\\[
\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
We need to find its eigenvalues. To do this, we usually form the characteristic equation<\/p>\n
\\[
\n\\det(A-\\lambda\\cdot I)=0,
\n\\]<\/p>\n
where \\( I \\) is the identity matrix, and \\( \\lambda \\) is an unknown number.<\/p>\n
The roots of this equation<\/p>\n
\\[
\n\\lambda_1,\\lambda_2,\\dots,\\lambda_n
\n\\]<\/p>\n
are the eigenvalues of the matrix \\( A \\).<\/p>\n
However, directly expanding the determinant<\/p>\n
\\[
\n\\det(A-\\lambda\\cdot I)
\n\\]<\/p>\n
is not always convenient. If the matrix has a high order, the calculations quickly become cumbersome. That is why numerical methods<\/a> use approaches that help simplify the construction of the characteristic polynomial.<\/p>\n The Danilevsky method works exactly in this way. First, it transforms the matrix \\( A \\) into a special form. Then, using this form, we write the characteristic equation. In other words, the method does not change the problem itself. It simply makes it more convenient for calculations.<\/p>\n In the Danilevsky method, the original matrix \\( A \\) is transformed into the Frobenius normal form. It has the following form:<\/p>\n \\[ This matrix has a very important feature. Its characteristic polynomial can be easily written using the elements of the first row. That is exactly why this form is useful.<\/p>\n Let us write the characteristic determinant of the matrix<\/a> \\( P \\):<\/p>\n \\[ After simplification, we get<\/p>\n \\[ The factor \\( (-1)^n \\) does not affect the roots of the equation. Therefore, the characteristic equation can be written as follows:<\/p>\n \\[ So, after transforming the matrix \\( A \\) into Frobenius form, the coefficients of the characteristic equation are already essentially available. We can simply read them from the first row of the matrix \\( P \\).<\/p>\n Now it is important to understand why we can replace the original matrix \\( A \\) with the matrix \\( P \\). After all, we are looking for the eigenvalues of the matrix \\( A \\), right?<\/p>\n The Danilevsky method uses similarity transformations. A matrix \\( B \\) is called similar to a matrix \\( A \\) if it can be written in the form<\/p>\n \\[ where \\( M \\) is a nonsingular matrix.<\/p>\n To see why the eigenvalues do not change, it is enough to check the characteristic determinant. We have<\/p>\n \\[ Since<\/p>\n \\[ we get<\/p>\n \\[ From this, we obtain<\/p>\n \\[ Therefore,<\/p>\n \\[ Using the property of the determinant of a product of matrices, we have<\/p>\n \\[ But<\/p>\n \\[ So,<\/p>\n \\[ This means that similar matrices have the same characteristic equations. And if the characteristic equations are the same, then their roots are also the same.<\/p>\n That is why the Danilevsky method does not change the eigenvalues of a matrix. It only transforms the matrix into a form from which the characteristic polynomial can be written more conveniently.<\/p>\n Now let us move on to the mechanism of the method itself. How exactly is the matrix \\( A \\) gradually transformed into a Frobenius matrix?<\/p>\n Denote the initial matrix as<\/p>\n \\[ Next, we perform the transformations one after another:<\/p>\n \\[ After all transformations, we obtain<\/p>\n \\[ Here, one indexing system is used: at each step, the value of \\( k \\) decreases from \\( n-1 \\) to \\( 1 \\). Thanks to this, the matrix is gradually transformed into Frobenius form from bottom to top.