\\[
\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
To find the eigenvalues of the matrix \\( A \\), we need to construct its characteristic polynomial<\/a><\/p>\n \\[ where \\( I \\) is the identity matrix of order \\( n \\).<\/p>\n In its general form, this polynomial can be written as<\/p>\n \\[ Then the eigenvalues of the matrix \\( A \\) are found from the equation<\/p>\n \\[ So, the main task of the method is to find the coefficients<\/p>\n \\[ Of course, we could directly expand the determinant<\/a> \\( \\det(\\lambda\\cdot I-A) \\). But is this always convenient? For higher-order matrices, this approach quickly becomes bulky. That is why the Leverrier method offers a different approach.<\/p>\n First, let us look at what the coefficients \\( \\sigma_1,\\sigma_2,\\dots,\\sigma_n \\) mean.<\/p>\n Let \\( \\lambda_1,\\lambda_2,\\dots,\\lambda_n \\) be the eigenvalues of the matrix \\( A \\), written with their algebraic multiplicities taken into account. Then the characteristic polynomial can be represented as<\/p>\n \\[ After expanding the brackets, we get the following polynomial:<\/p>\n \\[ But this same characteristic polynomial was already written in the form<\/p>\n \\[ Now let us compare the coefficients of the same powers of \\( \\lambda \\). This comparison shows that the coefficients \\( \\sigma_1,\\sigma_2,\\dots,\\sigma_n \\) are the elementary symmetric functions of the eigenvalues:<\/p>\n \\[ But there is an important point here. The eigenvalues themselves are still unknown. Therefore, we cannot find the coefficients directly using these formulas. We need a way to calculate them through the matrix \\( A \\) itself. This is exactly where the Leverrier method comes in.<\/p>\n The Leverrier method does not use the individual eigenvalues themselves. Instead, it uses their power sums. Let us denote<\/p>\n \\[ For example,<\/p>\n \\[ At first glance, it may seem that these sums also cannot be found without knowing the eigenvalues. But here an important property of matrices comes into play.<\/p>\n The sum of the eigenvalues of a matrix is equal to its trace. Therefore,<\/p>\n \\[ Since the trace of a matrix is the sum of the elements on its main diagonal, we have<\/p>\n \\[ And what happens for powers of the matrix? The eigenvalues of the matrix \\( A^k \\) are the numbers<\/p>\n \\[ So, the sum of these numbers is equal to the trace of the matrix \\( A^k \\). Therefore,<\/p>\n \\[ To write this in more detail, let us denote the elements of the matrix \\( A^k \\) as follows:<\/p>\n \\[ Then, for the second power of the matrix, we have<\/p>\n \\[ Similarly, for the third power,<\/p>\n \\[ In its general form,<\/p>\n \\[ Here we need to be careful. For example, \\( \\operatorname{tr}(A^2) \\) is the trace of the matrix \\( A^2 \\). First, we need to find the product \\( A^2=A\\cdot A \\), and only after that add the elements on the main diagonal of this new matrix.<\/p>\n In other words,<\/p>\n \\[ in the general case. This is a common mistake, so it is worth paying special attention to it.<\/p>\n Thus, through the matrix \\( A \\), we can find the values<\/p>\n \\[ And then, using them, we can move on to the coefficients of the characteristic polynomial.