
\nl_{11}^{0}=a_{11}^{(0)}=4,\\\\[4pt]
\nu_{12}^{0}=\\dfrac{a_{12}^{(0)}}{l_{11}^{0}}=\\dfrac{1}{4}=0.25,\\\\[4pt]
\nl_{21}^{0}=a_{21}^{(0)}=2,\\\\[4pt]
\nl_{22}^{0}=a_{22}^{(0)}-l_{21}^{0}\\cdot u_{12}^{0}
\n=3-2\\cdot 0.25=2.5.
\n\\end{gathered}
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\nL_0=
\n\\begin{pmatrix}
\n4 & 0\\\\
\n2 & 2.5
\n\\end{pmatrix},
\n\\qquad
\nU_0=
\n\\begin{pmatrix}
\n1 & 0.25\\\\
\n0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n

Now we switch the factors:<\/p>\n

\\[
\nA_1=U_0\\cdot L_0.
\n\\]<\/p>\n

Substitute the matrices we found:<\/p>\n

\\[
\nA_1=
\n\\begin{pmatrix}
\n1 & 0.25\\\\
\n0 & 1
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n4 & 0\\\\
\n2 & 2.5
\n\\end{pmatrix}.
\n\\]<\/p>\n

After the calculations, we get<\/p>\n

\\[
\nA_1=
\n\\begin{pmatrix}
\n4.5 & 0.625\\\\
\n2 & 2.5
\n\\end{pmatrix}.
\n\\]<\/p>\n

Check the stopping condition:<\/p>\n

\\[
\n\\max\\left(|4.5-4|,\\ |2.5-3|\\right)=0.5.
\n\\]<\/p>\n

Since<\/p>\n

\\[
\n0.5>0.1,
\n\\]<\/p>\n

we continue the iterative process.<\/p>\n

Now we decompose the matrix<\/p>\n

\\[
\nA_1=
\n\\begin{pmatrix}
\n4.5 & 0.625\\\\
\n2 & 2.5
\n\\end{pmatrix}.
\n\\]<\/p>\n

Perform the LU decomposition:<\/p>\n

\\[
\nA_1=L_1\\cdot U_1.
\n\\]<\/p>\n

Calculate the elements of the matrices \\( L_1 \\) and \\( U_1 \\):<\/p>\n

\\[
\n\\begin{gathered}
\nl_{11}^{1}=a_{11}^{(1)}=4.5,\\\\[4pt]
\nu_{12}^{1}=\\dfrac{a_{12}^{(1)}}{l_{11}^{1}}
\n=\\dfrac{0.625}{4.5}=0.139,\\\\[4pt]
\nl_{21}^{1}=a_{21}^{(1)}=2,\\\\[4pt]
\nl_{22}^{1}=a_{22}^{(1)}-l_{21}^{1}\\cdot u_{12}^{1}
\n=2.5-2\\cdot 0.139=2.222.
\n\\end{gathered}
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\nL_1=
\n\\begin{pmatrix}
\n4.5 & 0\\\\
\n2 & 2.222
\n\\end{pmatrix},
\n\\qquad
\nU_1=
\n\\begin{pmatrix}
\n1 & 0.139\\\\
\n0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n

Construct the next matrix:<\/p>\n

\\[
\nA_2=U_1\\cdot L_1.
\n\\]<\/p>\n

Substitute \\( U_1 \\) and \\( L_1 \\):<\/p>\n

\\[
\nA_2=
\n\\begin{pmatrix}
\n1 & 0.139\\\\
\n0 & 1
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n4.5 & 0\\\\
\n2 & 2.222
\n\\end{pmatrix}.
\n\\]<\/p>\n

After the calculations:<\/p>\n

\\[
\nA_2=
\n\\begin{pmatrix}
\n4.778 & 0.309\\\\
\n2 & 2.222
\n\\end{pmatrix}.
\n\\]<\/p>\n

Check the change in the diagonal elements:<\/p>\n

\\[
\n\\max\\left(|4.778-4.5|,\\ |2.222-2.5|\\right)=0.278.
\n\\]<\/p>\n

Since<\/p>\n

\\[
\n0.278>0.1,
\n\\]<\/p>\n

one more iteration is needed.<\/p>\n

We decompose the matrix<\/p>\n

\\[
\nA_2=
\n\\begin{pmatrix}
\n4.778 & 0.309\\\\
\n2 & 2.222
\n\\end{pmatrix}.
\n\\]<\/p>\n

