MHB Is the Product of Two Diagonalizable Matrices Always Diagonalizable?

  • Thread starter Thread starter Yankel
  • Start date Start date
  • Tags Tags
    Matrix
Yankel
Messages
390
Reaction score
0
Hello

I have a little question

if A and B are both invertible and diagonalizable matrices (from the same order), is A*B a diagonalizable matrix ? why ?

I have not got a clue...

thanks !
 
Physics news on Phys.org
Yankel said:
if A and B are both invertible and diagonalizable matrices (from the same order), is A*B a diagonalizable matrix ? why ?

It is not true. Verify that $A=\begin{bmatrix}1&{\;\;1}\\{2}&{-1}\end{bmatrix}$ and $ B=\begin{bmatrix}{1}&{2}\\{2}&{1}\end{bmatrix}$ are invertible and diagonalzable matrices on $\mathbb{R}$, however $AB=\begin{bmatrix}{3}&{3}\\{0}&{3}\end{bmatrix}$ it is not diagonalizable.
 
Yankel said:
Hello

I have a little question

if A and B are both invertible and diagonalizable matrices (from the same order), is A*B a diagonalizable matrix ? why ?

I have not got a clue...

thanks !

Not always. But when A,B are interchangable, i.e., AB=BA, then AB IS diagonalizable since then A and B are simultaneously diagonalizable, i.e., we can find the same invertible matrix
S such that $A=S^{-1} D_1 S, B=S^{-1} D_2 S$, where $D_1,D_2$ are both diagonal matrices. Thus we have $AB=S^{-1}D S$ with $D=D_1D_2$ a diagonal matrix.
 
The world of 2\times 2 complex matrices is very colorful. They form a Banach-algebra, they act on spinors, they contain the quaternions, SU(2), su(2), SL(2,\mathbb C), sl(2,\mathbb C). Furthermore, with the determinant as Euclidean or pseudo-Euclidean norm, isu(2) is a 3-dimensional Euclidean space, \mathbb RI\oplus isu(2) is a Minkowski space with signature (1,3), i\mathbb RI\oplus su(2) is a Minkowski space with signature (3,1), SU(2) is the double cover of SO(3), sl(2,\mathbb C) is the...
Back
Top