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tzt
2004 Oct 19
3
matrix of eigenvalues
I thought that the function
eigen(A)
will return a matrix with eigenvectors that are independent of each
other (thus forming a base and the matrix being invertible). This
seems not to be the case in the following example
A=matrix(c(1,2,0,1),nrow=2,byrow=T)
eigen(A) ->ev
solve(ev$vectors)
note that I try to get the upper triangular form with eigenvalues on
the diagonal and (possibly) 1 just