How to calculate Eigen Value and Eigen Vector of a matrix?

Question

I need a simple and clear method explained to find the eigen value and eigen vectors of a square matrix.Can anybody explain in detail or any sites which explains the procedure with examples in a clear way?

Answer 1

Let your matrix be A. Compute the polynomial
p(x)=det(A-xI).
i.e. the determinant of the matrix obtained by subtracting x off of
every diagonal element of A. If A is an nxn matrix, then p(x) will be a polynomial of degree n.

The eigenvalues of A will be the roots of p(x)=0.

Now, if x=a is an eigenvalue, look at the matrix equation
(A-aI)v=0.
This can be solved by elementary row operations. The vectors v obtained in this way are the eigenvectors corresponding to the eigenvalue a. There is a whole vector space of solutions, so it is enough to just find a basis for the solution space.

Now do this for every eigenvalue you found to find all the eigenvectors.

Answer 2

Here are some links below with very simple examples. It's quite surprising that there is not much out there that shows step-by-step how to find eigenvalues and eigenvectors.

The general idea is to find the eigenvalues first. This is easy because you simply form the characteristic polynomial and find it's roots. The roots are the eigenvalues.

Finding the eigenvectors requires some extra work, i.e. you still end up having to do matrix manipulations depending on how complex your matrix is.

Do you know how to find the eigenvalues and eigenvectors of a 2x2 matrix? If not, I suggest you start with this so you can get the general idea. Then you can try computing the same for 3x3 matrices. Eventually you can start thinking about general algorithms. Do all this manually (by hand) and only use software like Matlab to check answers. Beware of Matlab's rref function in early student versions.

Answer 3

Let A be an nxn matrix, we want to find the nx1 vector x such that:

Ax = (lambda)x is true, where lambda is an unknown number.

We do this by solving the equation:

Ax -(lamda)(Identy_Matrix)x = 0

Now it is always the case that the zero vector will solve the problem but that is trivial, and does not lead to what we want.

(A-Idenity_Matrix*lambda)x=0

has a non trivial solution if the determinate of this matrix is set to zero, and we solve for lamba (several values may be obtained, at most n).

We then plug in for each value of lambda and solve for the vector x associated with the individual value of lambda.

You should consult a good linear algebra text, from memory the one written by Howard Anton is good and no - nonsense.

Answer 4

If your matrix is A then it can be expressed as:

A = X D X(inverse) {D is diagonal}

X contains eigenvectors and D has the eigenvalues on its diagonal.
A must be square but it can be asymmetric or even complex.

Its a pain to do it manually I use a computer program

Answer 5

you need to start with a non-singular square matrix.... along the diagonal, subtract alpha from each term.... with alpha's in place...figure the determinate of matrix to form a polynomial with alpha terms... factor poly to figure EIGENVALUE.....

put EValue(s... one at a time) into matrix w/ alpha's to determine new singular matrix (the purpose of EValue is to turn non-singular into singular)
this singular matrix is then broken down so that one or more of the variables is a function of the other...
you come to a solution like x1= -x3 then the EVector is <1,0,-1>


hope this helps...

Answer 6

Av = labda.v <= just solve this.

another way is to wite the matrix as QDQ^-1
with D a diagonal matrix of eigenvalues.

Answer 7

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Answer 8

r u a 1st year student. ?
i request u to buy T.Veeragan's book for studying or u can go to a library.

Answer 9

there is a set pattren which is obvious
concern a library if net is hell