We can’t ﬁnd it … You may check the examples above. 9] If A is a n×n{\displaystyle n\times n}n×n matrix and {λ1,…,λk}{\displaystyle \{\lambda _{1},\ldots ,\lambda _{k}\}}{λ1​,…,λk​} are its eigenvalues, then the eigenvalues of matrix I + A (where I is the identity matrix) are {λ1+1,…,λk+1}{\displaystyle \{\lambda _{1}+1,\ldots ,\lambda _{k}+1\}}{λ1​+1,…,λk​+1}. Find the eigenvalues and eigenvectors of A and A2 and A-1 and A +41: = [-} -2] and A2 2 -[ 5 - 4 -4 5 Get more help from Chegg Get 1:1 help now from expert Calculus tutors Solve it with our calculus problem solver and calculator Eigenvalues and Eigenvectors, More Direction Fields and Systems of ODEs First let us speak a bit about eigenvalues. FINDING EIGENVALUES AND EIGENVECTORS EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix A = 1 −3 3 3 −5 3 6 −6 4 . -2 & 2 & 0 We can solve for the eigenvalues by finding the characteristic equation (note the "+" sign in the determinant rather than the "-" sign, because of the opposite signs of λ and ω2). 5] If A is invertible, then the eigenvalues of A−1A^{-1}A−1 are 1λ1,…,1λn{\displaystyle {\frac {1}{\lambda _{1}}},…,{\frac {1}{\lambda _{n}}}}λ1​1​,…,λn​1​ and each eigenvalue’s geometric multiplicity coincides. \end{bmatrix} = 0 \)Row reduce to echelon form gives$$\begin{bmatrix} More: Diagonal matrix Jordan decomposition Matrix exponential. \end{bmatrix} \ \begin{bmatrix} tr(A)=∑i=1naii=∑i=1nλi=λ1+λ2+⋯+λn. x_2 \\ Problem 9 Prove that. Every square matrix has special values called eigenvalues. x_2 \\ Hopefully you got the following: What do you notice about the product? x_3 }$$ This polynomial has a single root $$\lambda = 3$$ with eigenvector $$\mathbf v = (1, 1)\text{. SOLUTION: • In such problems, we ﬁrst ﬁnd the eigenvalues of the matrix. x_1 \\ A = 10−1 2 −15 00 2 λ =2, 1, or − 1 λ =2 = null(A − 2I) = span −1 1 1 eigenvectors of A for λ = 2 are c −1 1 1 for c =0 = set of all eigenvectors of A for λ =2 ∪ {0} Solve (A − 2I)x = 0. 0 What are these? Hence, A has eigenvalues 0, 3, −3 precisely when a = 1. In Mathematica the Dsolve[] function can be used to bypass the calculations of eigenvalues and eigenvectors to give the solutions for the differentials directly. 1 - \lambda & 0 & -1 \\ \[ A = \begin{bmatrix} Eigenvalues and eigenvectors. This video has not been made yet. Let A = " 2 0 2 3 #. 1 & 0 & -1 \\ {\displaystyle \det(A)=\prod _{i=1}^{n}\lambda _{i}=\lambda _{1}\lambda _{2}\cdots \lambda _{n}.}det(A)=i=1∏n​λi​=λ1​λ2​⋯λn​. 0 & 0 & 1 \\ Consider a square matrix n × n. If X is the non-trivial column vector solution of the matrix equation AX = λX, where λ is a scalar, then X is the eigenvector of matrix A and the corresponding value … Similarly, we can ﬁnd eigenvectors associated with the eigenvalue λ = 4 by solving Ax = 4x: 2x 1 +2x 2 5x 1 −x 2 = 4x 1 4x 2 ⇒ 2x 1 +2x 2 = 4x 1 and 5x 1 −x 2 = 4x 2 ⇒ x 1 = x 2. Let A be an n × n square matrix. {\displaystyle \lambda _{1}^{k},…,\lambda _{n}^{k}}.