<\/p>\n Now we have everything we need for the main part of the method. The power sums are known:<\/p>\n \\[ and we need to find the coefficients<\/p>\n \\[ For this, we use Newton\u2019s identities, also called the Newton\u2013Girard formulas. They connect the power sums of the eigenvalues \\( S_k \\) with the coefficients of the characteristic polynomial \\( \\sigma_k \\).<\/p>\n For our form of the characteristic polynomial, these formulas look like this:<\/p>\n \\[ The formulas continue in the same pattern. In general form,<\/p>\n \\[ Using these equalities, we can find the coefficients of the characteristic polynomial step by step.<\/p>\n From the first formula, we have<\/p>\n \\[ Since<\/p>\n \\[ we get<\/p>\n \\[ From the second formula,<\/p>\n \\[ Therefore,<\/p>\n \\[ From the third formula,<\/p>\n \\[ So,<\/p>\n \\[ From the fourth formula, in the same way, we get<\/p>\n \\[ Each next coefficient is calculated through the coefficients already found and the traces of powers of the matrix. That is why the Leverrier method is sequential: first \\( \\sigma_1 \\), then \\( \\sigma_2 \\), then \\( \\sigma_3 \\), and so on up to \\( \\sigma_n \\).<\/p>\n In its general form, the recurrence formula is<\/p>\n \\[ where<\/p>\n \\[ After finding all the coefficients, we construct the characteristic polynomial<\/p>\n \\[ Then the eigenvalues of the matrix are found as the roots of the characteristic equation<\/p>\n \\[ So, the Leverrier method reduces the process of finding eigenvalues to three main actions: calculate the traces of powers of the matrix, use them to find the coefficients of the characteristic polynomial, and solve the characteristic equation.<\/p>\n Now let us see how the Leverrier method is applied in practice. In each example, we will work step by step: first, we will find the traces of powers of the matrix, then calculate the coefficients of the characteristic polynomial, and after that find its roots.<\/p>\n \\[ For a \\( 2\\times2 \\) matrix, the characteristic polynomial has the form<\/p>\n \\[ So, we need to find the coefficients \\( \\sigma_1,\\sigma_2 \\).<\/p>\n First, we calculate the first power sum:<\/p>\n \\[ Therefore,<\/p>\n \\[ Next, we find \\( A^2 \\):<\/p>\n \\[ Now we calculate the trace of the matrix \\( A^2 \\):<\/p>\n \\[ Using the Leverrier formula, we have<\/p>\n \\[ Substitute the values we found:<\/p>\n \\[ So, the characteristic polynomial is<\/p>\n \\[ Now let us find its roots:<\/p>\n \\[ Factor the quadratic polynomial:<\/p>\n \\[ Therefore,<\/p>\n \\[ Therefore, we get the eigenvalues:<\/p>\n \\[ \\[ For a \\( 3\\times3 \\) matrix, we write the characteristic polynomial as<\/p>\n \\[ So, we need to find three coefficients \\( \\sigma_1,\\sigma_2,\\sigma_3 \\).