Combining projectors Consider an inner product space with two projection operators $P_1$ and $P_2$. I would like to construct a projection operator $P_1 \oplus P_2$ that projects onto the span of the union of the 1-eigensubspaces of $P_1$ and $P_2$.
In case these two subspaces are orthogonal (i.e. $P_1\circ P_2=P_2\circ P_1=0$), the new projector is simply
$$ P_1\oplus P_2 = P_1 + P_2$$
My question is, if there is an algebraic expression for the joint projection operator in the general case. 
If $P_1$ and $P_2$ commute as in $P_1\circ P_2=P_2\circ P_1$, then the combined projector must be
$$P_1 \oplus P_2 = P_1 + P_2 - P_1 \circ P_2$$
I have tried to come up with corrections for the non-commuting case, but have not been able to make it work yet. I would welcome any ideas or suggestions for how to approach this.
 A: Here is my own attempt at an answer. I have not yet worked out the details, so it's more of an idea and I would be happy if others help to fill in the missing parts and verify the approach.
The operator $P_1 + P_2$ is non-negative and symmetric, so it can be diagonalised and the sum of all eigensubspaces with eigenvalues $>0$ forms the desired eigensubspace of $P_1 \oplus P_2$.
So if we can map all eigenvalues $\lambda>0$ to $1$ while preserving the kernel, we have constructed the desired projector.
The analytic continuation of $f_t(x)=1-\exp(-t x )$ applied to $P_1+P_2$ realises this mapping for $t\to\infty$.
The series expansion of $f_t(x)$ is
$$f_t(x) = -\sum_{n=1}^\infty (-1)^n t^n \frac{x^n}{n!}$$
and therefore
$$f_t(P_1 + P_2) = - \sum_{n=1}^\infty (-1)^n t^n \frac{(P_1+P_2)^n}{n!}$$
Binomial expansion with $P^{k>0}=P$ and ($n\ge 1$) as well as, for now, the assumption of $\left[P_1,P_2\right]=0$ gives 
$$(P_2+P_2)^n = P_1 + P_2 + \sum_{k=1}^{n-1} {n\choose k} P_1 P_2=P_1+P_2+ (2^n-2)P_1 P_2$$
which makes
$$f_t(P_1+P_2) = -\sum_{n=1}^\infty \frac{(-t)^n}{n!}\left(P_1+P_2+ (2^n-2)P_1 P_2 \right)$$
Regrouping and evaluating the sum simplifies this to
$$f_t(P_1+P_2) = \left(1-\exp(-t)\right)\times \left(P_1+P_2 - 2 P_1 P_2\right)+\left(1-\exp(-2t)\right)\times P_1 P_2 $$
and finally 
$$P_1\oplus P_2 = \lim_{t\to\infty}f_t(P_1+P_2)=P_1+P_2-P_1 P_2$$
This is the correct result for $P_1$ and $P_2$ commuting. 
This is how far I got for now. The non-commuting binomial expansion makes things way more complicated. If anyone sees a shortcut I'd be very thankful to hear it. Also, any comments are highly welcome.
Edit: Some more thoughts
The non-commutative expansion is 
$$(P_1+P_2)^n = \sum_{k=0}^{n-1} {n-1\choose k} \left( P_{(1,2)}^k + P_{(2,1)}^k \right)$$
where 
$$P_{(a,b)}^k = \prod_{m=1}^k \left\{ \begin{array}{rl}
 P_a &\mbox{ if $m$ odd} \\
 P_b &\mbox{ if $m$ even}
       \end{array} \right. $$
Like above, plugging this result into the series of the saturation function gives
$$f_t(P_1+P_2) = -\sum_{n=1}^\infty \frac{(-t)^n}{n!} \sum_{k=0}^{n-1} {n-1\choose k} \left( P_{(1,2)}^k + P_{(2,1)}^k \right)$$
Takig the limit $t\to\infty$ we arrive at
$$P_1 \oplus P_2 = -\sum_{k=1}^\infty (-1)^k \left( P_{(1,2)}^k + P_{(2,1)}^k \right)$$
Checking this result by assuming that $P_1$ and $P_2$ commute and therefore $P_{(1,2)}^k = P_{(2,1)}^k = P_1 P_2$ if $k>1$ we get the special case
$$P_1 \oplus P_2 = P_1 + P_2 - 2  \sum_{k=1}^\infty (-1)^k P_1 P_2$$
which does only converge in the cauchy principal value sense to the known result for commuting projectors. So the general result does not always converge nicely and is not suitable for approximate series evaluations. 
I can conclude that there doesn't seem to be a nice closed form algebraic expression for the combined projector, but a more in depth analysis of the convergence of the result found has to be performed. If convergence can be guaranteed in the non-commuting case and with the simple special case  result it would still be useful.
Edit 2:
It seems that the limit can be regulated and 
$$P_1 \oplus P_2 = - \lim_{\alpha<1\to 1}\sum_{k=1}^\infty (-1)^k \alpha^k \left( P_{(1,2)}^k + P_{(2,1)}^k \right)$$
exists and is well behaved.