Perform the LU decomposition:<\/p>\n

\\[
\nA_2=L_2\\cdot U_2.
\n\\]<\/p>\n

Calculate the elements of the matrices \\( L_2 \\) and \\( U_2 \\):<\/p>\n

\\[
\n\\begin{gathered}
\nl_{11}^{2}=a_{11}^{(2)}=4.778,\\\\[4pt]
\nu_{12}^{2}=\\dfrac{a_{12}^{(2)}}{l_{11}^{2}}
\n=\\dfrac{0.309}{4.778}=0.065,\\\\[4pt]
\nl_{21}^{2}=a_{21}^{(2)}=2,\\\\[4pt]
\nl_{22}^{2}=a_{22}^{(2)}-l_{21}^{2}\\cdot u_{12}^{2}
\n=2.222-2\\cdot 0.065=2.093.
\n\\end{gathered}
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\nL_2=
\n\\begin{pmatrix}
\n4.778 & 0\\\\
\n2 & 2.093
\n\\end{pmatrix},
\n\\qquad
\nU_2=
\n\\begin{pmatrix}
\n1 & 0.065\\\\
\n0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n

Construct<\/p>\n

\\[
\nA_3=U_2\\cdot L_2.
\n\\]<\/p>\n

Substitute the matrices we found:<\/p>\n

\\[
\nA_3=
\n\\begin{pmatrix}
\n1 & 0.065\\\\
\n0 & 1
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n4.778 & 0\\\\
\n2 & 2.093
\n\\end{pmatrix}.
\n\\]<\/p>\n

After the calculations:<\/p>\n

\\[
\nA_3=
\n\\begin{pmatrix}
\n4.907 & 0.135\\\\
\n2 & 2.093
\n\\end{pmatrix}.
\n\\]<\/p>\n

Check the stopping criterion:<\/p>\n

\\[
\n\\max\\left(|4.907-4.778|,\\ |2.093-2.222|\\right)=0.129.
\n\\]<\/p>\n

Since<\/p>\n

\\[
\n0.129>0.1,
\n\\]<\/p>\n

we do not stop the iterations yet.<\/p>\n

We decompose the matrix<\/p>\n

\\[
\nA_3=
\n\\begin{pmatrix}
\n4.907 & 0.135\\\\
\n2 & 2.093
\n\\end{pmatrix}.
\n\\]<\/p>\n

Perform the LU decomposition:<\/p>\n

\\[
\nA_3=L_3\\cdot U_3.
\n\\]<\/p>\n

Calculate the elements of the matrices \\( L_3 \\) and \\( U_3 \\):<\/p>\n

\\[
\n\\begin{gathered}
\nl_{11}^{3}=a_{11}^{(3)}=4.907,\\\\[4pt]
\nu_{12}^{3}=\\dfrac{a_{12}^{(3)}}{l_{11}^{3}}
\n=\\dfrac{0.135}{4.907}=0.028,\\\\[4pt]
\nl_{21}^{3}=a_{21}^{(3)}=2,\\\\[4pt]
\nl_{22}^{3}=a_{22}^{(3)}-l_{21}^{3}\\cdot u_{12}^{3}
\n=2.093-2\\cdot 0.028=2.038.
\n\\end{gathered}
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\nL_3=
\n\\begin{pmatrix}
\n4.907 & 0\\\\
\n2 & 2.038
\n\\end{pmatrix},
\n\\qquad
\nU_3=
\n\\begin{pmatrix}
\n1 & 0.028\\\\
\n0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n

Construct<\/p>\n

\\[
\nA_4=U_3\\cdot L_3.
\n\\]<\/p>\n

Substitute the matrices:<\/p>\n

\\[
\nA_4=
\n\\begin{pmatrix}
\n1 & 0.028\\\\
\n0 & 1
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n4.907 & 0\\\\
\n2 & 2.038
\n\\end{pmatrix}.
\n\\]<\/p>\n

After the calculations, we have<\/p>\n

\\[
\nA_4=
\n\\begin{pmatrix}
\n4.962 & 0.056\\\\
\n2 & 2.038
\n\\end{pmatrix}.
\n\\]<\/p>\n

Now<\/p>\n

\\[
\n\\max\\left(|4.962-4.907|,\\ |2.038-2.093|\\right)=0.055.
\n\\]<\/p>\n

Since<\/p>\n

\\[
\n0.055\\le 0.1,
\n\\]<\/p>\n

the stopping condition is satisfied.<\/p>\n

Therefore, the required eigenvalues are<\/p>\n

\\[
\n\\lambda_1=4.962,
\n\\qquad
\n\\lambda_2=2.038.
\n\\]<\/p>\n

Example 2. Find the eigenvalues of the matrix using the LU decomposition method with tolerance \\( \\varepsilon=0.1 \\):
\n\\[
\nA=
\n\\begin{pmatrix}
\n6 & 1 & 0\\\\
\n1 & 4 & 1\\\\
\n0 & 1 & 2
\n\\end{pmatrix}.
\n\\]<\/h3>\n

