# Is the Markov matrix always symmetrical

## Eigenvalues and Eigenvectors

Mathematics for Computer Scientists pp 385-410 | Cite as

### abstract

We have a lot of

*n*linearly independent vectors**u**_{1}, . . . ,**u**_{n}∈ ℝ^{n}(or ℂ^{n}) is called the base, since every vector x**u**_{n}∈ ℝ^{n}as a linear combination$$ {\ text {x =}} \ sum \ limits_ {j {\ text {= 1}}} ^ n {y_j u_j} $$

lets write. If we consider these base vectors to be fixed, the vector x can be given both by its **Coordinates***x*_{1}, . . . ,*x*_{n}regarding the standard base e_{1},. . . , e_{n}, as well as by its coordinates*y*_{1}, . . . ,*y*_{n}regarding the new base**u**_{1}, . . . ,**u**_{n}to be discribed. If we use the basis vectors**u**_{j}as columns of a matrix$$ U = (u_1 u_2 ... u_n) $$

then we can easily calculate back and forth between the different coordinates: $$ x = Uy, y = U ^ {- 1} x. $$

In particular, the relationship between the basis vectors is through $$ u_j = Ue_j, j = 1, ..., n $$

given, i.e., the matrix *U*is the matrix of that linear map that contains the old base e_{1},. . . , e_{n}in the new base**u**_{1}, . . . ,**u**_{n}convicted. This linear mapping or the associated matrix*U*becomes**Base transformation**or**Coordinate transformation**called.This is a preview of subscription content, log in to check access.

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© Springer-Verlag Berlin Heidelberg 2008

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