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Find a linear map $T$ from $\mathbb{R}^2$ to $\mathbb{R}^2$ such that $T$ is neither $0$ nor the identity map, but $T^2 = T$.

I have no idea how to even start this, so help would be appreciated. Thanks!

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What happens if you send every point $(x,y)$ to the point on the $x$ axis that is directly under/over it? –  Arturo Magidin Mar 20 '12 at 3:46
    
Consider the matrix representation. How does multiplication work and what does that mean for concatenation of functions? –  Desiato Mar 20 '12 at 3:50
    
I love your handle, Linear Algebra Failure. –  Antonio Vargas Mar 20 '12 at 17:47

3 Answers 3

I'll spill the beans:

A linear transformation $T\colon\mathbf{V}\to\mathbf{V}$ that satisfies $T^2=T$ is called a projection. Essentially, they arise from expressing $\mathbf{V}$ as a direct sum of two spaces, $\mathbf{V}=\mathbf{U}\oplus\mathbf{W}$, which means that every vector in $\mathbf{V}$ can be written uniquely as $u+w$ for some $u\in\mathbf{U}$ and $w\in\mathbf{W}$, and defining $T(u+w) = u$.

The simplest case of projections occur in $\mathbb{R}^2$, and they are the "orthogonal projections onto the axes." Every element of $\mathbb{R}^2$ can be described by a pair of cartesian coordinates, $(a,b)$. The projection map onto $x$, $P_x$, simply tells you the $x$-position of the vector and "forgets" the $y$-position. That is, the map $P_x$ is given by $P_x(a,b) = (a,0)$. Imagine a flashlight shining down (and up) from the infinity $y$-direction towards the $x$-axis; $P_x(a,b)$ is the "shadow" that the point $(a,b)$ casts on the $x$-axis. Similarly, the projection map onto the $y$-axis, $P_y$, gives you the "shadow" of the point $(a,b)$ on the $y$-axis, if you have beams of light shining off in the infinite $x$-direction onto the $y$-axis, $P_y(a,b) = (0,b)$.

The shadow of the shadow is just the shadow itself, so applying $P_x$ (or $P_y$) twice is the same as applying it just once.

You can do something similar in $\mathbb{R}^3$, by projecting any vector $(a,b,c)$ to its "shadow" on, say, the "wall" made by the $yz$-plane; this would be the map $P_{yz}(a,b,c) = (0,b,c)$. Note that $P_{yz}(P_{yz}(a,b,c)) = P_{yz}(0,b,c) = (0,b,c)$, so $(P_{yz})^2 = P_{yz}$. Or you can project onto the $y$-axis, by taking $P_y(a,b,c) = (0,b,0)$.

Of course, there are other kinds of projections you can make: in $\mathbb{R}^2$, instead of projecting onto the $x$-axis or the $y$-axis, we could project, for example, onto the line $x=y$: take, say, $(a,b)$ to $(a,a)$. Or to the line $x=2y$, using $(a,b)\mapsto (a,2a)$.

Each of these projections has a "direction" along which we are projecting; this is given by the vectors that map to $0$. In the map $(a,b)\mapsto (a,0)$, the vectors that map to $0$ are the $y$-axis: all vectors $(0,b)$. You may remember that I described the projection as being the "shadow" from a flashlight that was flashing at infinity in the $y$-direction, down on the $x$-axis. The "light rays" that create the shadows would be vertical lines, parallel to the line $\{(0,b)\mid b\in\mathbb{R}\}$.

The projection $P(a,b) = (a,a)$ maps the vectors $(0,b)$ to $(0,0)$; if you draw what this map is doing, it is again given as "shadows" cast by vertical light rays, but now they are cast on the line $x=y$.

On the other hand, we could cast a "shadow" on the line $x=y$ by mapping $(a,b)$ to $(a+b,a+b)$. In this case, what goes to $(0,0)$? The vectors of the form $(r,-r)$. These vectors describe the line $x=-y$. If you draw what this projection is doing, now the "light rays" are parallel to the line $x=-y$, and we are casting the shadows on the line $x=y$. The map $(a,b)\longmapsto (0,a+b)$ is a projection onto the $y$-axis, now in the direction on $x=-y$ (the vectors $(r,-r)$ map to $(0,0)$ again.

More generally, $\mathbb{R}^2$, pick two distinct lines $L_1$ and $L_2$ through the origin; these two lines define a natural basis for $\mathbb{R}^2$: nonzero vectors $\mathbf{u}_0\in L_1$ and $\mathbf{w}_0\in L_2$ lying on the two lines. Then every vector of $\mathbb{R}^2$ can be written uniquely as $\alpha \mathbf{u}_0 + \beta\mathbf{w}_0$ for some scalars $\alpha$ and $\beta$, and we can define $T$ to be the "projection onto $L_1$ along $L_2$, $T(\alpha\mathbf{u}_0+\beta\mathbf{w}_0) = \alpha\mathbf{u}_0$.

Imagine using $L_1$ and $L_2$ to create a (possibly slanted) "grid" on the plane; then given any point $(a,b)\in\mathbb{R}^2$, draw the line $\mathcal{L}$ that is parallel to $L_2$ that goes through $(a,b)$; we map $(a,b)$ to the intersection of $\mathcal{L}$ with $L_1$. We are essentially casing shadows on $L_1$, using "light rays" that are parallel to $L_2$.

The map azarel suggests uses this idea with the $x$- and $y$-axes playing the roles of $L_1$ and $L_2$.

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I like to think of the projection map as "wiping out" a certain coordinate (like the linear transformation given by Azarel below). Perhaps that may be useful to the OP. –  user38268 Mar 20 '12 at 5:12
    
I feel even more lost with that. But to be fair, I don't understand projection in general; I'm currently using Linear Algebra and Applications by Cheney and Kincaid, and there isn't any thorough information on it. Perhaps you could explain a bit more about projections, in a pictorial sense. For instance, I was asked to describe (.5,.5)(.5, .5) geometrically. Errm, I don't know how to write a matrix on here, but it's a 2x2 matrix with .5 for all of them. When I drew it out, I had x1 as the x axis, x2 as the y axis, but then graphed out essentially...a dot. Something feels off. Thanks a lot! –  Linear Algebra Failure Mar 20 '12 at 5:18
    
@LinearAlgebraFailure: The matrix $\left(\begin{array}{cc}0.5&0.5\\0.5&0.5\end{array}\right)$ maps $(a,b)$ to $(\frac{1}{2}(a+b),\frac{1}{2}(a+b))$; this is also a projection, onto the line $x=y$, in the direction of $x=-y$. I've added some more explanations, maybe that will help. –  Arturo Magidin Mar 20 '12 at 17:10

You can always define first the linear transformation in a basis of the domain and after extend it to all the space. Define for example: T(1,0)=(1,0) T(0,1)=(0,0) You know this definition gives a unique linear transformation . Then T(T(1,0))=T(1,0) T(T(0,1))=T(0,0)=(0,0)=T(0,1). (T(0,0)=(0,0) because T it is a linear transf). If T=T^2 in a basis , they are the sema transformation

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What about $ \begin{bmatrix}{1}&{0}\\{0}&{0}\end{bmatrix}$?

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Is $ \begin{bmatrix}{1}&{0}\\{0}&{0}\end{bmatrix} \in \mathbb R^2$? –  Hassan Muhammad Mar 20 '12 at 6:03
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@HassanMuhammad That is the matrix associated to the linear map $T(x,y)=(x,0)$. –  azarel Mar 20 '12 at 6:24

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