Denote the initial matrix by<\/p>\n

\\[
\nA_0=A=
\n\\begin{pmatrix}
\n6 & 1 & 0\\\\
\n1 & 4 & 1\\\\
\n0 & 1 & 2
\n\\end{pmatrix}.
\n\\]<\/p>\n

Perform the decomposition<\/p>\n

\\[
\nA_0=L_0\\cdot U_0.
\n\\]<\/p>\n

Calculate the elements of the first column and the first row:<\/p>\n

\\[
\n\\begin{gathered}
\nl_{11}^{0}=a_{11}^{(0)}=6,\\\\[4pt]
\nu_{12}^{0}=\\dfrac{a_{12}^{(0)}}{l_{11}^{0}}
\n=\\dfrac{1}{6}=0.167,\\\\[4pt]
\nu_{13}^{0}=\\dfrac{a_{13}^{(0)}}{l_{11}^{0}}
\n=\\dfrac{0}{6}=0,\\\\[4pt]
\nl_{21}^{0}=a_{21}^{(0)}=1,\\qquad
\nl_{31}^{0}=a_{31}^{(0)}=0.
\n\\end{gathered}
\n\\]<\/p>\n

Next, find the elements of the second column and the second row:<\/p>\n

\\[
\n\\begin{gathered}
\nl_{22}^{0}=a_{22}^{(0)}-l_{21}^{0}\\cdot u_{12}^{0}
\n=4-1\\cdot 0.167=3.833,\\\\[4pt]
\nu_{23}^{0}=\\dfrac{a_{23}^{(0)}-l_{21}^{0}\\cdot u_{13}^{0}}{l_{22}^{0}}
\n=\\dfrac{1-1\\cdot 0}{3.833}=0.261,\\\\[4pt]
\nl_{32}^{0}=a_{32}^{(0)}-l_{31}^{0}\\cdot u_{12}^{0}
\n=1-0\\cdot 0.167=1.
\n\\end{gathered}
\n\\]<\/p>\n

The last diagonal element of the matrix \\( L_0 \\) is<\/p>\n

\\[
\nl_{33}^{0}=a_{33}^{(0)}-l_{31}^{0}\\cdot u_{13}^{0}-l_{32}^{0}\\cdot u_{23}^{0}
\n=2-0\\cdot 0-1\\cdot 0.261=1.739.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\nL_0=
\n\\begin{pmatrix}
\n6 & 0 & 0\\\\
\n1 & 3.833 & 0\\\\
\n0 & 1 & 1.739
\n\\end{pmatrix},
\n\\qquad
\nU_0=
\n\\begin{pmatrix}
\n1 & 0.167 & 0\\\\
\n0 & 1 & 0.261\\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n

Construct<\/p>\n

\\[
\nA_1=U_0\\cdot L_0.
\n\\]<\/p>\n

Substitute the matrices we found:<\/p>\n

\\[
\nA_1=
\n\\begin{pmatrix}
\n1 & 0.167 & 0\\\\
\n0 & 1 & 0.261\\\\
\n0 & 0 & 1
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n6 & 0 & 0\\\\
\n1 & 3.833 & 0\\\\
\n0 & 1 & 1.739
\n\\end{pmatrix}.
\n\\]<\/p>\n

After the calculations:<\/p>\n

\\[
\nA_1=
\n\\begin{pmatrix}
\n6.167 & 0.639 & 0\\\\
\n1 & 4.094 & 0.454\\\\
\n0 & 1 & 1.739
\n\\end{pmatrix}.
\n\\]<\/p>\n

Check the stopping condition:<\/p>\n

\\[
\n\\max\\left(|6.167-6|,\\ |4.094-4|,\\ |1.739-2|\\right)=0.261.
\n\\]<\/p>\n

Since<\/p>\n

\\[
\n0.261>0.1,
\n\\]<\/p>\n

we continue the calculations.<\/p>\n

Now decompose the matrix<\/p>\n

\\[
\nA_1=
\n\\begin{pmatrix}
\n6.167 & 0.639 & 0\\\\
\n1 & 4.094 & 0.454\\\\
\n0 & 1 & 1.739
\n\\end{pmatrix}.
\n\\]<\/p>\n