λ1k​,…,λnk​.. 4] The matrix A is invertible if and only if every eigenvalue is nonzero. Definition: Eigenvector and Eigenvalues Hence the set of eigenvectors associated with λ = 4 is spanned by u 2 = 1 1 . 0& - 2 & 0 \\ Matrix A is singular if and only if \( \lambda = 0$$ is an eigenvalue value of matrix A. So, let’s do that. 0 & 0 & -1 \\ 0 & e & f \\ Find all values of ‘a’ which will prove that A has eigenvalues 0, 3, and −3. In this session we learn how to find the eigenvalues and eigenvectors of a matrix. As for when, well this is a huge project and has taken me at least 10 years just to get this far, so you will have to be patient. {\displaystyle {tr} (A)=\sum _{i=1}^{n}a_{ii}=\sum _{i=1}^{n}\lambda _{i}=\lambda _{1}+\lambda _{2}+\cdots +\lambda _{n}.}tr(A)=i=1∑n​aii​=i=1∑n​λi​=λ1​+λ2​+⋯+λn​. -1/2 \\ \end{bmatrix}\)Write the characteristic equation.$$Det(A - \lambda I) = (1-\lambda)(-\lambda(1-\lambda)) - 1(2 - 2\lambda) = 0$$factor and rewrite the equation as$$(1 - \lambda)(\lambda - 2)(\lambda+1) = 0$$which gives 3 solutions$$\lambda = - 1 , \lambda = 1 , \lambda = 2$$eval(ez_write_tag([[728,90],'analyzemath_com-large-mobile-banner-1','ezslot_7',700,'0','0']));Find EigenvectorsEigenvectors for $$\lambda = - 1$$Substitute $$\lambda$$ by - 1 in the matrix equation $$(A - \lambda I) X = 0$$ with $$X = \begin{bmatrix} x_1 \\ Example Find eigenvalues and corresponding eigenvectors of A. The characteristic polynomial of the inverse is the reciprocal polynomial of the original, the eigenvalues share the same algebraic multiplicity. Find the eigenvalues and eigenvectors of A and A2 and A-1 and A +41: = [-} -2] and A2 2 -[ 5 - 4 -4 5 Get more help from Chegg Get 1:1 help now from expert Calculus tutors Solve it with our calculus problem solver and calculator 2] The determinant of A is the product of all its eigenvalues, 5] If A is invertible, then the eigenvalues of, 8] If A is unitary, every eigenvalue has absolute value, Eigenvalues And Eigenvectors Solved Problems, Find all eigenvalues and corresponding eigenvectors for the matrix A if, CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, NCERT Solutions Class 11 Business Studies, NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions For Class 6 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions for Class 8 Social Science, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, JEE Main Chapter Wise Questions And Solutions. Let A be an n × n matrix. Section 5.1 Eigenvalues and Eigenvectors ¶ permalink Objectives. Display decimals, number of significant digits: Clean. Suppose the matrix equation is written as A X – λ X = 0. If there exist a non trivial (not all zeroes) column vector X solution to the matrix equation, is called the eigenvector of matrix A and the corresponding value of, be the n × n identity matrix and substitute, is expanded, it is a polynomial of degree n and therefore, let us find the eigenvalues of matrix \( A = \begin{bmatrix} 0 & 2 & 1 \\ x_3 * all eigenvalues and no eigenvectors (a polynomial root solver) * some eigenvalues and some corresponding eigenvectors * all eigenvalues and all corresponding eigenvectors. 