<\/p>\n We start with the trace of the matrix \\( A \\):<\/p>\n \\[ Therefore,<\/p>\n \\[ Next, we calculate \\( A^2 \\):<\/p>\n \\[ Find the second power sum:<\/p>\n \\[ Now we find the second coefficient:<\/p>\n \\[ Substitute the values:<\/p>\n \\[ Now we move on to the third coefficient. To do this, we need to find \\( A^3 \\). Multiply the matrix \\( A^2 \\) by the matrix \\( A \\):<\/p>\n \\[ We have<\/p>\n \\[ Therefore,<\/p>\n \\[ Now we use the formula<\/p>\n \\[ Substitute the values we found:<\/p>\n \\[ Calculate:<\/p>\n \\[ So,<\/p>\n \\[ Therefore, the characteristic polynomial is<\/p>\n \\[ Now let us find the roots of the equation<\/p>\n \\[ Factor the polynomial:<\/p>\n \\[ Therefore,<\/p>\n \\[ Therefore,<\/p>\n \\[ \\[ For a \\( 4\\times4 \\) matrix, the characteristic polynomial has the form<\/p>\n \\[ So, we need to find the coefficients \\( \\sigma_1,\\sigma_2,\\sigma_3,\\sigma_4 \\).<\/p>\n Because of the zero elements, matrix multiplication here is relatively convenient, but the algorithm remains the same.<\/p>\n We start with the first power sum:<\/p>\n \\[ Therefore,<\/p>\n \\[ Next, we find \\( A^2 \\):<\/p>\n \\[ Therefore,<\/p>\n \\[ Find the second coefficient:<\/p>\n \\[ Substitute the values we found:<\/p>\n \\[ Now we calculate \\( A^3 \\). To do this, multiply the matrix \\( A^2 \\) by the matrix \\( A \\):<\/p>\n \\[ We get<\/p>\n \\[ Therefore,<\/p>\n \\[ We find the third coefficient using the formula<\/p>\n \\[ Substitute the values:<\/p>\n \\[ So,<\/p>\n \\[ It remains to find the fourth coefficient. For this, we calculate \\( A^4 \\). Multiply the matrix \\( A^3 \\) by the matrix \\( A \\):<\/p>\n \\[ We have<\/p>\n \\[ Therefore,<\/p>\n \\[ For the fourth coefficient, we use the formula<\/p>\n \\[ Substitute:<\/p>\n \\[ Calculate:<\/p>\n \\[ So, we have found the coefficients:<\/p>\n \\[ Therefore, the characteristic polynomial is<\/p>\n \\[ Now let us find its roots:<\/p>\n \\[ Factor the polynomial:<\/p>\n \\[ Therefore,<\/p>\n \\[ Therefore, we get the eigenvalues:<\/p>\n \\[ After the Leverrier method, it is worth comparing it with other approaches to finding eigenvalues. This makes it easier to see how different algorithms differ from one another and when each of them is convenient to use.<\/p>\n If you enjoy programming, the Leverrier method can be viewed not only as theoretical material, but also as a ready-to-implement algorithm. The flowchart below helps you see the full logic of the program: from entering the elements of the matrix to constructing the characteristic polynomial and finding the eigenvalues.<\/p>\n
\n\\chi(\\lambda)=\\det(\\lambda\\cdot I-A),
\n\\]<\/p>\n
\n\\chi(\\lambda)
\n=
\n\\lambda^n
\n-\\sigma_1\\cdot \\lambda^{n-1}
\n+\\sigma_2\\cdot \\lambda^{n-2}
\n-\\dots
\n+(-1)^n\\cdot\\sigma_n.
\n\\]<\/p>\n
\n\\chi(\\lambda)=0.
\n\\]<\/p>\n
\n\\sigma_1,\\sigma_2,\\dots,\\sigma_n.
\n\\]<\/p>\n
\n\\chi(\\lambda)
\n=
\n(\\lambda-\\lambda_1)\\cdot
\n(\\lambda-\\lambda_2)\\cdot
\n\\dots\\cdot
\n(\\lambda-\\lambda_n).
\n\\]<\/p>\n
\n\\begin{aligned}
\n\\chi(\\lambda)
\n=\\lambda^n
\n-(\\lambda_1+\\lambda_2+\\dots+\\lambda_n)\\cdot \\lambda^{n-1}
\n+\\left(\\sum_{1\\le i<j\\le n}\\lambda_i\\cdot\\lambda_j\\right)\\cdot \\lambda^{n-2} \\\\
\n-\\left(\\sum_{1\\le i<j<k\\le n}\\lambda_i\\cdot\\lambda_j\\cdot\\lambda_k\\right)\\cdot \\lambda^{n-3}
\n+\\dots
\n+(-1)^n\\cdot \\lambda_1\\cdot\\lambda_2\\cdot\\dots\\cdot\\lambda_n.
\n\\end{aligned}
\n\\]<\/p>\n
\n\\chi(\\lambda)
\n=
\n\\lambda^n
\n-\\sigma_1\\cdot \\lambda^{n-1}
\n+\\sigma_2\\cdot \\lambda^{n-2}
\n-\\sigma_3\\cdot \\lambda^{n-3}
\n+\\dots
\n+(-1)^n\\cdot\\sigma_n.
