0
$\begingroup$

I am having a hard time trying to prove the following:

Let $M$ be the space of all $2\times2$ complex matrices,

which is skew-hermitian ( i.e. $\bar{X}^t = -X$ ).

We consider M as a vector space over R,

For any $A\in M$,

define the operator $\text{ad}_A : M\to M $ by $\text{ad}_A (X) = AX – XA$.

Show that $\text{ad}_A$ is diagonalizable.

Let $$A=\left[\begin{array}{cc}i & 1\\-1 & i\end{array}\right].$$ How should I compute the eigenvalues of the operator $\text{ad}_A$ ?

(Maybe we should show there is a basis of $M$ consisting of eigenvectors of $\text{ad}_A$ ?)

Thanks.

  • 0
    @AlexeiAverchenko My comment was before the OP edited his question. I guess there's no need for it now. I've deleted it.2012-11-08
  • 0
    $X$ is also a real (and possibly skew-symmetric) matrix?2012-11-08
  • 1
    If we're talking about *real* matrices, we say "skew-symmetric", rather than "skew-hermitian". $M_2(\Bbb F)$ is standard notation for the ring of **all** $2\times2$ matrices with entries in a field $\Bbb F$, so that's a bit confusing. Also, note that the matrix $A$ you defined *isn't* in $M_2(\Bbb R)$, but *is* in $M_2(\Bbb C)$.2012-11-08
  • 0
    I am not sure if this is wat u wanted, note that $(ad_{A}(X))^{H}=XA-AX=-1*ad_{A}(X)$, so you always get a skew hermitian matrix, and since every skew hermitian matrix is diagonalizable, $ad_{A}(X)$ should also be diagonalizable.2012-11-08
  • 0
    I am trying to use generators and relations here. Let M ≤ S_5 be the subgroup generated by two transpositions t_1= (12) and t_2= (34). Let N = {g ∈S_5| gMg^(-1) = M} be the normalizer of M in S_5. How should I describe N by generators and relations? How should I show that N is a semidirect product of two Abelian groups? How to compute |N|? How many subgroups conjugate to M are there in S_5 ? Why? (I think Sylow's theorems should be used here.) @dineshdileep Please help!!!2012-11-08
  • 0
    @Jack: to ask a new question, use the "Ask Question" link. http://math.stackexchange.com/questions/ask2012-11-09
  • 0
    I reached the limit of no. of questions for today2012-11-09
  • 0
    So I can only update problem this way2012-11-09
  • 0
    Do you know how to solve it ?2012-11-09
  • 0
    @Jack: Consider whether being banned is worth this.2012-11-09
  • 0
    It is not worth talking to you about this2012-11-09

2 Answers 2