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Thread: Projection

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    Projection

    Let $\displaystyle H$ be a Hilbert Space and $\displaystyle P$ and $\displaystyle Q$ be the projections on closed linear subspaces $\displaystyle M$ and $\displaystyle N$ of $\displaystyle H$.
    Prove that if $\displaystyle Q-P$ is a projection,then its range is $\displaystyle N\cap M^\perp$.
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    Quote Originally Posted by problem View Post
    Let $\displaystyle H$ be a Hilbert Space and $\displaystyle P$ and $\displaystyle Q$ be the projections on closed linear subspaces $\displaystyle M$ and $\displaystyle N$ of $\displaystyle H$.
    Prove that if $\displaystyle Q-P$ is a projection,then its range is $\displaystyle N\cap M^\perp$.
    For a start, $\displaystyle I-P$ is a projection, with range $\displaystyle M^\perp$. So if $\displaystyle x\in N\cap M^\perp$ then $\displaystyle (I-P)x=x$, $\displaystyle Px=0$, $\displaystyle Qx=x$ and hence $\displaystyle (P-Q)x=x$. Thus if $\displaystyle P-Q$ is a projection then its range contains $\displaystyle N\cap M^\perp$.

    For the converse inclusion, if $\displaystyle P-Q$ is a projection then
    $\displaystyle (1)\quad (Q-P)^2 = Q-P$,
    $\displaystyle (2)\quad Q - PQ - QP + P = Q-P$ (since $\displaystyle P^2=P$ and $\displaystyle Q^2 = Q$),
    $\displaystyle (3)\quad PQ+QP=2P$.
    Multiply both sides of (3) on the left to get
    $\displaystyle (4)\quad QPQ=QP$.
    Multiply both sides of (3) on the right to get
    $\displaystyle (5)\quad QPQ=PQ$.
    From (4), (5) and (3),
    $\displaystyle (6)\quad PQ=QP=P$.
    Therefore
    $\displaystyle (7)\quad (I-P)Q=Q(I-P)=Q-P$.
    It follows from (7) that the range of $\displaystyle Q-P$ is contained in the range of $\displaystyle (I-P)$ and in the range of $\displaystyle Q$, and thus in $\displaystyle N\cap M^\perp$.
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