\n\\]<\/p>\n
\n\\begin{gathered}
\n\\sigma_1=\\lambda_1+\\lambda_2+\\dots+\\lambda_n, \\\\[4pt]
\n\\sigma_2=\\sum_{1\\le i<j\\le n}\\lambda_i\\cdot\\lambda_j, \\\\[4pt]
\n\\sigma_3=\\sum_{1\\le i<j<k\\le n}\\lambda_i\\cdot\\lambda_j\\cdot\\lambda_k, \\\\[4pt]
\n\\dots \\\\[4pt]
\n\\sigma_n=\\lambda_1\\cdot\\lambda_2\\cdot\\dots\\cdot\\lambda_n.
\n\\end{gathered}
\n\\]<\/p>\nTraces of Matrix Powers: How to Get the Needed Sums<\/h2>\n
\nS_k=
\n\\lambda_1^k+\\lambda_2^k+\\dots+\\lambda_n^k,
\n\\qquad
\nk=1,2,\\dots,n.
\n\\]<\/p>\n
\n\\begin{gathered}
\nS_1=\\lambda_1+\\lambda_2+\\dots+\\lambda_n, \\\\[4pt]
\nS_2=\\lambda_1^2+\\lambda_2^2+\\dots+\\lambda_n^2, \\\\[4pt]
\nS_3=\\lambda_1^3+\\lambda_2^3+\\dots+\\lambda_n^3.
\n\\end{gathered}
\n\\]<\/p>\n
\nS_1=\\operatorname{tr}(A).
\n\\]<\/p>\n
\nS_1
\n=
\n\\operatorname{tr}(A)
\n=
\na_{11}+a_{22}+\\dots+a_{nn}
\n=
\n\\sum_{i=1}^{n}a_{ii}.
\n\\]<\/p>\n
\n\\lambda_1^k,\\lambda_2^k,\\dots,\\lambda_n^k.
\n\\]<\/p>\n
\nS_k=\\operatorname{tr}(A^k).
\n\\]<\/p>\n
\nA^k=
\n\\begin{pmatrix}
\na_{11}^{(k)} & a_{12}^{(k)} & \\dots & a_{1n}^{(k)} \\\\
\na_{21}^{(k)} & a_{22}^{(k)} & \\dots & a_{2n}^{(k)} \\\\
\n\\vdots & \\vdots & \\ddots & \\vdots \\\\
\na_{n1}^{(k)} & a_{n2}^{(k)} & \\dots & a_{nn}^{(k)}
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nS_2
\n=
\n\\operatorname{tr}(A^2)
\n=
\na_{11}^{(2)}+a_{22}^{(2)}+\\dots+a_{nn}^{(2)}
\n=
\n\\sum_{i=1}^{n}a_{ii}^{(2)}.
\n\\]<\/p>\n
\nS_3
\n=
\n\\operatorname{tr}(A^3)
\n=
\na_{11}^{(3)}+a_{22}^{(3)}+\\dots+a_{nn}^{(3)}
\n=
\n\\sum_{i=1}^{n}a_{ii}^{(3)}.
\n\\]<\/p>\n
\nS_k
\n=
\n\\operatorname{tr}(A^k)
\n=
\na_{11}^{(k)}+a_{22}^{(k)}+\\dots+a_{nn}^{(k)}
\n=
\n\\sum_{i=1}^{n}a_{ii}^{(k)},
\n\\qquad
\nk=1,2,\\dots,n.
\n\\]<\/p>\n
\n\\operatorname{tr}(A^2)
\n\\neq
\na_{11}^2+a_{22}^2+\\dots+a_{nn}^2
\n\\]<\/p>\n
\nS_1,S_2,\\dots,S_n.
\n\\]<\/p>\nLeverrier Method: Sequential Calculation of the Coefficients<\/h2>\n
\nS_1,S_2,\\dots,S_n,
\n\\]<\/p>\n
\n\\sigma_1,\\sigma_2,\\dots,\\sigma_n.
