Eigenvalues and Eigenvectors Problems and Solutions

Introduction To Eigenvalues And Eigenvectors

A rectangular arrangement of numbers in the form of rows and columns is known as a matrix. In this article, we will discuss Eigenvalues and Eigenvectors Problems and Solutions.

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.

Suppose the matrix equation is written as A X – λ X = 0. Let I be the n × n identity matrix.

If I X is substituted by X in the equation above, we obtain

A X – λ I X = 0.

The equation is rewritten as (A – λ I) X = 0.

The equation above consists of non-trivial solutions, if and only if, the determinant value of the matrix is 0. The characteristic equation of A is Det (A – λ I) = 0. ‘A’ being an n × n matrix, if (A – λ I) is expanded, (A – λ I) will be the characteristic polynomial of A because it’s degree is n.

Properties Of Eigenvalues

Let A be a matrix with eigenvalues λ1,,λn{\displaystyle \lambda _{1},…,\lambda _{n}}

The following are the properties of eigenvalues.

1] The trace of A, defined as the sum of its diagonal elements, is also the sum of all eigenvalues,

tr(A)=i=1naii=i=1nλi=λ1+λ2++λn.{\displaystyle {tr} (A)=\sum _{i=1}^{n}a_{ii}=\sum _{i=1}^{n}\lambda _{i}=\lambda _{1}+\lambda _{2}+\cdots +\lambda _{n}.}

2] The determinant of A is the product of all its eigenvalues, det(A)=i=1nλi=λ1λ2λn.{\displaystyle \det(A)=\prod _{i=1}^{n}\lambda _{i}=\lambda _{1}\lambda _{2}\cdots \lambda _{n}.}

3] The eigenvalues of the kthk^{th} power of A; that is the eigenvalues of AkA^{k}, for any positive integer k, are λ1k,,λnk.{\displaystyle \lambda _{1}^{k},…,\lambda _{n}^{k}}..

4] The matrix A is invertible if and only if every eigenvalue is nonzero.

5] If A is invertible, then the eigenvalues of A1A^{-1} are 1λ1,,1λn{\displaystyle {\frac {1}{\lambda _{1}}},…,{\frac {1}{\lambda _{n}}}} and each eigenvalue’s geometric multiplicity coincides. The characteristic polynomial of the inverse is the reciprocal polynomial of the original, the eigenvalues share the same algebraic multiplicity.

6] If A is equal to its conjugate transpose, or equivalently if A is Hermitian, then every eigenvalue is real. The same is true of any symmetric real matrix.

7] If A is not only Hermitian but also positive-definite, positive-semidefinite, negative-definite, or negative-semidefinite, then every eigenvalue is positive, non-negative, negative, or non-positive, respectively.

8] If A is unitary, every eigenvalue has absolute value λi=1{\displaystyle |\lambda _{i}|=1}.

9] If A is a n×n{\displaystyle n\times n} matrix and {λ1,,λk}{\displaystyle \{\lambda _{1},\ldots ,\lambda _{k}\}} 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\}}.

Also Read:

Eigenvectors of a Matrix

Adjoint and Inverse of a Matrix

Normalized and Decomposition of Eigenvectors

Eigenvalues And Eigenvectors Solved Problems

Example 1: Find the eigenvalues and eigenvectors of the following matrix.

Solution:

Eigenvalues And Eigenvectors Example

Example 2: Find all eigenvalues and corresponding eigenvectors for the matrix A if

(230250003)\begin{pmatrix}2&-3&0\\ \:\:2&-5&0\\ \:\:0&0&3\end{pmatrix}

Solution: 

det((230250003)λ(100010001))(230250003)λ(100010001)λ(100010001)=(λ000λ000λ)=(230250003)(λ000λ000λ)=(2λ3025λ0003λ)=det(2λ3025λ0003λ)=(2λ)det(5λ003λ)(3)det(2003λ)+0det(25λ00)=(2λ)(λ2+2λ15)(3)2(λ+3)+00=λ3+13λ12λ3+13λ12=0(λ1)(λ3)(λ+4)=0Theeigenvaluesare:λ=1,λ=3,λ=4Eigenvectorsforλ=1(230250003)1(100010001)=(130260002)(A1I)(xyz)=(130001000)(xyz)=(000){x3y=0z=0}Isolate{z=0x=3y}Pluginto(xyz)η=(3yy0)  y0Lety=1(310)SimilarlyEigenvectorsforλ=3:(001)Eigenvectorsforλ=4:(120)Theeigenvectorsfor(230250003)=(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}\\

Example 3: Consider the matrix

Solving Eigen Values and Eigen Vactors

for some variable ‘a’. Find all values of ‘a’ which will prove that A has eigenvalues 0, 3, and −3.

Solution:

Let p (t) be the characteristic polynomial of A, i.e. let p (t) = det (A − tI) = 0. By expanding along the second column of A − tI, we can obtain the equation

How to Solve EigenValues and EigenVactors of Matrix

= (3 − t) [(−2 −t) (−1 − t) − 4] + 2[(−2 − t) a + 5]

= (3 − t) (2 + t + 2t + t2 −4) + 2 (−2a − ta + 5)

= (3 − t) (t2 + 3t − 2) + (−4a −2ta + 10)

= 3t2 + 9t − 6 − t3 − 3t2 + 2t − 4a − 2ta + 10

= −t3 + 11t − 2ta + 4 − 4a

= −t3 + (11 − 2a) t + 4 − 4a

For the eigenvalues of A to be 0, 3 and −3, the characteristic polynomial p (t) must have roots at t = 0, 3, −3. This implies p (t) = –t (t − 3) (t + 3) =–t(t2 − 9) = –t3 + 9t

Therefore, −t3 + (11 − 2a) t + 4 − 4a = −t3 + 9t.

For this equation to hold, the constant terms on the left and right-hand sides of the above equation must be equal. This means that 4 − 4a = 0, which implies a = 1.

Hence, A has eigenvalues 0, 3, −3 precisely when a = 1.

Example 4: Find the eigenvalues and eigenvectors of (200034049)\begin{pmatrix}2&0&0\\ \:0&3&4\\ \:0&4&9\end{pmatrix}

Solution: 

det((200034049)λ(100010001))(200034049)λ(100010001)λ(100010001)=(λ000λ000λ)=(200034049)(λ000λ000λ)=(2λ0003λ4049λ)=det(2λ0003λ4049λ)=(2λ)det(3λ449λ)0det(0409λ)+0det(03λ04)=(2λ)(λ212λ+11)00+00=λ3+14λ235λ+22λ3+14λ235λ+22=0(λ1)(λ2)(λ11)=0Theeigenvaluesare:λ=1,λ=2,λ=11Eigenvectorsforλ=1(200034049)1(100010001)=(100024048)(A1I)(xyz)=(100012000)(xyz)=(000){x=0y+2z=0}Isolate{x=0y=2z}Pluginto(xyz)η=(02zz)  z0Letz=1(021)SimilarlyEigenvectorsforλ=2:(100)Eigenvectorsforλ=11:(012)Theeigenvectorsfor(200034049)=(021),(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}\\