I have a problem in a following form
$$\mathrm{C = A\underbrace{X(YX)^{+}Y}_{\neq I}B}$$
where $+$ indicates pseudoinverse.
$\mathrm{dim(C) = (L, 1) \\ dim(A) = (L, N) \\ dim(X)=(N, M) \\ dim(Y)=(M, N) \\ dim(B)=(N, 1)}$
and ${1<M<N<L}$
I have verified numerically that $$\mathrm{C=AB}$$ How do I prove it analytically?
Update:
$\mathrm{X = \matrix{x_{11} & z & z & \cdots & z & z \\ z & x_{22} & z & \cdots & z & z \\ z & z & x_{33} & \cdots & z & z \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \\ z & z & z & \cdots & x_{M-1, M-1 } & z \\ z & z & z & \cdots & z & z \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \\ z & z & z & \cdots & z & z}}$
with $\mathrm{ z > 0}$
Matrices are complex-valued and have full rank
Based on the conditions you gave, I have the following counterexample:
$$A=\begin{bmatrix}0 & 1 &0\\1 & 0 & 1\\1 & 0 & 1\\0 & 1 & 1\end{bmatrix},\ B=\begin{bmatrix}0\\1\\0\end{bmatrix},\ X=\begin{bmatrix}2 & 1\\1&2\\1&1\end{bmatrix},\ \mathrm{and}\ Y=\begin{bmatrix}1 & 0 & 0\\1&1&1\end{bmatrix}.$$
Then, we have $$AX(YX)^{+}YB=\begin{bmatrix}0.75\\0.25\\0.25\\1\end{bmatrix}\ \mathrm{and}\ AB=\begin{bmatrix}1\\0\\0\\1\end{bmatrix}.$$
So, I guess there is a need for extra assumptions.
It can be shown that the matrix $X(YX)^{+}Y$ is similar to the matrix
$$\begin{bmatrix}I & 0\\ 0 & 0\end{bmatrix}.$$
So let $M:=X(YX)^{+}Y$ and define $V$ to be a basis of the right-eigenspace of $M$ associated with the eigenvalue 1 (i.e. $MV=V$) and let $U$ be a basis of the left-eigenspace of $M$ associated with the eigenvalue 1 (i.e. $UM=U$). Define also $U_0$ to be a basis of the left null-space of $M$ and $V_0$ to be a basis of the right null-space of $M$. Those matrices can be chosen such that
$$\begin{bmatrix}U\\U_0\end{bmatrix}\begin{bmatrix}V & V_0\end{bmatrix}=I.$$
Now we can claim that $AMB=AB$ if and only if $A$ and $B$ are such that $A=ZU+Z_0U_0$ and $B=VQ+V_0Q_0$ for some $(Q,Q_0,Z,Z_0)$ where $Z_0Q_0=0$
$$AMB=(ZU+Z_0U_0)M(VQ+V_0Q_0)=ZUVQ=ZQ$$ and $$AB=(ZU+Z_0U_0)(VQ+V_0Q_0)=ZQ+Z_0Q_0.$$
The result then follows.