Question on uniqueness of projections in linear algebra Suppose A and B are projection matrices and suppose that for some vector x in the column space of B, BAx=x. Can I say that Ax=x?
Since x is in the column space of B, I know that Bx=x it projects itself in its own column space. I guess my question is, is x the only vector that can give the outcome of x when premultiplied by the projection matrix B?
 A: Let $A = \begin{pmatrix}0&1\\0&1\end{pmatrix},$ $B = \begin{pmatrix}0&0\\0&1\end{pmatrix},$ and $x = \begin{pmatrix}0\\1\end{pmatrix}.$ Then $A^2 = A$ and $B^2 = B,$ so $A$ and $B$ are indeed projection matrices. On the other hand, 
$$
\begin{pmatrix}0&0\\0&1\end{pmatrix}\begin{pmatrix}0&1\\0&1\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}0&0\\0&1\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}= \begin{pmatrix}0\\1\end{pmatrix},
$$
so that $BAx = x,$ but $Ax\neq x.$
A: For orthogonal projections yes, because you have $\Vert Ax\Vert\leq\Vert x\Vert$, with equality sign obtained if and only if $x$ is in the eigenspace of $A$ with eigenvalue $1$. Then $\Vert BAx\Vert\leq\Vert Ax\Vert\leq\Vert x\Vert$. For all these inequalities to take equal sign, $BAx=Ax=x$.
But for non-orthogonal projections, no. You can construct a 2D example using parallelograms. Project the vector $x$ first onto two non-orthogonal straight lines to obtain $\vec{x}=\vec{a}+\vec{b}$. Let's say what $A$ does is project $\vec{x}$ onto $\vec{b}$, so $\vec{b}=A\vec{x}$. Then we can choose $\vec{c}=\vec{b}-\vec{x}$. This decomposes $\vec{b}=\vec{c}+\vec{x}$. So it's then possible for $B$ to project the vector $\vec{b}=A\vec{x}\,$ back to $\vec{x}=B\vec{b}=BA\vec{x}$.