Perform the decomposition<\/p>\n

\\[
\nA_1=L_1\\cdot U_1.
\n\\]<\/p>\n

Calculate the elements of the first column and the first row:<\/p>\n

\\[
\n\\begin{gathered}
\nl_{11}^{1}=a_{11}^{(1)}=6.167,\\\\[4pt]
\nu_{12}^{1}=\\dfrac{a_{12}^{(1)}}{l_{11}^{1}}
\n=\\dfrac{0.639}{6.167}=0.104,\\\\[4pt]
\nu_{13}^{1}=\\dfrac{a_{13}^{(1)}}{l_{11}^{1}}
\n=\\dfrac{0}{6.167}=0,\\\\[4pt]
\nl_{21}^{1}=a_{21}^{(1)}=1,\\qquad
\nl_{31}^{1}=a_{31}^{(1)}=0.
\n\\end{gathered}
\n\\]<\/p>\n

Next, find the elements of the second column and the second row:<\/p>\n

\\[
\n\\begin{gathered}
\nl_{22}^{1}=a_{22}^{(1)}-l_{21}^{1}\\cdot u_{12}^{1}
\n=4.094-1\\cdot 0.104=3.99,\\\\[4pt]
\nu_{23}^{1}=\\dfrac{a_{23}^{(1)}-l_{21}^{1}\\cdot u_{13}^{1}}{l_{22}^{1}}
\n=\\dfrac{0.454-1\\cdot 0}{3.99}=0.114,\\\\[4pt]
\nl_{32}^{1}=a_{32}^{(1)}-l_{31}^{1}\\cdot u_{12}^{1}
\n=1-0\\cdot 0.104=1.
\n\\end{gathered}
\n\\]<\/p>\n

The last diagonal element of the matrix \\( L_1 \\) is<\/p>\n

\\[
\nl_{33}^{1}=a_{33}^{(1)}-l_{31}^{1}\\cdot u_{13}^{1}-l_{32}^{1}\\cdot u_{23}^{1}
\n=1.739-0\\cdot 0-1\\cdot 0.114=1.625.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\nL_1=
\n\\begin{pmatrix}
\n6.167 & 0 & 0\\\\
\n1 & 3.99 & 0\\\\
\n0 & 1 & 1.625
\n\\end{pmatrix},
\n\\qquad
\nU_1=
\n\\begin{pmatrix}
\n1 & 0.104 & 0\\\\
\n0 & 1 & 0.114\\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n

Construct<\/p>\n

\\[
\nA_2=U_1\\cdot L_1.
\n\\]<\/p>\n

Substitute \\( U_1 \\) and \\( L_1 \\):<\/p>\n

\\[
\nA_2=
\n\\begin{pmatrix}
\n1 & 0.104 & 0\\\\
\n0 & 1 & 0.114\\\\
\n0 & 0 & 1
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n6.167 & 0 & 0\\\\
\n1 & 3.99 & 0\\\\
\n0 & 1 & 1.625
\n\\end{pmatrix}.
\n\\]<\/p>\n

After the calculations:<\/p>\n

\\[
\nA_2=
\n\\begin{pmatrix}
\n6.27 & 0.413 & 0\\\\
\n1 & 4.104 & 0.185\\\\
\n0 & 1 & 1.625
\n\\end{pmatrix}.
\n\\]<\/p>\n

Check the criterion:<\/p>\n

\\[
\n\\max\\left(|6.27-6.167|,\\ |4.104-4.094|,\\ |1.625-1.739|\\right)=0.114.
\n\\]<\/p>\n

Since<\/p>\n

\\[
\n0.114>0.1,
\n\\]<\/p>\n

one more iteration is needed.<\/p>\n

We decompose the matrix<\/p>\n

\\[
\nA_2=
\n\\begin{pmatrix}
\n6.27 & 0.413 & 0\\\\
\n1 & 4.104 & 0.185\\\\
\n0 & 1 & 1.625
\n\\end{pmatrix}.
\n\\]<\/p>\n

Perform the decomposition<\/p>\n

\\[
\nA_2=L_2\\cdot U_2.
\n\\]<\/p>\n

Calculate the elements of the first column and the first row:<\/p>\n

\\[
\n\\begin{gathered}
\nl_{11}^{2}=a_{11}^{(2)}=6.27,\\\\[4pt]
\nu_{12}^{2}=\\dfrac{a_{12}^{(2)}}{l_{11}^{2}}
\n=\\dfrac{0.413}{6.270}=0.066,\\\\[4pt]
\nu_{13}^{2}=\\dfrac{a_{13}^{(2)}}{l_{11}^{2}}
\n=\\dfrac{0}{6.27}=0,\\\\[4pt]
\nl_{21}^{2}=a_{21}^{(2)}=1,\\qquad
\nl_{31}^{2}=a_{31}^{(2)}=0.
\n\\end{gathered}
\n\\]<\/p>\n

