You cannot diagonalize the matrix because the 2nd row is a scalar multiple of the first row so there is no way to reduce the first and second row to diagonal form.
Hi guys.
I have the following matrix:
its characteristic polynomial is:
I've been asked if the above matrix has diagonal matrix over , I wrote "no, since the geometric multiplicity of the eigenvalue 1 is 1 (and thus smaller then the algebraic multiplicity).
and then, I've been asked if the above matrix has diagonal matrix over .
I'm guessing no, but I came here to ask, just to make sure I'm not missing anything here:
why does it make a difference if we're over or in this case?
Is it possible to get a different geometric multiplicity in different fields, for the same roots?
Hi Shakarri, thanks for the help.
I know that I cannot diagnolize it, but that's not what I'm getting at here.
The question is, does it make a difference if A is over or over , when I need to determine whether or not A has a similar diagonal matrix?
BTW, are you saying that every matrix A such that one of its rows is a scalar multiple of another cannot be diagonalized?
OK, that's confusing.
take for example the following transformation:
T could be represnted by:
and then, according to what you said, A cannot be diagonalized, since the first row is a scalar multiple of the second, but:
so there exist P such that
(D is a matrix in which the entries outside the main diagonal are all zero thus by definition diagonal), and therfore A can be diagonalized.
Algebraic multipicity is number of identical roots of characteristic equation. Geometric multiplicity is the number of independent eigenvectors corresponding to any multiple root. If algebraic multiplicity equals geometric multiplicity, A can be diagonalized.
In the OP 0 is a root of algebraic multiplicity 2 and 1 is a root of algebraic multiplicity 2
If algebraic multiplicity is k, geometric multiplicity is k if
Rank | A-Iλ_{k} | is n-k (Mirsky, pg 204),
or just try to find k independent eigenvectors corresponding to λ_{k}. If, for example, k=2 and you can only find one independent eigenvector, geometric multiplicity is less than algebraic multiplicity and you can’t diagonalize A.
again, I know all that.
I know what algebraic and geometric multiplicity are.
that was not what I asked.
The question is, does it make a difference if A is over \mathbb{R} or over \mathbb{C}, when I need to determine whether or not A has a similar diagonal matrix?
Yes. You might not get all the roots in R.
In general, a polynomial can only be factored into factors of the form (x-a) if you allow complex roots. There are complex eigen vectors corresponding to complex roots so rule about algebraic and geometric multiplicity remains the same, ie, in general only over C.
That has nothing to do with "diagonalizing". The fact that one row is a scalar multiple of another means that the matrix is not invertible, so has determinant 0 and so has 0 as an eigenvalue. Which we already knew.
Stormey, a matrix that has all real eigenvalues is or is not "diagonalizable" over the real number or complex numbers simultaneously. That is if it is not diagonalizable over the real numbers, it is not diagonalizable over the complex numbers.