<\/p>\n At the step with number \\( k \\), we work with the current matrix \\( A^{(n-k-1)} \\). To construct the matrix \\( M_k \\), we use the elements of the \\( (k+1) \\)-th row of this current matrix:<\/p>\n \\[ The element \\( a_{k+1,k}^{(n-k-1)} \\) is especially important. It is called the pivot element. To perform the transformation, we need \\( a_{k+1,k}^{(n-k-1)} \\neq 0 \\). If this element is equal to zero, division by zero occurs. In this case, an additional interchange of rows and the corresponding columns is performed in order to preserve the similarity transformation.<\/p>\n The matrix \\( M_k \\) almost coincides with the identity matrix. Only its \\( k \\)-th row is different. Its elements are defined as follows:<\/p>\n \\[ Here, \\( \\delta_{ij} \\) is the Kronecker delta:<\/p>\n \\[ So, all rows of the matrix \\( M_k \\), except the \\( k \\)-th row, remain the same as in the identity matrix. The \\( k \\)-th row is changed in a special way so that, after the transformation, the required row of the current matrix takes the form of a row of the Frobenius matrix.<\/p>\n The inverse matrix<\/a> \\( M_k^{-1} \\) also has a simple form. It coincides with the identity matrix in all rows except the \\( k \\)-th row:<\/p>\n \\[ Thus, one step of the Danilevsky method has a clear structure. First, we take the current matrix. Then, using its \\( (k+1) \\)-th row, we construct the matrices \\( M_k \\) and \\( M_k^{-1} \\). After that, we perform the similarity transformation<\/p>\n \\[ As a result, the matrix becomes closer to the Frobenius normal form.<\/p>\n The general idea of the method can be shown as follows:<\/p>\n \\[ So, the Danilevsky method first transforms the original matrix into Frobenius form. Then, using the first row of the obtained matrix, we write the characteristic equation. And the roots of this equation are the eigenvalues of the original matrix.<\/p>\n Now let us look at how the Danilevsky method works with specific matrices. In each example, we will transform the given matrix into Frobenius form. Then, using its first row, we will write the characteristic equation. We will move step by step so that it is clear where each transformation matrix comes from.<\/p>\n \\[ We have a \\( 2\\times2 \\) matrix, so we need to perform only one step of the Danilevsky method.<\/p>\n Denote<\/p>\n \\[ For a \\( 2\\times2 \\) matrix, we work with the second row. The pivot element is \\( a_{21}^{(0)}=3 \\). Since \\( a_{21}^{(0)}\\neq 0 \\), we can construct the matrices \\( M_1 \\) and \\( M_1^{-1} \\).<\/p>\n The matrix \\( M_1 \\) has the form<\/p>\n \\[ Substitute the values:<\/p>\n \\[ The inverse matrix has the form<\/p>\n \\[ Therefore,<\/p>\n \\[ Now we perform the similarity transformation:<\/p>\n \\[ Substitute the matrices:<\/p>\n \\[ After multiplication, we get<\/p>\n \\[ This is already a Frobenius matrix:<\/p>\n \\[ For a Frobenius matrix of order two, the characteristic equation has the form<\/p>\n \\[ In our case,<\/p>\n \\[ So,<\/p>\n \\[ Now factor the quadratic trinomial:<\/p>\n \\[ Therefore,<\/p>\n \\[ From this, we get<\/p>\n \\[ \\[ We have a \\( 3\\times3 \\) matrix. However, let us pay attention to its third row:<\/p>\n \\[ It already has the form of the corresponding row of a Frobenius matrix. Indeed, for a \\( 3\\times3 \\) matrix, Frobenius form looks like this:<\/p>\n \\[ So, the bottom row is already ready. Therefore, the step involving the construction of \\( M_2 \\) does not need to be performed. We only need to perform one step of the Danilevsky method to transform the second row into the form<\/p>\n \\[ Denote<\/p>\n \\[ We work with the second row of the matrix \\( A^{(0)} \\). The pivot element is \\( a_{21}^{(0)}=1 \\). Since \\( a_{21}^{(0)}\\neq 0 \\), we can construct the matrices \\( M_1 \\) and \\( M_1^{-1} \\).