Next, find the elements of the second column and the second row:<\/p>\n

\\[
\n\\begin{gathered}
\nl_{22}^{2}=a_{22}^{(2)}-l_{21}^{2}\\cdot u_{12}^{2}
\n=4.104-1\\cdot 0.066=4.038,\\\\[4pt]
\nu_{23}^{2}=\\dfrac{a_{23}^{(2)}-l_{21}^{2}\\cdot u_{13}^{2}}{l_{22}^{2}}
\n=\\dfrac{0.185-1\\cdot 0}{4.038}=0.046,\\\\[4pt]
\nl_{32}^{2}=a_{32}^{(2)}-l_{31}^{2}\\cdot u_{12}^{2}
\n=1-0\\cdot 0.066=1.
\n\\end{gathered}
\n\\]<\/p>\n

The last diagonal element of the matrix \\( L_2 \\) is<\/p>\n

\\[
\nl_{33}^{2}=a_{33}^{(2)}-l_{31}^{2}\\cdot u_{13}^{2}-l_{32}^{2}\\cdot u_{23}^{2}
\n=1.625-0\\cdot 0-1\\cdot 0.046=1.58.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\nL_2=
\n\\begin{pmatrix}
\n6.270 & 0 & 0\\\\
\n1 & 4.038 & 0\\\\
\n0 & 1 & 1.58
\n\\end{pmatrix},
\n\\qquad
\nU_2=
\n\\begin{pmatrix}
\n1 & 0.066 & 0\\\\
\n0 & 1 & 0.046\\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n

Construct<\/p>\n

\\[
\nA_3=U_2\\cdot L_2.
\n\\]<\/p>\n

Substitute the matrices:<\/p>\n

\\[
\nA_3=
\n\\begin{pmatrix}
\n1 & 0.066 & 0\\\\
\n0 & 1 & 0.046\\\\
\n0 & 0 & 1
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n6.27 & 0 & 0\\\\
\n1 & 4.038 & 0\\\\
\n0 & 1 & 1.58
\n\\end{pmatrix}.
\n\\]<\/p>\n

After the calculations:<\/p>\n

\\[
\nA_3=
\n\\begin{pmatrix}
\n6.336 & 0.266 & 0\\\\
\n1 & 4.084 & 0.072\\\\
\n0 & 1 & 1.58
\n\\end{pmatrix}.
\n\\]<\/p>\n

Check the stopping condition:<\/p>\n

\\[
\n\\max\\left(|6.336-6.27|,\\ |4.084-4.104|,\\ |1.58-1.625|\\right)=0.066.
\n\\]<\/p>\n

Since<\/p>\n

\\[
\n0.066\\le 0.1,
\n\\]<\/p>\n

we stop the iterative process.<\/p>\n

Therefore, the required eigenvalues are<\/p>\n

\\[
\n\\lambda_1=6.336,
\n\\qquad
\n\\lambda_2=4.084,
\n\\qquad
\n\\lambda_3=1.58.
\n\\]<\/p>\n

Example 3. Find the eigenvalues of the matrix using the LU decomposition method with tolerance \\( \\varepsilon=0.1 \\):
\n\\[
\nA=
\n\\begin{pmatrix}
\n7 & 1 & 0 & 0\\\\
\n1 & 5 & 1 & 0\\\\
\n0 & 1 & 3 & 1\\\\
\n0 & 0 & 1 & 1
\n\\end{pmatrix}.
\n\\]<\/h3>\n

Denote the initial matrix by<\/p>\n

\\[
\nA_0=A=
\n\\begin{pmatrix}
\n7 & 1 & 0 & 0\\\\
\n1 & 5 & 1 & 0\\\\
\n0 & 1 & 3 & 1\\\\
\n0 & 0 & 1 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n

Perform the decomposition<\/p>\n

\\[
\nA_0=L_0\\cdot U_0.
\n\\]<\/p>\n

Calculate the first column and the first row:<\/p>\n

\\[
\n\\begin{gathered}
\nl_{11}^{0}=a_{11}^{(0)}=7,\\\\[4pt]
\nu_{12}^{0}=\\dfrac{a_{12}^{(0)}}{l_{11}^{0}}
\n=\\dfrac{1}{7}=0.143,\\\\[4pt]
\nu_{13}^{0}=\\dfrac{a_{13}^{(0)}}{l_{11}^{0}}
\n=\\dfrac{0}{7}=0,\\\\[4pt]
\nu_{14}^{0}=\\dfrac{a_{14}^{(0)}}{l_{11}^{0}}
\n=\\dfrac{0}{7}=0,\\\\[4pt]
\nl_{21}^{0}=a_{21}^{(0)}=1,\\qquad
\nl_{31}^{0}=a_{31}^{(0)}=0,\\qquad
\nl_{41}^{0}=a_{41}^{(0)}=0.
\n\\end{gathered}
\n\\]<\/p>\n

