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I know what a projection operator is, but I am unable to explain it in words without using mathematical symbols. Can anyone help me?

I don't need examples or the definition - I want to know why and how its need arose, and what is the idea behind it?

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    $\begingroup$ Did you read the Wikipedia exposition? $\endgroup$ – Bill Dubuque May 29 '15 at 13:36
  • $\begingroup$ maths.kisogo.com/index.php?title=Projector there are canonical (natural) projections too, that means they can be stated without reference to basis, eg $P_V:V\oplus W\rightarrow V$ given by $P:(v,w)\rightarrow v$ for example. Also have you ever written "taking the x coordinate" - that's a projection. $\endgroup$ – Alec Teal May 30 '15 at 6:09
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    $\begingroup$ What's wrong with the current answers? Sure, the most popular one is a bit tongue in cheek (simply describing what idempotence means in a more colloquial setting), but the answers below it do describe projection mathematically in pretty clear detail. $\endgroup$ – Dustan Levenstein Jun 6 '15 at 13:38
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This is how I used to imagine projections:

If a mouse:

gets run over by a steamroller:

enter image description here

It will look like this:

enter image description here

Now if it gets run over by a steamroller another time, it will still look like this:

enter image description here

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    $\begingroup$ Wow, I've been here for three years now, and still I want to give you all my rep points, this is too good!!! +1! I hope you get that green check! $\endgroup$ – Patrick Da Silva May 29 '15 at 20:39
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    $\begingroup$ Hahaha that is one of the the most amazing answers I've seen on here so visual :P Bust seriously great stuff I might use this If i ever teach linear lol $\endgroup$ – AIM_BLB May 30 '15 at 4:23
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    $\begingroup$ the forth picture is really enlightenning $\endgroup$ – A. Chu Jun 3 '15 at 13:10
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    $\begingroup$ Reminds me of that adage "A picture is worth 100 words" $\endgroup$ – Cloverr Jun 6 '15 at 14:43
  • $\begingroup$ Glad you appreciate it !! $\endgroup$ – user50618 Jun 22 '15 at 20:30
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I do think the wiki's explanation is pretty good:

A projection is a mapping of a set (or other mathematical structure) into a subset 
(or sub-structure), which is equal to its square for mapping composition (or, 
in other words, which is idempotent).

Even more simply: it is an idempotent homomorphism.

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    $\begingroup$ "idempotent homomorphism"? Try using that kind of terminology on a student fresh out of high school. $\endgroup$ – Mehrdad May 30 '15 at 7:42
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    $\begingroup$ The question has been changed so an answer like this makes less sense. Originally it was pretty vague. $\endgroup$ – muaddib May 30 '15 at 7:56
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    $\begingroup$ @Mehrdad It's perfect for that kind of situation. When you say, "It's an idempotent homomorphism," they involuntarily look away from you while they try to remember if they've ever heard either of the words "idempotent" and "homomorphism". And then you can make your escape! That's the goal, right? $\endgroup$ – David Richerby May 30 '15 at 21:44
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    $\begingroup$ @DavidRicherby Linear Algebra students know what idempotence is before they learn projections I think? $\endgroup$ – BCLC Jun 6 '15 at 13:53
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If you want a practical application of it I have one, it is indeed as they say that you take a higher dimensional object and put it into a lower dimensional one. A real world example is computer games. They are almost all generated in a 3D enviorement where they interact with polygons and all that computer stuff. However our screens are only 2D so all that 3D delicious game stuff must be projected onto a 2D surface to be displayed.

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I always think of it as mapping an object to its shadow (like a 3D person "projected" onto the plane of the ground a 2D image). You can get different projections by adjusting the location of the light source. Of course this is just 3D-2D "intuition" but it has worked for me...

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Let $C$ be a nonempty subset of a normed space $X$ and let $x \in X$. An element $y_0 \in C$ is said to be a best approximation to $x$ if $\Vert x − y_0\Vert = d(x,C)$, where \begin{equation} d(x,C) = \lbrace inf_y\in C \Vert x-y\Vert\rbrace \end{equation} . The number$ d(x,C)$ is called the distance from x to C . The (possibly empty) set of all best approximations from $x$to $C$is denoted by \begin{equation} P_C(x) = \lbrace y \in C : \Vert x − y\Vert = d(x,C)\rbrace \end{equation} This defines a mapping $P_C$ from $X$ into $2^C$ and is called the metric projection onto $C$. The metric projection mapping is also known as the nearest point projection mapping, proximity mapping, and best approximation operator

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It projects the input onto a lower dimensional subset.

One example of a projection is the function

\begin{align} P:\mathbb{R}^3&\to\mathbb{R}^3\\ (x,y,z)&\mapsto (x,y,0) \end{align}

That is $P(x,y,z) = (x,y,0)$. You can visualize this function as it takes an input (a point in $3$D-space) and maps the point to the point below it on the $(x,y)$-plane (which is $2$-dimensional).

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  • $\begingroup$ That's a lot of mathematical symbols!! $\endgroup$ – user135988 May 29 '15 at 13:38
  • $\begingroup$ Does this concept apply to quotient space? $\endgroup$ – sleeve chen May 29 '15 at 13:39
  • $\begingroup$ @user135988 Not the first sentence :-). I tried to find a nice figure, but I couldn't. $\endgroup$ – Eff May 29 '15 at 13:40
  • $\begingroup$ @sleevechen I'm not sure, actually. $\endgroup$ – Eff May 29 '15 at 13:40
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The key is to choose the right frame of reference, and all will become clear. Consider the operator that takes points (x, y, z) in 3-dimensional space and maps them to (x, y, 0). Geometrically, this is what user50618 suggested with the steamroller above: the x and y positions remain the same, while the z gets compressed down to zero.

The matrix representing this projection is:

1   0   0
0   1   0
0   0   0

If you put the column vector [ x y z ] to the right of that matrix, you see the multiplication:

1   0   0       x      x
0   1   0   *   y  =   y
0   0   0       z      0

If you do it again, no more change occurs. z remains zero. This is called being "idempotent."

And if you look at the diagonal of the matrix, you see two 1's for the directions that stay the same, and a single 0 for the direction that gets collapsed.

Now, ANY projection of 3-space onto a 2-dimensional subspace has EXACTLY that same geometry. It's just much harder to see when the direction of collapse is not lined up nicely with one of the coordinate axes.

The process of diagonalizing a matrix represents finding the natural directions of the linear transformation, and rewriting the matrix with respect to those natural directions. When written in a basis of its eigenvectors (the name for the natural directions), the matrix will be diagonal. In the case of a projection, the diagonal entries will all be ones or zeros, representing the directions which are preserved or collapsed.


Here is an example (not a projection), which is easy to write:

1   -1
-1  1

It is not immediately obvious what this linear transformation does, because its action is not aligned nicely with the coordinate axes. But think about what it does to the vector (1, 1). It collapses it to zero. And think about what it does to the vector (1, -1). That becomes (2, -2). This tells us that the matrix represents the same geometric action as this matrix:

0   0
0   2

That is, one axis gets stretched to double its length, and the other gets squashed down to zero. This becomes clear when you use the "right" basis - the basis of eigenvectors - to write the matrix.


Note that some matrices cannot be diagonalized, and some that can require complex numbers. But projections always can be diagonalized using only real numbers.

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A projection map $f: A \to B$ is something you often need to simplify the space $A$. It is build so that differences which occur in the space $A$ will often no longer occur anymore in the space $B$. Then, if you know enough about the map $f$ and find out a bit about the space $B$ (which is easier as their are less differences) then you may find out interesting things about $A$.

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