<\/p>\n The matrix \\( M_1 \\) differs from the identity matrix only in the first row:<\/p>\n \\[ Substitute the values from the second row of the matrix \\( A^{(0)} \\):<\/p>\n \\[ Thus,<\/p>\n \\[ The inverse matrix has the form<\/p>\n \\[ Therefore,<\/p>\n \\[ Now we perform the similarity transformation:<\/p>\n \\[ Substitute the matrices:<\/p>\n \\[ After multiplication, we get<\/p>\n \\[ So, the matrix already has Frobenius form:<\/p>\n \\[ Now we write the characteristic equation:<\/p>\n \\[ In our case,<\/p>\n \\[ Therefore,<\/p>\n \\[ Now factor the polynomial:<\/p>\n \\[ From this, we get the eigenvalues:<\/p>\n \\[ \\[ We have a \\( 3\\times3 \\) matrix, so in general we need to perform two steps of the Danilevsky method.<\/p>\n Denote<\/p>\n \\[ First, we take \\( k=2 \\). We work with the third row of the matrix \\( A^{(0)} \\). The pivot element is \\( a_{32}^{(0)}=-1 \\). Since \\( a_{32}^{(0)}\\neq 0 \\), we can construct the matrices \\( M_2 \\) and \\( M_2^{-1} \\).<\/p>\n The matrix \\( M_2 \\) has the form<\/p>\n \\[ Substitute the values:<\/p>\n \\[ After simplification, we have<\/p>\n \\[ The inverse matrix is<\/p>\n \\[ Therefore,<\/p>\n \\[ Now we perform the first transformation:<\/p>\n \\[ Substitute the matrices:<\/p>\n \\[ After multiplication, we get<\/p>\n \\[ The last row already has the required form:<\/p>\n \\[ Now we perform the second step. We take \\( k=1 \\). We work with the second row of the matrix \\( A^{(1)} \\). The pivot element is \\( a_{21}^{(1)}=-4 \\). Since \\( a_{21}^{(1)}\\neq 0 \\), we construct the matrices \\( M_1 \\) and \\( M_1^{-1} \\).<\/p>\n The matrix \\( M_1 \\) has the form<\/p>\n \\[ Substitute the values:<\/p>\n \\[ So,<\/p>\n \\[ The inverse matrix is<\/p>\n \\[ Therefore,<\/p>\n \\[ Now we perform the second transformation:<\/p>\n \\[ Substitute the matrices:<\/p>\n \\[ After multiplication, we have<\/p>\n \\[ Now both the second and third rows coincide with the corresponding rows of the Frobenius matrix. Therefore, we have obtained the Frobenius matrix:<\/p>\n \\[ Now we write the characteristic equation:<\/p>\n \\[ In our case,<\/p>\n \\[ Therefore,<\/p>\n \\[ So,<\/p>\n \\[ From this, we get<\/p>\n \\[ Note.<\/strong> We can see that one eigenvalue is repeated. This is not a mistake, because a characteristic equation may have multiple roots.<\/p><\/blockquote>\n The Danilevsky method clearly shows how the eigenvalues of a matrix can be found through matrix transformations. But this topic does not end here. To better understand the connection between different numerical approaches, it is worth exploring a few related directions.<\/p>\nFrobenius Form: What Form the Matrix Is Transformed Into<\/h2>\n
\nP =
\n\\begin{pmatrix}
\np_1 & p_2 & p_3 & \\dots & p_{n-1} & p_n \\\\
\n1 & 0 & 0 & \\dots & 0 & 0 \\\\
\n0 & 1 & 0 & \\dots & 0 & 0 \\\\
\n0 & 0 & 1 & \\dots & 0 & 0 \\\\
\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots & \\vdots \\\\
\n0 & 0 & 0 & \\dots & 1 & 0
\n\\end{pmatrix}.
\n\\]<\/p>\n
\n\\chi_P(\\lambda)=
\n\\begin{vmatrix}
\np_1-\\lambda & p_2 & p_3 & \\dots & p_n \\\\
\n1 & -\\lambda & 0 & \\dots & 0 \\\\
\n0 & 1 & -\\lambda & \\dots & 0 \\\\
\n0 & 0 & 1 & \\dots & 0 \\\\
\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\
\n0 & 0 & 0 & \\dots & -\\lambda
\n\\end{vmatrix}.
