- #1
JanClaesen
- 56
- 0
If a n x n matrix A has an eigenvalue decomposition, so if it has n different eigenvalues, by the way, is it correct that a n x n matrix that doesn't have n different eigenvalues can't be decomposed? Are the more situations in which it can't be decomposed? Why can't I just put the same eigenvalue more than once in the diagonal matrix D, perhaps P is not anymore invertible?
A=PDP^-1
Ax = PDP^-1 * x
The vector x holds the weights for combining the vectors in the column space of P^-1,
so P^-1 * x are the coordinates of the vector x if unit vectors are used as base and x itself are the coordinates of the vector x if the column space of P^-1 is used as base.
D scales every component of this transformed vector, and finally the coordinates of this deformed vector are expressed again in terms of P^-1.
If a matrix A doesn't have an eigenvalue decomposition, does this imply that the vector x isn't just scaled, but is scaled/rotated/translated?
A=PDP^-1
Ax = PDP^-1 * x
The vector x holds the weights for combining the vectors in the column space of P^-1,
so P^-1 * x are the coordinates of the vector x if unit vectors are used as base and x itself are the coordinates of the vector x if the column space of P^-1 is used as base.
D scales every component of this transformed vector, and finally the coordinates of this deformed vector are expressed again in terms of P^-1.
If a matrix A doesn't have an eigenvalue decomposition, does this imply that the vector x isn't just scaled, but is scaled/rotated/translated?