Next, find the elements of the second column and the second row:<\/p>\n

\\[
\n\\begin{gathered}
\nl_{22}^{0}=a_{22}^{(0)}-l_{21}^{0}\\cdot u_{12}^{0}
\n=5-1\\cdot 0.143=4.857,\\\\[4pt]
\nu_{23}^{0}=\\dfrac{a_{23}^{(0)}-l_{21}^{0}\\cdot u_{13}^{0}}{l_{22}^{0}}
\n=\\dfrac{1-1\\cdot 0}{4.857}=0.206,\\\\[4pt]
\nu_{24}^{0}=\\dfrac{a_{24}^{(0)}-l_{21}^{0}\\cdot u_{14}^{0}}{l_{22}^{0}}
\n=\\dfrac{0-1\\cdot 0}{4.857}=0,\\\\[4pt]
\nl_{32}^{0}=a_{32}^{(0)}-l_{31}^{0}\\cdot u_{12}^{0}
\n=1-0\\cdot 0.143=1,\\\\[4pt]
\nl_{42}^{0}=a_{42}^{(0)}-l_{41}^{0}\\cdot u_{12}^{0}
\n=0-0\\cdot 0.143=0.
\n\\end{gathered}
\n\\]<\/p>\n

Now calculate the elements of the third column and the third row:<\/p>\n

\\[
\n\\begin{gathered}
\nl_{33}^{0}=a_{33}^{(0)}-l_{31}^{0}\\cdot u_{13}^{0}-l_{32}^{0}\\cdot u_{23}^{0}
\n=3-0\\cdot 0-1\\cdot 0.206=2.794,\\\\[4pt]
\nu_{34}^{0}=\\dfrac{a_{34}^{(0)}-l_{31}^{0}\\cdot u_{14}^{0}-l_{32}^{0}\\cdot u_{24}^{0}}{l_{33}^{0}}
\n=\\dfrac{1-0\\cdot 0-1\\cdot 0}{2.794}=0.358,\\\\[4pt]
\nl_{43}^{0}=a_{43}^{(0)}-l_{41}^{0}\\cdot u_{13}^{0}-l_{42}^{0}\\cdot u_{23}^{0}
\n=1-0\\cdot 0-0\\cdot 0.206=1.
\n\\end{gathered}
\n\\]<\/p>\n

The last diagonal element of the matrix \\( L_0 \\) is<\/p>\n

\\[
\nl_{44}^{0}=a_{44}^{(0)}-l_{41}^{0}\\cdot u_{14}^{0}-l_{42}^{0}\\cdot u_{24}^{0}-l_{43}^{0}\\cdot u_{34}^{0}
\n=1-0\\cdot 0-0\\cdot 0-1\\cdot 0.358=0.642.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\nL_0=
\n\\begin{pmatrix}
\n7 & 0 & 0 & 0\\\\
\n1 & 4.857 & 0 & 0\\\\
\n0 & 1 & 2.794 & 0\\\\
\n0 & 0 & 1 & 0.642
\n\\end{pmatrix},
\n\\qquad
\nU_0=
\n\\begin{pmatrix}
\n1 & 0.143 & 0 & 0\\\\
\n0 & 1 & 0.206 & 0\\\\
\n0 & 0 & 1 & 0.358\\\\
\n0 & 0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n

Construct<\/p>\n

\\[
\nA_1=U_0\\cdot L_0.
\n\\]<\/p>\n

Substitute the matrices we found:<\/p>\n

\\[
\nA_1=
\n\\begin{pmatrix}
\n1 & 0.143 & 0 & 0\\\\
\n0 & 1 & 0.206 & 0\\\\
\n0 & 0 & 1 & 0.358\\\\
\n0 & 0 & 0 & 1
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n7 & 0 & 0 & 0\\\\
\n1 & 4.857 & 0 & 0\\\\
\n0 & 1 & 2.794 & 0\\\\
\n0 & 0 & 1 & 0.642
\n\\end{pmatrix}.
\n\\]<\/p>\n

After the calculations, we have<\/p>\n

\\[
\nA_1=
\n\\begin{pmatrix}
\n7.143 & 0.694 & 0 & 0\\\\
\n1 & 5.063 & 0.575 & 0\\\\
\n0 & 1 & 3.152 & 0.23\\\\
\n0 & 0 & 1 & 0.642
\n\\end{pmatrix}.
\n\\]<\/p>\n

Check the stopping condition:<\/p>\n

\\[
\n\\max\\left(|7.143-7|,\\ |5.063-5|,\\ |3.152-3|,\\ |0.642-1|\\right)=0.358.
\n\\]<\/p>\n