\n\\]<\/p>\n
\n\\chi_P(\\lambda)
\n=
\n(-1)^n\\cdot
\n\\left(
\n\\lambda^n
\n-p_1\\cdot \\lambda^{n-1}
\n-p_2\\cdot \\lambda^{n-2}
\n-p_3\\cdot \\lambda^{n-3}
\n-\\dots
\n-p_n
\n\\right).
\n\\]<\/p>\n
\n\\lambda^n
\n-p_1\\cdot \\lambda^{n-1}
\n-p_2\\cdot \\lambda^{n-2}
\n-p_3\\cdot \\lambda^{n-3}
\n-\\dots
\n-p_n=0.
\n\\]<\/p>\nSimilarity Transformations: Why Eigenvalues Do Not Change<\/h2>\n
\nB=M^{-1}\\cdot A \\cdot M,
\n\\]<\/p>\n
\n\\det(B-\\lambda\\cdot I)
\n=
\n\\det(M^{-1}\\cdot A\\cdot M-\\lambda\\cdot I).
\n\\]<\/p>\n
\n\\lambda\\cdot I=M^{-1}\\cdot (\\lambda\\cdot I)\\cdot M,
\n\\]<\/p>\n
\nM^{-1}\\cdot A\\cdot M-\\lambda\\cdot I
\n=M^{-1}\\cdot A\\cdot M-M^{-1}\\cdot (\\lambda\\cdot I)\\cdot M.
\n\\]<\/p>\n
\nB-\\lambda\\cdot I
\n=
\nM^{-1}\\cdot (A-\\lambda\\cdot I)\\cdot M.
\n\\]<\/p>\n
\n\\det(B-\\lambda\\cdot I)
\n=
\n\\det\\left(M^{-1}\\cdot (A-\\lambda\\cdot I)\\cdot M\\right).
\n\\]<\/p>\n
\n\\det(B-\\lambda\\cdot I)
\n=
\n\\det(M^{-1})\\cdot \\det(A-\\lambda\\cdot I)\\cdot \\det(M).
\n\\]<\/p>\n
\n\\det(M^{-1})\\cdot \\det(M)=1.
\n\\]<\/p>\n
\n\\det(B-\\lambda\\cdot I)=\\det(A-\\lambda\\cdot I).
\n\\]<\/p>\nTransformation Matrices: The Main Mechanism of the Danilevsky Method<\/h2>\n
\nA^{(0)}=A.
\n\\]<\/p>\n
\nA^{(n-k)}
\n=
\nM_k^{-1}\\cdot A^{(n-k-1)}\\cdot M_k,
\n\\qquad
\nk=n-1,n-2,\\dots,1.
\n\\]<\/p>\n
\nP=A^{(n-1)}.
\n\\]<\/p>\n
\na_{k+1,1}^{(n-k-1)},
\na_{k+1,2}^{(n-k-1)},
\n\\dots,
\na_{k+1,n}^{(n-k-1)}.
\n\\]<\/p>\n
\nM_k:\\;
\n\\left\\{
\n\\begin{array}{lll}
\nm_{ij}=\\delta_{ij}, & i=1,2,\\dots,n,\\; i\\neq k, & j=1,2,\\dots,n,\\\\[6pt]
\nm_{kj}=-\\dfrac{a_{k+1,j}^{(n-k-1)}}{a_{k+1,k}^{(n-k-1)}}, & j=1,2,\\dots,n,\\; j\\neq k, & \\\\[12pt]
\nm_{kk}=\\dfrac{1}{a_{k+1,k}^{(n-k-1)}}. & &
\n\\end{array}
\n\\right.
\n\\]<\/p>\n
\n\\delta_{ij}=
\n\\begin{cases}
\n1, & i=j, \\\\
\n0, & i\\neq j.
\n\\end{cases}
\n\\]<\/p>\n
\nM_k^{-1}:\\;
\n\\left\\{
\n\\begin{array}{lll}
\nm_{ij}=\\delta_{ij}, & i=1,2,\\dots,n,\\; i\\neq k, & j=1,2,\\dots,n,\\\\[6pt]
\nm_{kj}=a_{k+1,j}^{(n-k-1)}, & j=1,2,\\dots,n. &
\n\\end{array}
\n\\right.
