The Hermitian operator can have many outer product representations? Let $H$ be a Hermitian operator, so it can be written as
$$H=\lambda_1P_1+\lambda_2P_2........\lambda_kP_k,$$
where $\lambda_i$  are eigen values and $P_i$ corresponding projector operators
for the corresponding eigenspace where $1 \leq i \leq k$. Also if $i \ne j$, $\;$ $P_iP_j=0$.
I am not sure about the above but if its true then I am struggling with the question that the projector operator for a space can be written in infinitely many ways in bra-ket outer product form (by choosing any set of orthonormal basis for that space). So all the $P_i$ above independently can be written in many ways, thus $H$ can be written in many ways, are all these representations for the same $H$, am I missing something?

EDIT
Let the eigen space corresponding to eigen value $\lambda_i$ have dimension $d_i$. Also let an orthonormal basis corresponding to it be ${|j_{i1}\rangle,|j_{i2}\rangle,.....|j_{id_i}\rangle}$. Thus $H$ can be written as
$$H=\sum_{i=1}^{i=k} \lambda_i \sum_{l=1}^{l=d_i}|j_{il}\rangle \langle j_{il}|$$
Now if I take a different orthonormal basis for each eigen space then also it is the same
operator, ie say
$$H^{'}=\sum_{i=1}^{i=k} \lambda_i \sum_{l=1}^{l=d_i}|m_{il}\rangle \langle m_{il}|$$
$$H=H^{'}......(1)$$
where ${|m_{i1}\rangle,|m_{i2}\rangle,.....|m_{id_i}\rangle}$ is the new orthonormal basis for space corresponding to eigen value $\lambda_i$. But one orthonormal basis can be obtained from another by some unitary transformation thus
$$H^{'}=UHU^{\dagger}.....(2)$$
for some unitary operator $U$. But $(1)$ and $(2)$ are contradictory, because using both we get $H=UHU^{\dagger}$, but $H=H$.Where am I going wrong ?
 A: You are right. Think of the limit case of the Heritian identity operator $I$, it can be written in infinitely many ways, one for each Hilbertian basis of the Hilbert space.
ADDENDUM. Referring to the EDIT, if you explicitly perform computations you see that $U$ commutes with $H$ and thus it is not effective as it must be. 
A: When you give the form of an operator with respect to a certain basis, it would be good to explicitly state which basis you are using. The fact that you get $H' = UHU^*$ is giving you a way to represent the same operator in a different basis. Let's say you have bases $B$ and $B'$ for a Hilbert space. What you have is the operator $H$ in the basis $B$, and we shall denote this by $[H]_B$. Now the same operator in the basis $B'$ is related to $B$ by a unitary transformation, so
$$[H]_{B'} = U[H]_BU^*.$$
So an operator must not be confused with its representation, because operator algebras can be given in the abstract through, say, generators and relations alone, and there is, in principle, no representation Hilbert space.
A: I) Let us for simplicity assume that the Hilbert space $H$ is finite-dimensional. Let $A:H\to H$ be a normal operator [and hence orthogonally diagonalizable] with eigenvalues
$${\rm Spec}(A)~=~\{\lambda_1, \ldots, \lambda_n\}.$$
We can uniquely decompose the Hilbert space
$$H~=~\oplus_{i=1}^n H_i,$$
in orthogonal eigenspaces 
$$H_i~:=~  {\rm ker}(A-\lambda_i{\bf 1}_H) ~\subseteq~H$$
for $A$. There exist unique corresponding projection operators $P_i:H\to H$, so that
$$ P_i^2~=~P_i, \qquad P_i^{\dagger}~=~P_i,\qquad {\rm Im}(P_i)=H_i.$$
Then we may uniquely write 
$$A~=~\sum_{i=1}^n \lambda_iP_i. $$
II) We see that it is enough to analyze and restrict attention to a single eigenspace $H_i$ for a fixed index $i\in\{1,\ldots,n\}$. Let us therefore drop the index $i$ from the notation, i.e. call $H_i$ for $H$. In other words, we may assume without loss of generality that the operator $A:H\to H$ is proportional to the identity
$$ A~=~\lambda{\bf 1}_H. $$
We can now pick an orthonormal basis $\left(|e_k\rangle\right)_k$ for $H$ in infinitely many ways. However the "outer product representation" of the unit operator
$${\bf 1}_H~=~\sum_k |e_k\rangle\otimes \langle e_k|$$
will be independent of the choice of orthonormal basis $\left(|e_k\rangle\right)_k$. Moreover it will commute with any other operator. In particular, unitary operators.
