I suggest that you check out the two following references to answer your question:

Dear Loopinder Sood,

The eigenvalues of a matrix A are first calculated with the characteristic equaton |A-lambda*I|=0 and then the eigenvectors {Phi} are obtained with [A-lambda*I]*{Phi}={0} for each lambda

I suggest a SCHUR descomposition, in MatLab:

https://www.mathworks.com/help/matlab/math/eigenvalues.html?requestedDomain=www.mathworks.com

My answer was wrong (I am surprised to read what I wrote :))), I am deleting it.

Dear Viera

I think what you are saying is true for Diagonal Matrices, but not true for Triangular Matrices.

Dear Loopinder, let us verify (see the attached file) for the eigenvalue a.

All other verifications are similar.

It's known that if  we have a triangular matrix [A], the eigenvalues are simply the values of the main diagonal.

For the matrix [ a, 0, 0 ;,d, e, 0 ; g, h, i ] the eigengenvalues are : a , e , i

The eigenvector for lambda1=a is :

{Phi}={ 1 , d/(a-e) , (-g+h.d/(e-a))/(i-a) }

Proceed with these similar calculations to obtain the other eigenvectors.