Since<\/p>\n

\\[
\n0.358>0.1,
\n\\]<\/p>\n

one more iteration is needed.<\/p>\n

We decompose the matrix<\/p>\n

\\[
\nA_1=
\n\\begin{pmatrix}
\n7.143 & 0.694 & 0 & 0\\\\
\n1 & 5.063 & 0.575 & 0\\\\
\n0 & 1 & 3.152 & 0.23\\\\
\n0 & 0 & 1 & 0.642
\n\\end{pmatrix}.
\n\\]<\/p>\n

Perform the decomposition<\/p>\n

\\[
\nA_1=L_1\\cdot U_1.
\n\\]<\/p>\n

Calculate the first column and the first row:<\/p>\n

\\[
\n\\begin{gathered}
\nl_{11}^{1}=a_{11}^{(1)}=7.143,\\\\[4pt]
\nu_{12}^{1}=\\dfrac{a_{12}^{(1)}}{l_{11}^{1}}
\n=\\dfrac{0.694}{7.143}=0.097,\\\\[4pt]
\nu_{13}^{1}=\\dfrac{a_{13}^{(1)}}{l_{11}^{1}}
\n=\\dfrac{0}{7.143}=0,\\\\[4pt]
\nu_{14}^{1}=\\dfrac{a_{14}^{(1)}}{l_{11}^{1}}
\n=\\dfrac{0}{7.143}=0,\\\\[4pt]
\nl_{21}^{1}=a_{21}^{(1)}=1,\\qquad
\nl_{31}^{1}=a_{31}^{(1)}=0,\\qquad
\nl_{41}^{1}=a_{41}^{(1)}=0.
\n\\end{gathered}
\n\\]<\/p>\n

Next, find the elements of the second column and the second row:<\/p>\n

\\[
\n\\begin{gathered}
\nl_{22}^{1}=a_{22}^{(1)}-l_{21}^{1}\\cdot u_{12}^{1}
\n=5.063-1\\cdot 0.097=4.966,\\\\[4pt]
\nu_{23}^{1}=\\dfrac{a_{23}^{(1)}-l_{21}^{1}\\cdot u_{13}^{1}}{l_{22}^{1}}
\n=\\dfrac{0.575-1\\cdot 0}{4.966}=0.116,\\\\[4pt]
\nu_{24}^{1}=\\dfrac{a_{24}^{(1)}-l_{21}^{1}\\cdot u_{14}^{1}}{l_{22}^{1}}
\n=\\dfrac{0-1\\cdot 0}{4.966}=0,\\\\[4pt]
\nl_{32}^{1}=a_{32}^{(1)}-l_{31}^{1}\\cdot u_{12}^{1}
\n=1-0\\cdot 0.097=1,\\\\[4pt]
\nl_{42}^{1}=a_{42}^{(1)}-l_{41}^{1}\\cdot u_{12}^{1}
\n=0-0\\cdot 0.097=0.
\n\\end{gathered}
\n\\]<\/p>\n

Now calculate the elements of the third column and the third row:<\/p>\n

\\[
\n\\begin{gathered}
\nl_{33}^{1}=a_{33}^{(1)}-l_{31}^{1}\\cdot u_{13}^{1}-l_{32}^{1}\\cdot u_{23}^{1}
\n=3.152-0\\cdot 0-1\\cdot 0.116=3.036,\\\\[4pt]
\nu_{34}^{1}=\\dfrac{a_{34}^{(1)}-l_{31}^{1}\\cdot u_{14}^{1}-l_{32}^{1}\\cdot u_{24}^{1}}{l_{33}^{1}}
\n=\\dfrac{0.23-0\\cdot 0-1\\cdot 0}{3.036}=0.076,\\\\[4pt]
\nl_{43}^{1}=a_{43}^{(1)}-l_{41}^{1}\\cdot u_{13}^{1}-l_{42}^{1}\\cdot u_{23}^{1}
\n=1-0\\cdot 0-0\\cdot 0.116=1.
\n\\end{gathered}
\n\\]<\/p>\n

The last diagonal element of the matrix \\( L_1 \\) is<\/p>\n

\\[
\nl_{44}^{1}=a_{44}^{(1)}-l_{41}^{1}\\cdot u_{14}^{1}-l_{42}^{1}\\cdot u_{24}^{1}-l_{43}^{1}\\cdot u_{34}^{1}
\n=0.642-0\\cdot 0-0\\cdot 0-1\\cdot 0.076=0.566.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\nL_1=
\n\\begin{pmatrix}
\n7.143 & 0 & 0 & 0\\\\
\n1 & 4.966 & 0 & 0\\\\
\n0 & 1 & 3.036 & 0\\\\
\n0 & 0 & 1 & 0.566
\n\\end{pmatrix},
\n\\qquad
\nU_1=
\n\\begin{pmatrix}
\n1 & 0.097 & 0 & 0\\\\
\n0 & 1 & 0.116 & 0\\\\
\n0 & 0 & 1 & 0.076\\\\
\n0 & 0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n

