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Let there be a projection $P$ given.
Suppose that a Hilbert can be written as a direct sum of two orthogonal sets: $H = M \oplus M^{\perp}$.
I am to show that: $$\text{Ker}P_M = M^{\perp}$$ How can it be done? I do not know much about the properties of projections.

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    $\begingroup$ You have not stated the connection between $P$ and $M$. $\endgroup$ – DisintegratingByParts Jun 3 '18 at 22:29
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Assume $P$ projects onto $M$. Take an arbitrary $x\in H$. We know that $x=x_M + x_{M^{\perp}}$, so $P_M[x] = x_M \in M$. Now for any $x \in H$, we know $\text{ker}P_M = \{y: \langle P_M[x],y \rangle = 0\}$, so let's ask what functions/vectors $y \in H$ are orthogonal to $x_M$, i.e., we want $\{y: \langle x_M, y \rangle = 0\}$. But we see that by definition, any $y \in H$ can be written as $y = y_M + y_{M^{\perp}}$, so $\langle x_M, y \rangle = 0$ iff $y = y_{M^{\perp}}$, i.e., $y \in M^{\perp}$.

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You know that $\operatorname{Im} P_M = M$ and that $P_M^* = P_M$. Also in general for a bounded linear map $T : H \to H$ we have $\ker T = (\operatorname{Im }T^*)^\perp$.

Now it is simply$$\ker P_M = (\operatorname{Im} P_M^* )^\perp = (\operatorname{Im} P_M)^\perp = M^\perp$$

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