Can a non-symmetric projection matrix exist? Can a non-symmetric projection matrix exist? I am currently using the matrix $M = I - wi^T$ where $i$ is a vector of ones and $w$ is a vector of weights so $w^T i = 1$. Is $M$ a projection matrix? Is it possible for non-symmetric projection matrices  to exist (as I believe this matrix to be)? Thanks in advance. 
 A: Yes and yes.
If by projection matrix you mean that $P^2=P$, then e.g. $$\begin{pmatrix}1&1\\0&0\end{pmatrix}$$ satisfies this.
Your matrix $P=I-wi^T$, when expanded out in components, reads $P_{jk}=\delta_{jk}-w_j i_k$ (using $i$ as a vector is a somewhat unfortunate notation). Then you can check that $\left(P^2\right)_{jk}=P_{jl}P_{lk}=P_{jk}$ indeed holds, by virtue of your condition $w_j i_j=1$.
Update:
Your matrix $P$ acts in the following way:
It annihilates $w$, since $$Pw=\left(I-w i^T\right) w=w-w (i^Tw)=0\,.$$ On the other hand, it projects onto the space orthogonal to $i$, since for any $v$
$$i^T \, Pv=i^T \left(I-w i^T\right) v=i^T v - (i^T w) (i^T v)=0\,.$$
That means it does not project orthogonally, since $Pv\neq P\cdot(v+\lambda i)$ -- rather, it projects 'along $w$', i.e. $Pv=P\cdot(v+\lambda w)$. This bring us back to your first question: $P$ is generically not symmetric, 
$$P^T=\left(I-w i^T\right)^T=I- i w^T \neq P\,,$$ 
unless $w=\lambda i$. (The factor lambda is fixed by the normalisation $i^Tw=1$.) That is, excactly when $w$ and $i$ are proportional, $P$ is symmetric and projects orthogonally.
A: If $P$ is a square matrix such that $P^2=P$ ($P$ is idempotent), then it represents the projection onto $C(P)$ (column space of $P$) along $C(I-P)$ ($I$ the identity matrix), that is,
$$
v=Pv+(I-P)v
$$
is the unique way to write $v$ as the sum of a vector in $C(P)$ with a vector in $C(I-P)$. Indeed, if
$$
v=x+y
$$
with $x\in C(P)$ and $y\in C(I-P)$, then $x=Px'$ and $y=(I-P)y'$ for some $x'$ and $y'$, so
$$
Pv=P(Px'+(I-P)y')=P^2x'+(P-P^2)y'=Px'+0=x
$$
and, similarly, $(I-P)v=y$.
Consequently, $C(P)\cap C(I-P)=\{0\}$.
Conversely, if $U$ is any subspace of $\mathbb{R}^n$ and $\{v_1,v_2,\dots,v_k\}$ is a basis for $U$, if you complete the basis to a basis $\{v_1,\dots,v_k,v_{k+1},\dots,v_n\}$ of $\mathbb{R}^n$, the map
$$
v=\alpha_1v_1+\dots+\alpha_kv_k+\alpha_{k+1}v_{k+1}+\dots+\alpha_nv_n
\mapsto
\alpha_1v_1+\dots+\alpha_kv_k
$$
is linear and its matrix $P$ has the property that $P^2=P$.
On the other hand, if you add the assumption that $C(I-P)$ is the orthogonal complement to $C(P)$, then the matrix $P$ is symmetric (where the inner product is the standard one).
Indeed, we must have $\langle (I-P)v,Pw\rangle=0$, for every $v,w$, which translates to
$$
((I-P)v)^TPw=0
$$
that is
$$
vPw=vP^TPw
$$
for all $v,w$, and so $P=P^TP$, which implies $P^T$ is symmetric.
In your case with $P=I-wi^T$, symmetry reads
$$
I-iw^T=I-wi^T
$$
that is, $iw^T=wi^T$; multiplying by $i$ on the right,
$$
iw^Ti=wi^Ti
$$
so
$$
w=\frac{\langle w,i\rangle}{\langle i,i\rangle}i
$$
and $w$ is a scalar multiple of $i$. Conversely, if $w=\alpha i$, then $I-wi^T=I-\alpha ii^T$ is clearly symmetric.
Hence $I-wi^T$ is symmetric if and only if $w$ is a scalar multiple of $i$. But it is idempotent as soon as $\langle i,w\rangle=1$:
$$
(I-wi^T)(I-wi^T)=I-wi^T-wi^T+wi^Twi^T=
I-wi^T-wi^T+\langle i,w\rangle wi^T
$$