\n\\]<\/p>\n
\nA^{(n-k)}
\n=
\nM_k^{-1}\\cdot A^{(n-k-1)}\\cdot M_k.
\n\\]<\/p>\n
\nA
\n\\longrightarrow
\nP
\n\\longrightarrow
\n\\lambda^n
\n-p_1\\cdot \\lambda^{n-1}
\n-p_2\\cdot \\lambda^{n-2}
\n-p_3\\cdot \\lambda^{n-3}
\n-\\dots
\n-p_n=0
\n\\longrightarrow
\n\\lambda_1,\\lambda_2,\\dots,\\lambda_n.
\n\\]<\/p>\nEigenvalues of a Matrix: Practical Part With Examples<\/h2>\n
Example 1. Find the eigenvalues of the matrix<\/h3>\n
\nA=
\n\\begin{pmatrix}
\n2 & 1 \\\\
\n3 & 4
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nA^{(0)}=A=
\n\\begin{pmatrix}
\n2 & 1 \\\\
\n3 & 4
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_1=
\n\\begin{pmatrix}
\n\\dfrac{1}{a_{21}^{(0)}} & -\\dfrac{a_{22}^{(0)}}{a_{21}^{(0)}} \\\\
\n0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_1=
\n\\begin{pmatrix}
\n\\dfrac{1}{3} & -\\dfrac{4}{3} \\\\
\n0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_1^{-1}=
\n\\begin{pmatrix}
\na_{21}^{(0)} & a_{22}^{(0)} \\\\
\n0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_1^{-1}=
\n\\begin{pmatrix}
\n3 & 4 \\\\
\n0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nA^{(1)}=M_1^{-1}\\cdot A^{(0)}\\cdot M_1.
\n\\]<\/p>\n
\nA^{(1)}=
\n\\begin{pmatrix}
\n3 & 4 \\\\
\n0 & 1
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n2 & 1 \\\\
\n3 & 4
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n\\dfrac{1}{3} & -\\dfrac{4}{3} \\\\
\n0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nA^{(1)}=
\n\\begin{pmatrix}
\n6 & -5 \\\\
\n1 & 0
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nP=
\n\\begin{pmatrix}
\n6 & -5 \\\\
\n1 & 0
\n\\end{pmatrix}.
\n\\]<\/p>\n
\n\\lambda^2-p_1\\cdot\\lambda-p_2=0.
\n\\]<\/p>\n
\np_1=6,
\n\\qquad
\np_2=-5.
\n\\]<\/p>\n
\n\\lambda^2-6\\cdot\\lambda+5=0.
\n\\]<\/p>\n
\n\\lambda^2-6\\cdot\\lambda+5=
\n(\\lambda-1)\\cdot(\\lambda-5).
\n\\]<\/p>\n
\n(\\lambda-1)\\cdot(\\lambda-5)=0.
\n\\]<\/p>\n
\n\\lambda_1=1,
\n\\qquad
\n\\lambda_2=5.
\n\\]<\/p>\nExample 2. Find the eigenvalues of the matrix<\/h3>\n
\nA=
\n\\begin{pmatrix}
\n3 & 0 & 0 \\\\
\n1 & 1 & 2 \\\\
\n0 & 1 & 0
\n\\end{pmatrix}.
\n\\]<\/p>\n
\n0,\\quad 1,\\quad 0.
\n\\]<\/p>\n
\nP=
\n\\begin{pmatrix}
\np_1 & p_2 & p_3 \\\\
\n1 & 0 & 0 \\\\
\n0 & 1 & 0
\n\\end{pmatrix}.
\n\\]<\/p>\n
\n1,\\quad 0,\\quad 0.