0 & -2 & -1 \\ A is singular if and only if 0 is an eigenvalue of A. We call such a v an eigenvector of A corresponding to the eigenvalue λ. Well, let's start by doing the following matrix multiplication problem where we're multiplying a square matrix by a vector. x_1 \\ Given the above solve the following problems (answers to … Session Overview If the product A x points in the same direction as the vector x, we say that x is an eigenvector of A. Eigenvalues and eigenvectors describe what happens when a matrix is multiplied by a vector. x_2 \\ \end{bmatrix}$$$$\begin{bmatrix} 8] If A is unitary, every eigenvalue has absolute value ∣λi∣=1{\displaystyle |\lambda _{i}|=1}∣λi​∣=1. For the first eigenvector: which clearly has the solution: So we'll choose the first eigenvector (which can be multiplied by an arbitrary constant). Find a basis for this eigenspace. math; ... Find The Eigenvalues And Eigenvectors For The Matrix And Show A Calculation That Verifies Your Answer. If there exist a non trivial (not all zeroes) column vector X solution to the matrix equation A X = λ X ; where λ is a scalar, then X is called the eigenvector of matrix A and the corresponding value of λ is called the eigenvalue of matrix A. In Chemical Engineering they are mostly used to solve differential equations and to analyze the stability of a system. In this case we get complex eigenvalues which are definitely a fact of life with eigenvalue/eigenvector problems so get used to them. Solution for 1. The determinant of the triangular matrix − is the product down the diagonal, and so it factors into the product of the terms , −. Those are the “eigenvectors”. 14. -2 & 2 & -1 Find the eigenvalues of the matrix 2 2 1 3 and ﬁnd one eigenvector for each eigenvalue. eval(ez_write_tag([[300,250],'analyzemath_com-large-mobile-banner-2','ezslot_8',701,'0','0'])); Let A be an n × n square matrix. The l =2 eigenspace for the matrix 2 4 3 4 2 1 6 2 1 4 4 3 5 is two-dimensional. 13. \({\lambda _{\,1}} = - 1 + 5\,i$$ : -1/2 \\ Let I be the n × n identity matrix. 0 & 1 & 0 \\ To explain eigenvalues, we ﬁrst explain eigenvectors. This textbook survival guide was created for the textbook: Linear Algebra and Its Applications,, edition: 4. Consider a square matrix n × n. If X is the non-trivial column vector solution of the matrix equation AX = λX, where λ is a scalar, then X is the eigenvector of matrix A and the corresponding value of λ is the eigenvalue of matrix A. The eigenspace corresponding to is the null space of which is . Determining Eigenvalues and Eigenvectors. Eigenvalues and Eigenvectors 6.1 Introduction to Eigenvalues Linear equationsAx D bcomefrom steady stateproblems. In this section we will define eigenvalues and eigenfunctions for boundary value problems. Example 4: Find the eigenvalues and eigenvectors of (200 034 049)\begin{pmatrix}2&0&0\\ \:0&3&4\\ \:0&4&9\end{pmatrix}⎝⎜⎛​200​034​049​⎠⎟⎞​, det⁡((200034049)−λ(100010001))(200034049)−λ(100010001)λ(100010001)=(λ000λ000λ)=(200034049)−(λ000λ000λ)=(2−λ0003−λ4049−λ)=det⁡(2−λ0003−λ4049−λ)=(2−λ)det⁡(3−λ449−λ)−0⋅det⁡(0409−λ)+0⋅det⁡(03−λ04)=(2−λ)(λ2−12λ+11)−0⋅ 0+0⋅ 0=−λ3+14λ2−35λ+22−λ3+14λ2−35λ+22=0−(λ−1)(λ−2)(λ−11)=0The eigenvalues are:λ=1, λ=2, λ=11Eigenvectors for λ=1(200034049)−1⋅(100010001)=(100024048)(A−1I)(xyz)=(100012000)(xyz)=(000){x=0y+2z=0}Isolate{x=0y=−2z}Plug into (xyz)η=(0−2zz)   z≠ 0Let z=1(0−21)SimilarlyEigenvectors for λ=2:(100)Eigenvectors for λ=11:(012)The eigenvectors for (200034049)=(0−21), (100), (012)\det \left(\begin{pmatrix}2&0&0\\ 0&3&4\\ 0&4&9\end{pmatrix}-λ\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\right)\\\begin{pmatrix}2&0&0\\ 0&3&4\\ 0&4&9\end{pmatrix}-λ\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\\λ\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}=\begin{pmatrix}λ&0&0\\ 0&λ&0\\ 0&0&λ\end{pmatrix}\\=\begin{pmatrix}2&0&0\\ 0&3&4\\ 0&4&9\end{pmatrix}-\begin{pmatrix}λ&0&0\\ 0&λ&0\\ 0&0&λ\end{pmatrix}\\=\begin{pmatrix}2-λ&0&0\\ 0&3-λ&4\\ 0&4&9-λ\end{pmatrix}\\=\det \begin{pmatrix}2-λ&0&0\\ 0&3-λ&4\\ 0&4&9-λ\end{pmatrix}\\=\left(2-λ\right)\det \begin{pmatrix}3-λ&4\\ 4&9-λ\end{pmatrix}-0\cdot \det \begin{pmatrix}0&4\\ 0&9-λ\end{pmatrix}+0\cdot \det \begin{pmatrix}0&3-λ\\ 0&4\end{pmatrix}\\=\left(2-λ\right)\left(λ^2-12λ+11\right)-0\cdot \:0+0\cdot \:0\\=-λ^3+14λ^2-35λ+22\\-λ^3+14λ^2-35λ+22=0\\-\left(λ-1\right)\left(λ-2\right)\left(λ-11\right)=0\\\mathrm{The\:eigenvalues\:are:}\\λ=1,\:λ=2,\:λ=11\\\mathrm{Eigenvectors\:for\:}λ=1\\\begin{pmatrix}2&0&0\\ 0&3&4\\ 0&4&9\end{pmatrix}-1\cdot \begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}=\begin{pmatrix}1&0&0\\ 0&2&4\\ 0&4&8\end{pmatrix}\\\left(A-1I\right)\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}1&0&0\\ 0&1&2\\ 0&0&0\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}\\\begin{Bmatrix}x=0\\ y+2z=0\end{Bmatrix}\\Isolate\\\begin{Bmatrix}x=0\\ y=-2z\end{Bmatrix}\\\mathrm{Plug\:into\:}\begin{pmatrix}x\\ y\\ z\end{pmatrix}\\η=\begin{pmatrix}0\\ -2z\\ z\end{pmatrix}\space\space\:z\ne \:0\\\mathrm{Let\:}z=1\\\begin{pmatrix}0\\ -2\\ 1\end{pmatrix}\\Similarly\\\mathrm{Eigenvectors\:for\:}λ=2:\quad \begin{pmatrix}1\\ 0\\ 0\end{pmatrix}\\\mathrm{Eigenvectors\:for\:}λ=11:\quad \begin{pmatrix}0\\ 1\\ 2\end{pmatrix}\\\mathrm{The\:eigenvectors\:for\:}\begin{pmatrix}2&0&0\\ 0&3&4\\ 0&4&9\end{pmatrix}\\=\begin{pmatrix}0\\ -2\\ 1\end{pmatrix},\:\begin{pmatrix}1\\ 0\\ 0\end{pmatrix},\:\begin{pmatrix}0\\ 