\n\\]<\/p>\n
\n\\begin{gathered}
\nS_1-\\sigma_1=0, \\\\[4pt]
\nS_2-\\sigma_1\\cdot S_1+2\\cdot\\sigma_2=0, \\\\[4pt]
\nS_3-\\sigma_1\\cdot S_2+\\sigma_2\\cdot S_1-3\\cdot\\sigma_3=0, \\\\[4pt]
\nS_4-\\sigma_1\\cdot S_3+\\sigma_2\\cdot S_2-\\sigma_3\\cdot S_1+4\\cdot\\sigma_4=0.
\n\\end{gathered}
\n\\]<\/p>\n
\nS_k
\n-\\sigma_1\\cdot S_{k-1}
\n+\\sigma_2\\cdot S_{k-2}
\n-\\dots
\n+(-1)^{k-1}\\cdot\\sigma_{k-1}\\cdot S_1
\n+(-1)^k\\cdot k\\cdot\\sigma_k
\n=0.
\n\\]<\/p>\n
\n\\sigma_1=S_1.
\n\\]<\/p>\n
\nS_1=\\operatorname{tr}(A),
\n\\]<\/p>\n
\n\\sigma_1=\\operatorname{tr}(A).
\n\\]<\/p>\n
\nS_2-\\sigma_1\\cdot S_1+2\\cdot\\sigma_2=0.
\n\\]<\/p>\n
\n\\sigma_2=
\n\\frac{1}{2}\\cdot
\n(\\sigma_1\\cdot S_1-S_2).
\n\\]<\/p>\n
\nS_3-\\sigma_1\\cdot S_2+\\sigma_2\\cdot S_1-3\\cdot\\sigma_3=0.
\n\\]<\/p>\n
\n\\sigma_3=
\n\\frac{1}{3}\\cdot
\n(\\sigma_2\\cdot S_1-\\sigma_1\\cdot S_2+S_3).
\n\\]<\/p>\n
\n\\sigma_4=
\n\\frac{1}{4}\\cdot
\n(\\sigma_3\\cdot S_1-\\sigma_2\\cdot S_2+\\sigma_1\\cdot S_3-S_4).
\n\\]<\/p>\n
\n\\sigma_k=
\n\\frac{1}{k}\\cdot
\n\\left(
\n\\sigma_{k-1}\\cdot S_1
\n-\\sigma_{k-2}\\cdot S_2
\n+\\sigma_{k-3}\\cdot S_3
\n-\\dots
\n+(-1)^{k-1}\\cdot\\sigma_0\\cdot S_k
\n\\right),
\n\\]<\/p>\n
\nk=1,2,\\dots,n,
\n\\qquad
\n\\sigma_0=1.
\n\\]<\/p>\n
\n\\chi(\\lambda)
\n=
\n\\lambda^n
\n-\\sigma_1\\cdot \\lambda^{n-1}
\n+\\sigma_2\\cdot \\lambda^{n-2}
\n-\\dots
\n+(-1)^n\\cdot\\sigma_n.
\n\\]<\/p>\n
\n\\chi(\\lambda)=0.
\n\\]<\/p>\nPractical Part: How to Find Eigenvalues Using the Leverrier Method<\/h2>\n
Example 1. Find the eigenvalues of a matrix<\/h3>\n
\nA=
\n\\begin{pmatrix}
\n4 & 1\\\\
\n2 & 3
\n\\end{pmatrix}.
\n\\]<\/p>\n
\n\\chi(\\lambda)
\n=
\n\\lambda^2-\\sigma_1\\cdot\\lambda+\\sigma_2.
\n\\]<\/p>\n
\nS_1=\\operatorname{tr}(A)=4+3=7.
\n\\]<\/p>\n
\n\\sigma_1=S_1=7.
\n\\]<\/p>\n
\nA^2=
\n\\begin{pmatrix}
\n4 & 1\\\\
\n2 & 3
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n4 & 1\\\\
\n2 & 3
\n\\end{pmatrix}
\n=
\n\\begin{pmatrix}
\n18 & 7\\\\
\n14 & 11
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nS_2=\\operatorname{tr}(A^2)=18+11=29.
\n\\]<\/p>\n
\n\\sigma_2=
\n\\frac{1}{2}\\cdot(\\sigma_1\\cdot S_1-S_2).