\n\\]<\/p>\n
\nA^{(0)}=
\n\\begin{pmatrix}
\n3 & 0 & 0 \\\\
\n1 & 1 & 2 \\\\
\n0 & 1 & 0
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_1=
\n\\begin{pmatrix}
\n\\dfrac{1}{a_{21}^{(0)}} & -\\dfrac{a_{22}^{(0)}}{a_{21}^{(0)}} & -\\dfrac{a_{23}^{(0)}}{a_{21}^{(0)}} \\\\
\n0 & 1 & 0 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_1=
\n\\begin{pmatrix}
\n\\dfrac{1}{1} & -\\dfrac{1}{1} & -\\dfrac{2}{1} \\\\
\n0 & 1 & 0 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_1=
\n\\begin{pmatrix}
\n1 & -1 & -2 \\\\
\n0 & 1 & 0 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_1^{-1}=
\n\\begin{pmatrix}
\na_{21}^{(0)} & a_{22}^{(0)} & a_{23}^{(0)} \\\\
\n0 & 1 & 0 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_1^{-1}=
\n\\begin{pmatrix}
\n1 & 1 & 2 \\\\
\n0 & 1 & 0 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nA^{(1)}=M_1^{-1}\\cdot A^{(0)} \\cdot M_1.
\n\\]<\/p>\n
\nA^{(1)}=
\n\\begin{pmatrix}
\n1 & 1 & 2 \\\\
\n0 & 1 & 0 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n3 & 0 & 0 \\\\
\n1 & 1 & 2 \\\\
\n0 & 1 & 0
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n1 & -1 & -2 \\\\
\n0 & 1 & 0 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nA^{(1)}=
\n\\begin{pmatrix}
\n4 & -1 & -6 \\\\
\n1 & 0 & 0 \\\\
\n0 & 1 & 0
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nP=
\n\\begin{pmatrix}
\n4 & -1 & -6 \\\\
\n1 & 0 & 0 \\\\
\n0 & 1 & 0
\n\\end{pmatrix}.
\n\\]<\/p>\n
\n\\lambda^3-p_1\\cdot\\lambda^2-p_2\\cdot\\lambda-p_3=0.
\n\\]<\/p>\n
\np_1=4,
\n\\qquad
\np_2=-1,
\n\\qquad
\np_3=-6.
\n\\]<\/p>\n
\n\\lambda^3-4\\cdot\\lambda^2+\\lambda+6=0.
\n\\]<\/p>\n
\n\\lambda^3-4\\cdot\\lambda^2+\\lambda+6
\n=
\n(\\lambda-3)\\cdot (\\lambda-2)\\cdot (\\lambda+1).
\n\\]<\/p>\n
\n\\lambda_1=3,
\n\\qquad
\n\\lambda_2=2,
\n\\qquad
\n\\lambda_3=-1.
\n\\]<\/p>\nExample 3. Find the eigenvalues of the matrix<\/h3>\n
\nA=
\n\\begin{pmatrix}
\n-4 & -1 & -3 \\\\
\n4 & 3 & 6 \\\\
\n0 & -1 & -2
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nA^{(0)}=
\n\\begin{pmatrix}
\n-4 & -1 & -3 \\\\
\n4 & 3 & 6 \\\\
\n0 & -1 & -2
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_2=
\n\\begin{pmatrix}
\n1 & 0 & 0 \\\\
\n-\\dfrac{a_{31}^{(0)}}{a_{32}^{(0)}} & \\dfrac{1}{a_{32}^{(0)}} & -\\dfrac{a_{33}^{(0)}}{a_{32}^{(0)}} \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_2=
\n\\begin{pmatrix}
\n1 & 0 & 0 \\\\
\n-\\dfrac{0}{-1} & \\dfrac{1}{-1} & -\\dfrac{-2}{-1} \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_2=
\n\\begin{pmatrix}
\n1 & 0 & 0 \\\\
\n0 & -1 & -2 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_2^{-1}=
\n\\begin{pmatrix}
\n1 & 0 & 0 \\\\
\na_{31}^{(0)} & a_{32}^{(0)} & a_{33}^{(0)} \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_2^{-1}=
\n\\begin{pmatrix}
\n1 & 0 & 0 \\\\
\n0 & -1 & -2 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nA^{(1)}=M_2^{-1}\\cdot A^{(0)} \\cdot M_2.