1\\ 2\end{pmatrix}\\det⎝⎜⎛​⎝⎜⎛​200​034​049​⎠⎟⎞​−λ⎝⎜⎛​100​010​001​⎠⎟⎞​⎠⎟⎞​⎝⎜⎛​200​034​049​⎠⎟⎞​−λ⎝⎜⎛​100​010​001​⎠⎟⎞​λ⎝⎜⎛​100​010​001​⎠⎟⎞​=⎝⎜⎛​λ00​0λ0​00λ​⎠⎟⎞​=⎝⎜⎛​200​034​049​⎠⎟⎞​−⎝⎜⎛​λ00​0λ0​00λ​⎠⎟⎞​=⎝⎜⎛​2−λ00​03−λ4​049−λ​⎠⎟⎞​=det⎝⎜⎛​2−λ00​03−λ4​049−λ​⎠⎟⎞​=(2−λ)det(3−λ4​49−λ​)−0⋅det(00​49−λ​)+0⋅det(00​3−λ4​)=(2−λ)(λ2−12λ+11)−0⋅0+0⋅0=−λ3+14λ2−35λ+22−λ3+14λ2−35λ+22=0−(λ−1)(λ−2)(λ−11)=0Theeigenvaluesare:λ=1,λ=2,λ=11Eigenvectorsforλ=1⎝⎜⎛​200​034​049​⎠⎟⎞​−1⋅⎝⎜⎛​100​010​001​⎠⎟⎞​=⎝⎜⎛​100​024​048​⎠⎟⎞​(A−1I)⎝⎜⎛​xyz​⎠⎟⎞​=⎝⎜⎛​100​010​020​⎠⎟⎞​⎝⎜⎛​xyz​⎠⎟⎞​=⎝⎜⎛​000​⎠⎟⎞​{x=0y+2z=0​}Isolate{x=0y=−2z​}Pluginto⎝⎜⎛​xyz​⎠⎟⎞​η=⎝⎜⎛​0−2zz​⎠⎟⎞​  z​=0Letz=1⎝⎜⎛​0−21​⎠⎟⎞​SimilarlyEigenvectorsforλ=2:⎝⎜⎛​100​⎠⎟⎞​Eigenvectorsforλ=11:⎝⎜⎛​012​⎠⎟⎞​Theeigenvectorsfor⎝⎜⎛​200​034​049​⎠⎟⎞​=⎝⎜⎛​0−21​⎠⎟⎞​,⎝⎜⎛​100​⎠⎟⎞​,⎝⎜⎛​012​⎠⎟⎞​, Eigenvalues and Eigenvectors Problems and Solutions, Introduction To Eigenvalues And Eigenvectors. In fact, we could write our solution like this: Th… (solution: x = 1 or x = 5.) The solution of du=dt D Au is changing with time— growing or decaying or oscillating. 2] The determinant of A is the product of all its eigenvalues, det⁡(A)=∏i=1nλi=λ1λ2⋯λn. Eigenvalues and eigenvectors are used for: Computing prediction and confidence ellipses; Principal Components Analysis (later in the course) Factor Analysis (also later in this course) For the present we will be primarily concerned with eigenvalues and eigenvectors of the variance-covariance matrix. In this article, we will discuss Eigenvalues and Eigenvectors Problems and Solutions. In one example the best we will be able to do is estimate the eigenvalues as that is something that will happen on a fairly regular basis with these kinds of problems. For any x ∈ IR2, if x+Ax and x−Ax are eigenvectors of A ﬁnd the corresponding eigenvalue. =solution. \end{bmatrix} \ \begin{bmatrix} 6. Finding of eigenvalues and eigenvectors. Therefore, −t3 + (11 − 2a) t + 4 − 4a = −t3 + 9t. Eigenvalues and Eigenvectors on Brilliant, the largest community of math and science problem solvers. x_2 \\ Example 2: Find all eigenvalues and corresponding eigenvectors for the matrix A if, (2−30  2−50  003)\begin{pmatrix}2&-3&0\\ \:\:2&-5&0\\ \:\:0&0&3\end{pmatrix}⎝⎜⎛​220​−3−50​003​⎠⎟⎞​, det⁡((2−302−50003)−λ(100010001))(2−302−50003)−λ(100010001)λ(100010001)=(λ000λ000λ)=(2−302−50003)−(λ000λ000λ)=(2−λ−302−5−λ0003−λ)=det⁡(2−λ−302−5−λ0003−λ)=(2−λ)det⁡(−5−λ003−λ)−(−3)det⁡(2003−λ)+0⋅det⁡(2−5−λ00)=(2−λ)(λ2+2λ−15)−(−3)⋅ 2(−λ+3)+0⋅ 0=−λ3+13λ−12−λ3+13λ−12=0−(λ−1)(λ−3)(λ+4)=0The eigenvalues are:λ=1, λ=3, λ=−4Eigenvectors for λ=1(2−302−50003)−1⋅(100010001)=(1−302−60002)(A−1I)(xyz)=(1−30001000)(xyz)=(000){x−3y=0z=0}Isolate{z=0x=3y}Plug into (xyz)η=(3yy0)   y≠ 0Let y=1(310)SimilarlyEigenvectors for λ=3:(001)Eigenvectors for λ=−4:(120)The eigenvectors for (2−302−50003)=(310), (001), (120)\det \left(\begin{pmatrix}2&-3&0\\ 2&-5&0\\ 