proof of theorem for normal matrices
1) ( is normal)
Keeping in mind that every matrix commutes with its own powers, let’s compute
which shows to be normal.
2) ( is normal )
Let , be the distinct eigenvalues of A, and let ; then it’s possible to find a -degree polynomial such that , solving the linear Vandermonde system:
Since these eigenvalues are distinct, the Vandermonde matrix is full rank, and the linear system admits a unique solution; so a -degree polynomial can be found such that and therefore . Writing these equations in matrix form, we have
and since is normal we have .
Let’s evaluate .
But, keeping in mind that ,
which is the thesis.
Remark: note that this is a constructive proof, giving explicitly a way to find polynomial by solving Vandermonde system in the eigenvalues.
Let (which is easily checked to be normal),
with . Then and the Vandermonde system is
from which we find
A simple calculation yields
|Title||proof of theorem for normal matrices|
|Date of creation||2013-03-22 15:36:36|
|Last modified on||2013-03-22 15:36:36|
|Owner||Andrea Ambrosio (7332)|
|Last modified by||Andrea Ambrosio (7332)|
|Author||Andrea Ambrosio (7332)|