\n\\]<\/p>\n
\nA^{(1)}=
\n\\begin{pmatrix}
\n1 & 0 & 0 \\\\
\n0 & -1 & -2 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n-4 & -1 & -3 \\\\
\n4 & 3 & 6 \\\\
\n0 & -1 & -2
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n1 & 0 & 0 \\\\
\n0 & -1 & -2 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nA^{(1)}=
\n\\begin{pmatrix}
\n-4 & 1 & -1 \\\\
\n-4 & 1 & 0 \\\\
\n0 & 1 & 0
\n\\end{pmatrix}.
\n\\]<\/p>\n
\n0,\\quad 1,\\quad 0.
\n\\]<\/p>\n
\nM_1=
\n\\begin{pmatrix}
\n\\dfrac{1}{a_{21}^{(1)}} & -\\dfrac{a_{22}^{(1)}}{a_{21}^{(1)}} & -\\dfrac{a_{23}^{(1)}}{a_{21}^{(1)}} \\\\
\n0 & 1 & 0 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_1=
\n\\begin{pmatrix}
\n\\dfrac{1}{-4} & -\\dfrac{1}{-4} & -\\dfrac{0}{-4} \\\\
\n0 & 1 & 0 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_1=
\n\\begin{pmatrix}
\n-\\dfrac{1}{4} & \\dfrac{1}{4} & 0 \\\\
\n0 & 1 & 0 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_1^{-1}=
\n\\begin{pmatrix}
\na_{21}^{(1)} & a_{22}^{(1)} & a_{23}^{(1)} \\\\
\n0 & 1 & 0 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nM_1^{-1}=
\n\\begin{pmatrix}
\n-4 & 1 & 0 \\\\
\n0 & 1 & 0 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nA^{(2)}=M_1^{-1}\\cdot A^{(1)} \\cdot M_1.
\n\\]<\/p>\n
\nA^{(2)}=
\n\\begin{pmatrix}
\n-4 & 1 & 0 \\\\
\n0 & 1 & 0 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n-4 & 1 & -1 \\\\
\n-4 & 1 & 0 \\\\
\n0 & 1 & 0
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n-\\dfrac{1}{4} & \\dfrac{1}{4} & 0 \\\\
\n0 & 1 & 0 \\\\
\n0 & 0 & 1
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nA^{(2)}=
\n\\begin{pmatrix}
\n-3 & 0 & 4 \\\\
\n1 & 0 & 0 \\\\
\n0 & 1 & 0
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nP=
\n\\begin{pmatrix}
\n-3 & 0 & 4 \\\\
\n1 & 0 & 0 \\\\
\n0 & 1 & 0
\n\\end{pmatrix}.
\n\\]<\/p>\n
\n\\lambda^3-p_1\\cdot \\lambda^2-p_2\\cdot \\lambda-p_3=0.
\n\\]<\/p>\n
\np_1=-3,
\n\\qquad
\np_2=0,
\n\\qquad
\np_3=4.
\n\\]<\/p>\n
\n\\lambda^3+3\\cdot \\lambda^2-4=0.
\n\\]
\nNow factor the polynomial:
\n\\[
\n\\lambda^3+3\\cdot \\lambda^2-4
\n=
\n(\\lambda-1)\\cdot (\\lambda+2)^2.
\n\\]<\/p>\n
\n(\\lambda-1)\\cdot (\\lambda+2)^2=0.
\n\\]<\/p>\n
\n\\lambda_1=1,
\n\\qquad
\n\\lambda_2=-2,
\n\\qquad
\n\\lambda_3=-2.
\n\\]<\/p>\nWhat to Consider Next: Topics for Further Learning<\/h2>\n
\n
Eigenvalues of a Matrix: From a Flowchart to Your Own Code<\/h2>\n
![\"A](\"https:\/\/www.mathros.net.ua\/en\/wp-content\/uploads\/2026\/05\/eigenvalues-of-a-matrix-danilevsky-method1.jpg\")
<\/p>\n","protected":false},"excerpt":{"rendered":"