Find the eigenvalues and eigenvectors of the matrix

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, λ2,…., λn.

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,

\(\begin{array}{l}{\displaystyle {tr} (A)=\sum _{i=1}^{n}a_{ii}=\sum _{i=1}^{n}\lambda _{i}=\lambda _{1}+\lambda _{2}+\cdots +\lambda _{n}.}\end{array} \)

(2) The determinant of A is the product of all its eigenvalues,

\(\begin{array}{l}{\displaystyle \det(A)=\prod _{i=1}^{n}\lambda _{i}=\lambda _{1}\lambda _{2}\cdots \lambda _{n}.}\end{array} \)

(3) The eigenvalues of the kth power of A; that is the eigenvalues of Ak, for any positive integer k, are

\(\begin{array}{l}{\displaystyle \lambda _{1}^{k},…,\lambda _{n}^{k}}.\end{array} \)

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

(5) If A is invertible, then the eigenvalues of A-1 are

\(\begin{array}{l}{\displaystyle {\frac {1}{\lambda _{1}}},…,{\frac {1}{\lambda _{n}}}}\end{array} \)

(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.

(9) If A is a n×n matrix and {λ1, λ2,…., λk} are its eigenvalues, then the eigenvalues of the matrix I + A (where I is the identity matrix) are {λ1+ 1, λ2+1,…., λk+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:

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

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

Solution: 

\(\begin{array}{l}\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}\\\end{array} \)

Example 3: Consider the matrix

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

= (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

\(\begin{array}{l}\begin{pmatrix}2&0&0\\ \:0&3&4\\ \:0&4&9\end{pmatrix}\end{array} \)

Solution: 

\(\begin{array}{l}\det \left(\begin{pmatrix}2&0&0\\ 0&3&4\\ 0&4&9\end{pmatrix}-λ\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\right)\\\begin{pmatrix}2&0&0\\ 0&3&4\\ 0&4&9\end{pmatrix}-λ\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\\λ\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}=\begin{pmatrix}λ&0&0\\ 0&λ&0\\ 0&0&λ\end{pmatrix}\\=\begin{pmatrix}2&0&0\\ 0&3&4\\ 0&4&9\end{pmatrix}-\begin{pmatrix}λ&0&0\\ 0&λ&0\\ 0&0&λ\end{pmatrix}\\=\begin{pmatrix}2-λ&0&0\\ 0&3-λ&4\\ 0&4&9-λ\end{pmatrix}\\=\det \begin{pmatrix}2-λ&0&0\\ 0&3-λ&4\\ 0&4&9-λ\end{pmatrix}\\=\left(2-λ\right)\det \begin{pmatrix}3-λ&4\\ 4&9-λ\end{pmatrix}-0\cdot \det \begin{pmatrix}0&4\\ 0&9-λ\end{pmatrix}+0\cdot \det \begin{pmatrix}0&3-λ\\ 0&4\end{pmatrix}\\=\left(2-λ\right)\left(λ^2-12λ+11\right)-0\cdot \:0+0\cdot \:0\\=-λ^3+14λ^2-35λ+22\\-λ^3+14λ^2-35λ+22=0\\-\left(λ-1\right)\left(λ-2\right)\left(λ-11\right)=0\\\mathrm{The\:eigenvalues\:are:}\\λ=1,\:λ=2,\:λ=11\\\mathrm{Eigenvectors\:for\:}λ=1\\\begin{pmatrix}2&0&0\\ 0&3&4\\ 0&4&9\end{pmatrix}-1\cdot \begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}=\begin{pmatrix}1&0&0\\ 0&2&4\\ 0&4&8\end{pmatrix}\\\left(A-1I\right)\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}1&0&0\\ 0&1&2\\ 0&0&0\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}\\\begin{Bmatrix}x=0\\ y+2z=0\end{Bmatrix}\\Isolate\\\begin{Bmatrix}x=0\\ y=-2z\end{Bmatrix}\\\mathrm{Plug\:into\:}\begin{pmatrix}x\\ y\\ z\end{pmatrix}\\η=\begin{pmatrix}0\\ -2z\\ z\end{pmatrix}\space\space\:z\ne \:0\\\mathrm{Let\:}z=1\\\begin{pmatrix}0\\ -2\\ 1\end{pmatrix}\\Similarly\\\mathrm{Eigenvectors\:for\:}λ=2:\quad \begin{pmatrix}1\\ 0\\ 0\end{pmatrix}\\\mathrm{Eigenvectors\:for\:}λ=11:\quad \begin{pmatrix}0\\ 1\\ 2\end{pmatrix}\\\mathrm{The\:eigenvectors\:for\:}\begin{pmatrix}2&0&0\\ 0&3&4\\ 0&4&9\end{pmatrix}\\=\begin{pmatrix}0\\ -2\\ 1\end{pmatrix},\:\begin{pmatrix}1\\ 0\\ 0\end{pmatrix},\:\begin{pmatrix}0\\ 1\\ 2\end{pmatrix}\\\end{array} \)

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