\n\\]<\/p>\n
\n\\sigma_2=
\n\\frac{1}{2}\\cdot(7\\cdot7-29)
\n=
\n\\frac{1}{2}\\cdot(49-29)
\n=
\n10.
\n\\]<\/p>\n
\n\\chi(\\lambda)
\n=
\n\\lambda^2-7\\cdot\\lambda+10.
\n\\]<\/p>\n
\n\\lambda^2-7\\cdot\\lambda+10=0.
\n\\]<\/p>\n
\n\\lambda^2-7\\cdot\\lambda+10
\n=
\n(\\lambda-2)\\cdot(\\lambda-5).
\n\\]<\/p>\n
\n(\\lambda-2)\\cdot(\\lambda-5)=0.
\n\\]<\/p>\n
\n\\lambda_1=2,
\n\\qquad
\n\\lambda_2=5.
\n\\]<\/p>\nExample 2. Find the eigenvalues of a matrix<\/h3>\n
\nA=
\n\\begin{pmatrix}
\n2 & 1 & 0\\\\
\n1 & 2 & 0\\\\
\n0 & 0 & 3
\n\\end{pmatrix}.
\n\\]<\/p>\n
\n\\chi(\\lambda)
\n=
\n\\lambda^3
\n-\\sigma_1\\cdot\\lambda^2
\n+\\sigma_2\\cdot\\lambda
\n-\\sigma_3.
\n\\]<\/p>\n
\nS_1=\\operatorname{tr}(A)=2+2+3=7.
\n\\]<\/p>\n
\n\\sigma_1=S_1=7.
\n\\]<\/p>\n
\nA^2=
\n\\begin{pmatrix}
\n2 & 1 & 0\\\\
\n1 & 2 & 0\\\\
\n0 & 0 & 3
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n2 & 1 & 0\\\\
\n1 & 2 & 0\\\\
\n0 & 0 & 3
\n\\end{pmatrix}
\n=
\n\\begin{pmatrix}
\n5 & 4 & 0\\\\
\n4 & 5 & 0\\\\
\n0 & 0 & 9
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nS_2=\\operatorname{tr}(A^2)=5+5+9=19.
\n\\]<\/p>\n
\n\\sigma_2=
\n\\frac{1}{2}\\cdot(\\sigma_1\\cdot S_1-S_2).
\n\\]<\/p>\n
\n\\sigma_2=
\n\\frac{1}{2}\\cdot(7\\cdot7-19)
\n=
\n\\frac{1}{2}\\cdot(49-19)
\n=
\n15.
\n\\]<\/p>\n
\nA^3=A^2\\cdot A.
\n\\]<\/p>\n
\nA^3=
\n\\begin{pmatrix}
\n5 & 4 & 0\\\\
\n4 & 5 & 0\\\\
\n0 & 0 & 9
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n2 & 1 & 0\\\\
\n1 & 2 & 0\\\\
\n0 & 0 & 3
\n\\end{pmatrix}
\n=
\n\\begin{pmatrix}
\n14 & 13 & 0\\\\
\n13 & 14 & 0\\\\
\n0 & 0 & 27
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nS_3=\\operatorname{tr}(A^3)=14+14+27=55.
\n\\]<\/p>\n
\n\\sigma_3=
\n\\frac{1}{3}\\cdot
\n(\\sigma_2\\cdot S_1-\\sigma_1\\cdot S_2+S_3).
\n\\]<\/p>\n
\n\\sigma_3=
\n\\frac{1}{3}\\cdot
\n(15\\cdot7-7\\cdot19+55).
\n\\]<\/p>\n
\n\\sigma_3=
\n\\frac{1}{3}\\cdot
\n(105-133+55)
\n=
\n\\frac{27}{3}
\n=
\n9.
\n\\]<\/p>\n
\n\\sigma_1=7,
\n\\qquad
\n\\sigma_2=15,
\n\\qquad
\n\\sigma_3=9.