Construct<\/p>\n

\\[
\nA_2=U_1\\cdot L_1.
\n\\]<\/p>\n

Substitute \\( U_1 \\) and \\( L_1 \\):<\/p>\n

\\[
\nA_2=
\n\\begin{pmatrix}
\n1 & 0.097 & 0 & 0\\\\
\n0 & 1 & 0.116 & 0\\\\
\n0 & 0 & 1 & 0.076\\\\
\n0 & 0 & 0 & 1
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n7.143 & 0 & 0 & 0\\\\
\n1 & 4.966 & 0 & 0\\\\
\n0 & 1 & 3.036 & 0\\\\
\n0 & 0 & 1 & 0.566
\n\\end{pmatrix}.
\n\\]<\/p>\n

After the calculations, we have<\/p>\n

\\[
\nA_2=
\n\\begin{pmatrix}
\n7.24 & 0.482 & 0 & 0\\\\
\n1 & 5.082 & 0.352 & 0\\\\
\n0 & 1 & 3.112 & 0.043\\\\
\n0 & 0 & 1 & 0.566
\n\\end{pmatrix}.
\n\\]<\/p>\n

Check the stopping criterion:<\/p>\n

\\[
\n\\max\\left(|7.24-7.143|,\\ |5.082-5.063|,\\ |3.112-3.152|,\\ |0.566-0.642|\\right)=0.097.
\n\\]<\/p>\n

Since<\/p>\n

\\[
\n0.097\\le 0.1,
\n\\]<\/p>\n

we stop the iterative process.<\/p>\n

Therefore, the required eigenvalues are<\/p>\n

\\[
\n\\lambda_1=7.24,
\n\\qquad
\n\\lambda_2=5.082,
\n\\qquad
\n\\lambda_3=3.112,
\n\\qquad
\n\\lambda_4=0.566.
\n\\]<\/p>\n

What to Study Next: Other Numerical Methods<\/h2>\n

After LU decomposition, it is worth looking at other numerical approaches. They solve similar problems, but they work in different ways. This makes it easier to see which method is convenient in different situations.<\/p>\n

    \n
  1. Rotation Method: How a Matrix Is Gradually Simplified<\/a> \u2014 The article will explain how to find the eigenvalues of a symmetric matrix using successive rotations and the transition to diagonal form.<\/li>\n
  2. Power Method: How to Find the Largest Eigenvalue<\/a> \u2014 The article will show how the power method helps approximately find the eigenvalue of a matrix with the largest absolute value.<\/li>\n
  3. Deflation Method: How to Find the Next Eigenvalue<\/a> \u2014 The article will discuss the case when one eigenvalue is already known, and the next one is found using an additional transformation of the matrix.<\/li>\n<\/ol>\n

    Eigenvalues of a Matrix: Algorithm for Program Implementation<\/h2>\n

    If you enjoy programming, this method can easily be turned into a small educational project. The flowchart below shows the logic of an algorithm that performs the iterative process based on LU decomposition for a \\( 2\\times 2 \\) matrix, controls the accuracy of the calculations, and outputs approximate eigenvalues of the matrix.<\/p>\n

    Try to implement this algorithm in Pascal<\/em>, Python<\/em><\/a>, C++<\/em>, JavaScript<\/em>, or another language you are comfortable with. This way, you will not only understand the method better, but also see how a mathematical idea gradually turns into working program code.<\/p>\n

    \"Flowchart<\/p>\n","protected":false},"excerpt":{"rendered":"

    Eigenvalues of a matrix can be found not only through the characteristic polynomial. In numerical methods, another approach is often<\/p>\n","protected":false},"author":1,"featured_media":4044,"comment_status":"open","ping_status":"closed","sticky":false,"template":"template-centered.php","format":"standard","meta":{"footnotes":""},"categories":[532],"tags":[533,417,556,557,134],"class_list":["post-4020","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-eigenvalues-and-eigenvectors-of","tag-eigenvalues","tag-linear-algebra","tag-lu-decomposition","tag-lu-method","tag-numerical-methods"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/4020","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/comments?post=4020"}],"version-history":[{"count":22,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/4020\/revisions"}],"predecessor-version":[{"id":4045,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/4020\/revisions\/4045"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/media\/4044"}],"wp:attachment":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/media?parent=4020"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/categories?post=4020"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/tags?post=4020"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}