Generators in group theory

Question

When we consider rotation matrix along $z$ axis and take the infinitesimal value the parameter (rotation angle), we get corresponding generator of the rotation.It has the form shown in the equation.
$$J_3=begin{pmatrix}
0 &-i & 0 
i & 0 & 0 
0 & 0 & 0
end{pmatrix}$$
Now we say by taking exponential I.e. $e^{iJ_{3}alpha}$ we obtain rotation matrix corresponding to rotation angle $alpha$.
But my question is how can we take exponential of $J_3$? Because it is a singular matrix, 3rd column and 3rd row elements are zero.This matrix can't be diagonalizable. In the sense that if we use the expression $P D P^{-1}$. To determine $P^{-1}$ , we need the matrix  $J_3$ to be diagonalizable.

jacob1729 · Accepted Answer

For completeness (although the other answers are probably simpler) the matrix $J_3$ can be written as:
$$ J_3 = begin{pmatrix}1/sqrt{2} & 1/sqrt{2}& 0 i/sqrt{2} & -i sqrt{2}& 0 0 & 0 & 1end{pmatrix}
begin{pmatrix}1 & 0 & 0  0 & -1 & 0  0 & 0 & 0 end{pmatrix}
begin{pmatrix}1/sqrt{2} & -i/sqrt{2}& 0 1/sqrt{2} & i sqrt{2}& 0 0 & 0 & 1end{pmatrix}$$
and its matrix exponential can be computed by exponentiating the diagonal matrix to give:
$$ exp(alpha J_3) = begin{pmatrix}1/sqrt{2} & 1/sqrt{2}& 0 i/sqrt{2} & -i sqrt{2}& 0 0 & 0 & 1end{pmatrix}
begin{pmatrix}e^alpha & 0 & 0  0 & e^{-alpha} & 0  0 & 0 & 1 end{pmatrix}
begin{pmatrix}1/sqrt{2} & -i/sqrt{2}& 0 1/sqrt{2} & i sqrt{2}& 0 0 & 0 & 1end{pmatrix}$$
$$exp(alpha J_3) = begin{pmatrix}cos alpha & sinalpha & 0  -sin alpha & cos alpha & 0  0 & 0 & 1 end{pmatrix}$$
Note that invertibility has nothing to do with diagonalisability. In particular, matrices with determinant zero can usually be diagonalised.

JulianDeV · Answer

The exponential of a matrix is defined through its power series:
begin{equation}
e^A = sum_{n= 0}^infty frac{A^n}{n!}
end{equation}

Cosmas Zachos · Answer

You really don't need to overthink it. Just do it.
Note that
$$
J_3^2= operatorname {diag} ( 1,1,0) equiv K; ~~~~J_3^3=J_3.
$$
You then expand the exponential of the matrix, as normally defined,
$$
exp ( ialpha J_3) = I + ialpha J_3 + ( ialpha J_3 )^2 /2!+ ( ialpha J_3 )^3/3! +...
= operatorname{diag} ( 0,0,1) + cosalpha ~K + isinalpha ~J_3
=begin{pmatrix}
cosalpha & sinalpha & 0 
-sinalpha & cosalpha & 0 
0 & 0 & 1
end{pmatrix}.
$$
Now explore what various values for α  signify on the x-y plane: π/4, π/2, π, 2π. It is a bland rotation. The z direction is left alone.

Generators in group theory

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