\n\\]<\/p>\n
\n\\chi(\\lambda)
\n=
\n\\lambda^3-7\\cdot\\lambda^2+15\\cdot\\lambda-9.
\n\\]<\/p>\n
\n\\lambda^3-7\\cdot\\lambda^2+15\\cdot\\lambda-9=0.
\n\\]<\/p>\n
\n\\lambda^3-7\\cdot\\lambda^2+15\\cdot\\lambda-9
\n=
\n(\\lambda-1)\\cdot(\\lambda-3)^2.
\n\\]<\/p>\n
\n(\\lambda-1)\\cdot(\\lambda-3)^2=0.
\n\\]<\/p>\n
\n\\lambda_1=1,
\n\\qquad
\n\\lambda_2=3,
\n\\qquad
\n\\lambda_3=3.
\n\\]<\/p>\nExample 3. Find the eigenvalues of a matrix<\/h3>\n
\nA=
\n\\begin{pmatrix}
\n2 & 1 & 0 & 0\\\\
\n1 & 2 & 0 & 0\\\\
\n0 & 0 & 3 & 1\\\\
\n0 & 0 & 1 & 3
\n\\end{pmatrix}.
\n\\]<\/p>\n
\n\\chi(\\lambda)
\n=
\n\\lambda^4
\n-\\sigma_1\\cdot\\lambda^3
\n+\\sigma_2\\cdot\\lambda^2
\n-\\sigma_3\\cdot\\lambda
\n+\\sigma_4.
\n\\]<\/p>\n
\nS_1=\\operatorname{tr}(A)=2+2+3+3=10.
\n\\]<\/p>\n
\n\\sigma_1=S_1=10.
\n\\]<\/p>\n
\nA^2=
\n\\begin{pmatrix}
\n2 & 1 & 0 & 0\\\\
\n1 & 2 & 0 & 0\\\\
\n0 & 0 & 3 & 1\\\\
\n0 & 0 & 1 & 3
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n2 & 1 & 0 & 0\\\\
\n1 & 2 & 0 & 0\\\\
\n0 & 0 & 3 & 1\\\\
\n0 & 0 & 1 & 3
\n\\end{pmatrix}
\n=
\n\\begin{pmatrix}
\n5 & 4 & 0 & 0\\\\
\n4 & 5 & 0 & 0\\\\
\n0 & 0 & 10 & 6\\\\
\n0 & 0 & 6 & 10
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nS_2=\\operatorname{tr}(A^2)=5+5+10+10=30.
\n\\]<\/p>\n
\n\\sigma_2=
\n\\frac{1}{2}\\cdot(\\sigma_1\\cdot S_1-S_2).
\n\\]<\/p>\n
\n\\sigma_2=
\n\\frac{1}{2}\\cdot(10\\cdot10-30)
\n=
\n\\frac{1}{2}\\cdot70
\n=
\n35.
\n\\]<\/p>\n
\nA^3=A^2\\cdot A.
\n\\]<\/p>\n
\nA^3=
\n\\begin{pmatrix}
\n5 & 4 & 0 & 0\\\\
\n4 & 5 & 0 & 0\\\\
\n0 & 0 & 10 & 6\\\\
\n0 & 0 & 6 & 10
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n2 & 1 & 0 & 0\\\\
\n1 & 2 & 0 & 0\\\\
\n0 & 0 & 3 & 1\\\\
\n0 & 0 & 1 & 3
\n\\end{pmatrix}
\n=
\n\\begin{pmatrix}
\n14 & 13 & 0 & 0\\\\
\n13 & 14 & 0 & 0\\\\
\n0 & 0 & 36 & 28\\\\
\n0 & 0 & 28 & 36
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nS_3=\\operatorname{tr}(A^3)=14+14+36+36=100.
\n\\]<\/p>\n
\n\\sigma_3=
\n\\frac{1}{3}\\cdot
\n(\\sigma_2\\cdot S_1-\\sigma_1\\cdot S_2+S_3).
