Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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

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 $L_2$.

share|cite|improve this answer
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

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

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.