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Show that the orthogonal projection is linear.

Let $x_i=y_i+z_i$, where $x_i\in X$, $y_i\in Y$, $z_i\in Y^\perp$, and $\alpha,\beta$ be scalars. Then \begin{align}P(\alpha x_1+\beta x_2)&=P(\alpha(y_1+z_1)+\beta(y_2+z_2))\\&=P(\alpha y_1+\beta y_2+\alpha z_1+\beta z_2)\end{align}

Now I know $\alpha x_1 + \beta y_1 \in Y$ but not sure what to do here.

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  • $\begingroup$ I think you mean that $\alpha y_1 + \beta y_2 \in Y$. Can you show that $\alpha z_1 + \beta z_2 \in Y^{\perp}$? $\endgroup$ – Michael Joyce Feb 8 '16 at 17:09
  • $\begingroup$ yes I have shown that too $\endgroup$ – can2 Feb 8 '16 at 17:13
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The key here is that $X = Y\oplus Y^\perp$, i.e. for any $x\in X$ there are unique $y\in Y, z\in Y^\perp$ such that $x = y+z$. This is essential in order for $P(x) = P(y+z) = y$ to be well defined in the first place.

Now, what you showed is that $\alpha x_1 +\beta x_2$ can be uniquely written as $(\alpha y_1 + \beta y_2) + (\alpha z_1 + \beta z_2)$ where $\alpha y_1 + \beta y_2\in Y$, $\alpha z_1 + \beta z_2\in Y^\perp$. So, what is $P(\alpha x_1 + \beta x_2)$?

Edit:

Your definition of orthogonal projection assumes that for each $x\in X$ there is unique $y\in Y$ such that $x-y\in Y^\perp$. This is actually equivalent to stating that $Y\oplus Y^\perp = X$, i.e. there are unique $y\in Y$, $z\in Y^\perp$ such that $x = y+z$ (notice that $z = x-y$ from your definition).

So, assume that there is unique $y\in Y$ such that $x-y\in Y^\perp$. Then, $P(x) = P(y+(x-y)) = y$ is just restating my claim if you substitute $z = x-y$.

The bigger question is why such $y$ exists. In finite-dimensional case this follows immediately from existence of orthonormal basis for $Y$. Then you can define $y = \sum \langle x,e_i\rangle e_i$. In infinite-dimensional case we can use projection theorem for Hilbert spaces when we can find such $y \in Y$ that minimizes length $\|x-y\|$ (think of a point and a line: minimum distance between point and line is given by orthogonal projection).

I hope this clarifies things a bit.

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  • $\begingroup$ This is what I don't understand, why is P(y+z) = y? My definition of $P$ is the unique vector $y \in Y$ s.t. $y = Px$ and $x-y$ is orthogonal to the set $Y$ $\endgroup$ – can2 Feb 8 '16 at 17:31
  • $\begingroup$ @can2 : his $z$ is your $x-y$ (he wrote $y \in Y, z \in Y^{\perp} \implies z$ is orthogonal to $y$) $\endgroup$ – reuns Feb 8 '16 at 17:45
  • $\begingroup$ @can2, please see my edit. $\endgroup$ – Ennar Feb 8 '16 at 18:23
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Note that $P(ax_1+bx_2) = ay_1+by_2 = aP(x_1)+bP(x_2)$ by definition of a projection onto $Y$.

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  • $\begingroup$ please see my above comment $\endgroup$ – can2 Feb 8 '16 at 17:31

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