\n\\]<\/p>\n
\n\\sigma_3=
\n\\frac{1}{3}\\cdot
\n(35\\cdot10-10\\cdot30+100).
\n\\]<\/p>\n
\n\\sigma_3=
\n\\frac{1}{3}\\cdot
\n(350-300+100)
\n=
\n\\frac{150}{3}
\n=
\n50.
\n\\]<\/p>\n
\nA^4=A^3\\cdot A.
\n\\]<\/p>\n
\nA^4=
\n\\begin{pmatrix}
\n14 & 13 & 0 & 0\\\\
\n13 & 14 & 0 & 0\\\\
\n0 & 0 & 36 & 28\\\\
\n0 & 0 & 28 & 36
\n\\end{pmatrix}
\n\\cdot
\n\\begin{pmatrix}
\n2 & 1 & 0 & 0\\\\
\n1 & 2 & 0 & 0\\\\
\n0 & 0 & 3 & 1\\\\
\n0 & 0 & 1 & 3
\n\\end{pmatrix}
\n=
\n\\begin{pmatrix}
\n41 & 40 & 0 & 0\\\\
\n40 & 41 & 0 & 0\\\\
\n0 & 0 & 136 & 120\\\\
\n0 & 0 & 120 & 136
\n\\end{pmatrix}.
\n\\]<\/p>\n
\nS_4=\\operatorname{tr}(A^4)=41+41+136+136=354.
\n\\]<\/p>\n
\n\\sigma_4=
\n\\frac{1}{4}\\cdot
\n(\\sigma_3\\cdot S_1-\\sigma_2\\cdot S_2+\\sigma_1\\cdot S_3-S_4).
\n\\]<\/p>\n
\n\\sigma_4=
\n\\frac{1}{4}\\cdot
\n(50\\cdot10-35\\cdot30+10\\cdot100-354).
\n\\]<\/p>\n
\n\\sigma_4=
\n\\frac{1}{4}\\cdot
\n(500-1050+1000-354)
\n=
\n\\frac{96}{4}
\n=
\n24.
\n\\]<\/p>\n
\n\\sigma_1=10,
\n\\qquad
\n\\sigma_2=35,
\n\\qquad
\n\\sigma_3=50,
\n\\qquad
\n\\sigma_4=24.
\n\\]<\/p>\n
\n\\chi(\\lambda)
\n=
\n\\lambda^4
\n-10\\cdot\\lambda^3
\n+35\\cdot\\lambda^2
\n-50\\cdot\\lambda
\n+24.
\n\\]<\/p>\n
\n\\lambda^4
\n-10\\cdot\\lambda^3
\n+35\\cdot\\lambda^2
\n-50\\cdot\\lambda
\n+24
\n=0.
\n\\]<\/p>\n
\n\\lambda^4
\n-10\\cdot\\lambda^3
\n+35\\cdot\\lambda^2
\n-50\\cdot\\lambda
\n+24
\n=
\n(\\lambda-1)\\cdot(\\lambda-2)\\cdot(\\lambda-3)\\cdot(\\lambda-4).
\n\\]<\/p>\n
\n(\\lambda-1)\\cdot(\\lambda-2)\\cdot(\\lambda-3)\\cdot(\\lambda-4)=0.
\n\\]<\/p>\n
\n\\lambda_1=1,
\n\\qquad
\n\\lambda_2=2,
\n\\qquad
\n\\lambda_3=3,
\n\\qquad
\n\\lambda_4=4.
\n\\]<\/p>\nWhat to Explore Next: Methods That Complement This Topic<\/h2>\n
\n
Eigenvalues of a Matrix: Turn the Leverrier Method into Code<\/h2>\n
![\"Flowchart](\"https:\/\/www.mathros.net.ua\/en\/wp-content\/uploads\/2026\/06\/eigenvalues-of-a-matrix-leverrier-method1.jpg\")
<\/p>\n","protected":false},"excerpt":{"rendered":"