Eigenvalues and Eigenvectors

Student project for MA265

Author: Libo Dong

1.What is eigenvalue and eigenvector?

Let Linear transformation L:V->V be a linear transformation of an n-dimensional vector space V into itself. The number λ is called an 'eigenvalue'of L if there exist a non zero vector x in V such that L(X) = λ*x. Note that every non zero vector x satisfying this equation is then called an eigenvector of L associated with the eigenvalue λ.(Eigen means 'proper' in German).

-Referenced from Elementary Linear Algebra with Applications and Labs

In another way,eigenvector can be referred as a square matrix which parallel to the original vector after multiplied to the vector.

2.How to calculate eigenvalue and eigenvector?

a.from the definition it can be derived that if A is the original vector, A*x=λ*x. b.By subtracting both sides of the equation by λ*x, the equation will be : A*x-λ*x=0. c.By the definition of identity matrix, it is fine to add I to the λ*x term; A*x-λ*I*x=0.(Note: A*I=A for every matrix A,I is a matrix with ones on the main diagonal and zeros elsewhere.) d.By taking out the x matrix, the equaiton will be: (A-λ*I)*x=0.If there exists (A-λ*I)^-1 then both sides can be multiplied by it, to obtain x = 0.But if A − λI is not invertible,the determinant of the (A-λ*I) will be 0 and this λ can be calculated. e.By changing λ to the calculated value in step d. to step a. the corresponding eigenvector x can be obtained.

Alternative way: If a matlab software is available the steps stated above can be done by program.

a.first enter the matrix that you want to calculate: A=[a,b,c;d,e,f;g,h,i].(use space to separate each row elements and semicolon to separate rows) the screen will show:

$ A= \left(\begin{array}{cccc}a&b\\e&f\end{array}\right) $.

b.type in roots(poly(A)). This command order matlab to calculate the roots of the determinant equation (equals to zero)of the orignals matrix which subtracted by the identity matrix. poly(A) is the characteristic equation of the matirx A For example:

$ det(A-\lambda I)=\left(\begin{array}{cccc}a-\lambda&b&c\\d&e-\lambda&f\\g&h&i-\lambda\end{array}\right)=0. $

The results is the eigenvalue to the matrix.the result of the command above will display the possible λ values.

c.Using the command m=rref(A-(one of the value calculated above)*eye(size(A))).eye(size(A)) command create a identity matrix that as same size as A. By the definition of A it can be derived that A*x-λ*x-0. this command actually row reduce the matrix defined by the euqaiton above to obtain the corresponding x matrix, which is the corresponding eigen vector.


$ A= \left(\begin{array}{cccc}3&-3\\6&12\end{array}\right) $.

Solve for the eigenvalue/eigenvector.

By hand:

$ det(A-\lambda*I)= \left(\begin{array}{cccc}3-lambda&-3\\6&12-lambda\end{array}\right) =0. $

        λ=6 or 9

        $ \left(\begin{array}{cccc}3&-3\\6&12\end{array}\right)*x=\lambda *x. $
        case 1: λ=5
        $ \left(\begin{array}{cccc}3&-3\\6&12\end{array}\right) $$ \left(\begin{array}{cccc}x1\\x2\end{array}\right) $=$ \left(\begin{array}{cccc}6*x1\\6*x2\end{array}\right). $
        it can be derived that 3x1-3x2=6x1; 6x1+12x2=6x2,thus x1=-x2.Then 

$ x=\left(\begin{array}{cccc}1\\-1\end{array}\right) $

        case 2: λ=9
        $ \left(\begin{array}{cccc}3&-3\\6&12\end{array}\right) $$ \left(\begin{array}{cccc}x1\\x2\end{array}\right) $=$ \left(\begin{array}{cccc}9*x1\\9*x2\end{array}\right) $.
        it can be derived that 3x1-3x2=9x1; 6x1+12x2=9x2,thus -2*x1=x2.Then $ x= \left(\begin{array}{cccc}1\\-2\end{array}\right) $

By Matlab:

        Type in A=[3 -3;6 12]
        Display:A=$ \left(\begin{array}{cccc}3&-3\\6&12\end{array}\right) $
        Type in roots(poly(A))
        Display: ans =
        Type in m=rref(A-(9)*eye(size(A)))
        Display: m =
                    1.0000    0.5000
                         0         0
        This row reduction form clearly shows that x1 is 2*x2
        Type in m=rref(A-(6)*eye(size(A)))
        This shows x1=x2

Relation to other math courses: MA266

Back to MA265, Fall 2011, Walther

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