0&0&3\end{pmatrix}-λ\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\right)\\\begin{pmatrix}2&-3&0\\ 2&-5&0\\ 0&0&3\end{pmatrix}-λ\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\\λ\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}=\begin{pmatrix}λ&0&0\\ 0&λ&0\\ 0&0&λ\end{pmatrix}\\=\begin{pmatrix}2&-3&0\\ 2&-5&0\\ 0&0&3\end{pmatrix}-\begin{pmatrix}λ&0&0\\ 0&λ&0\\ 0&0&λ\end{pmatrix}\\=\begin{pmatrix}2-λ&-3&0\\ 2&-5-λ&0\\ 0&0&3-λ\end{pmatrix}\\=\det \begin{pmatrix}2-λ&-3&0\\ 2&-5-λ&0\\ 0&0&3-λ\end{pmatrix}\\=\left(2-λ\right)\det \begin{pmatrix}-5-λ&0\\ 0&3-λ\end{pmatrix}-\left(-3\right)\det \begin{pmatrix}2&0\\ 0&3-λ\end{pmatrix}+0\cdot \det \begin{pmatrix}2&-5-λ\\ 0&0\end{pmatrix}\\=\left(2-λ\right)\left(λ^2+2λ-15\right)-\left(-3\right)\cdot \:2\left(-λ+3\right)+0\cdot \:0\\=-λ^3+13λ-12\\-λ^3+13λ-12=0\\-\left(λ-1\right)\left(λ-3\right)\left(λ+4\right)=0\\\mathrm{The\:eigenvalues\:are:}\\λ=1,\:λ=3,\:λ=-4\\\mathrm{Eigenvectors\:for\:}λ=1\\\begin{pmatrix}2&-3&0\\ 2&-5&0\\ 0&0&3\end{pmatrix}-1\cdot \begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}=\begin{pmatrix}1&-3&0\\ 2&-6&0\\ 0&0&2\end{pmatrix}\\\left(A-1I\right)\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}1&-3&0\\ 0&0&1\\ 0&0&0\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}\\\begin{Bmatrix}x-3y=0\\ z=0\end{Bmatrix}\\Isolate\\\begin{Bmatrix}z=0\\ x=3y\end{Bmatrix}\\\mathrm{Plug\:into\:}\begin{pmatrix}x\\ y\\ z\end{pmatrix}\\η=\begin{pmatrix}3y\\ y\\ 0\end{pmatrix}\space\space\:y\ne \:0\\\mathrm{Let\:}y=1\\\begin{pmatrix}3\\ 1\\ 0\end{pmatrix}\\Similarly\\\mathrm{Eigenvectors\:for\:}λ=3:\quad \begin{pmatrix}0\\ 0\\ 1\end{pmatrix}\\\mathrm{Eigenvectors\:for\:}λ=-4:\quad \begin{pmatrix}1\\ 2\\ 0\end{pmatrix}\\\mathrm{The\:eigenvectors\:for\:}\begin{pmatrix}2&-3&0\\ 2&-5&0\\ 0&0&3\end{pmatrix}\\=\begin{pmatrix}3\\ 1\\ 0\end{pmatrix},\:\begin{pmatrix}0\\ 0\\ 1\end{pmatrix},\:\begin{pmatrix}1\\ 2\\ 0\end{pmatrix}\\det⎝⎜⎛​⎝⎜⎛​220​−3−50​003​⎠⎟⎞​−λ⎝⎜⎛​100​010​001​⎠⎟⎞​⎠⎟⎞​⎝⎜⎛​220​−3−50​003​⎠⎟⎞​−λ⎝⎜⎛​100​010​001​⎠⎟⎞​λ⎝⎜⎛​100​010​001​⎠⎟⎞​=⎝⎜⎛​λ00​0λ0​00λ​⎠⎟⎞​=⎝⎜⎛​220​−3−50​003​⎠⎟⎞​−⎝⎜⎛​λ00​0λ0​00λ​⎠⎟⎞​=⎝⎜⎛​2−λ20​−3−5−λ0​003−λ​⎠⎟⎞​=det⎝⎜⎛​2−λ20​−3−5−λ0​003−λ​⎠⎟⎞​=(2−λ)det(−5−λ0​03−λ​)−(−3)det(20​03−λ​)+0⋅det(20​−5−λ0​)=(2−λ)(λ2+2λ−15)−(−3)⋅2(−λ+3)+0⋅0=−λ3+13λ−12−λ3+13λ−12=0−(λ−1)(λ−3)(λ+4)=0Theeigenvaluesare:λ=1,λ=3,λ=−4Eigenvectorsforλ=1⎝⎜⎛​220​−3−50​003​⎠⎟⎞​−1⋅⎝⎜⎛​100​010​001​⎠⎟⎞​=⎝⎜⎛​120​−3−60​002​⎠⎟⎞​(A−1I)⎝⎜⎛​xyz​⎠⎟⎞​=⎝⎜⎛​100​−300​010​⎠⎟⎞​⎝⎜⎛​xyz​⎠⎟⎞​=⎝⎜⎛​000​⎠⎟⎞​{x−3y=0z=0​}Isolate{z=0x=3y​}Pluginto⎝⎜⎛​xyz​⎠⎟⎞​η=⎝⎜⎛​3yy0​⎠⎟⎞​  y​=0Lety=1⎝⎜⎛​310​⎠⎟⎞​SimilarlyEigenvectorsforλ=3:⎝⎜⎛​001​⎠⎟⎞​Eigenvectorsforλ=−4:⎝⎜⎛​120​⎠⎟⎞​Theeigenvectorsfor⎝⎜⎛​220​−3−50​003​⎠⎟⎞​=⎝⎜⎛​310​⎠⎟⎞​,⎝⎜⎛​001​⎠⎟⎞​,⎝⎜⎛​120​⎠⎟⎞​. 3 # its applications,, edition: 4 direction asAx harmonics problems, population models,.... 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