Question

Describe the kernel and range of (a) the orthogonal projection on the xz -plane (b) the orthogonal projection on the yz-plane (c) the orthogonal projection on the plane defined by the equation y=x

Answer 1

Alright, I suspect I can help you out here.

Answer 2

i dont really understant what it means by "orthogonal projection". If it was just a normal projection, then for a), since all the vectors are being projected onto the xz-plane, then a good basis would be the x axis and z axis, or rather, the vectors (1,0,0) and (0, 0, 1). Then null space (kernel) would be all the vectors that are on the y axis, because when you project them onto the xz plane, they will shrink onto the origin

Answer 3

Likewise, for b), since we are projecting onto the yz plane, a basis would be (0,1,0) and (0, 0,1), while the Null Space would be anything on the x axis. now for the last problem im still thinking about it...lol

Answer 4

An orthogonal projection is a projection in which the range U and null space (or kernel) V are orthogonal subspaces of the vector space. This means that if\[u \in U, v \in V\] then \[u*v = 0\](The dot product of u and v is 0).

Now, a projection onto the xz plane is achieved by taking the x and z portions of any vector in R^3 and eliminating the y component. Thus, we want to use the matrix\[P = \left[\begin{matrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{matrix}\right]\]as our projection. We see that this will, when multiplied times a vector in R^3, preserve the x and z components while eliminating the y. The range of this projection is the space containing all vectors a such that\[Pa \neq 0\] and the kernel is the vector space containing all vectors a such that \[Pa = 0\]. Clearly, the range will be all vectors with an x or z component, and the kernel will be all vectors with only a y component. Thus the range is \[C_1\left[\begin{matrix}1 \\ 0 \\ 0\end{matrix}\right] + C_2\left[\begin{matrix}0 \\ 0 \\ 1\end{matrix}\right], C_1,C_2 \neq 0\] and the kernel is \[C \left[\begin{matrix}0 \\ 1 \\ 0\end{matrix}\right], \forall C \in R\]

Answer 5

We can clearly see that since the dot product of two vectors is the summation of the products of their components, and vector in the range dotted with any vector in the null space will give us 0, so the two spaces are orthogonal.

Answer 6

alright, for the plane defined by y = x, that means all the vectors on the plane will be in the form v = (x,x,z), because the x and y coordinates must be equal, and there are no restrictions on z. this is really 2 independant vectors added together:
x(1,1,0)+z(0,0,1)
those two vectors make a basis for the projection onto the plane.

Answer 7

how would you figure out the basis for the null space of c) blacksteel? i can see the null space in my head (lol), but i cant think of a method to obtain the answer.

Answer 8

Oh, I'm sorry - i miswrote my earlier answer - the range is made up of all possible linear combinations of non-zero columns of the matrix. So the range would be
\[C_1\left[\begin{matrix}1 \\ 0 \\ 0\end{matrix}\right] + C_2\left[\begin{matrix}0 \\ 0 \\ 1\end{matrix}\right]\], no restrictions on the variables.

Answer 9

Nice :) its good to have a systematic way to do things, i lack that >.<

Answer 10

for the basis or the null space?

Answer 11

The whole thing. This image shows what we are doing geometrically - the red line is x = y and the blue line is the one we are projecting onto it. As you can see, the projection is not simply a straight drop down such that y = x.

Answer 12

Furthermore (and this is how I realized that this solution was wrong), an orthogonal projection matrix must be self-adjoint (ie, P = P-transpose).

Answer 13

you guys are super smart. i will just be a spectator for this event.

Answer 14

I dont see how my reasoning is wrong though >.< if we are on a plane where the only restriction is that y=x, then (1,1,0) and (0,0,1) have to be a basis for it. any vector on that plane is going to look like (a,a,b). if it doesnt look like that, its not on the plane.

Answer 15

the null space would be the line where y = -x, and z = 0, cause a basis for that line would be (1,-1,0), which is orthogonal to both the basis vectors.

Answer 16

So then solving it becomes a bit more complex. You are right that the z component will be unchanged. The line x = y has a slope of 1, so the line perpendicular to it intersecting the tip of our vector has a slope of -1and must contain the point (x1,y1) if our vector is (x1, y1, z1). Then the equation of the line must be y = -x + b, and y1 = -x1 + b, so b = y1 + x1. Thus, the line is y = -x + y1 + x1, and thus will intersect the line y = x at x = (x1 + y1)/2. (Since then y = -(x1 + y1)/2 + x1 + y1 = (x1 + y1)/2).

Thus our projection matrix should be\[P = \left[\begin{matrix}1/2 & 1/2 & 0 \\ 1/2 & 1/2 & 0 \\ 0 & 0 & 1\end{matrix}\right]\]We see that this matrix is self-adjoint and satisfies the problem, and we can solve for the null-space and range. The range is, of course, \[C_1\left[\begin{matrix}1/2 \\ 1/2 \\ 0\end{matrix}\right] + C_2\left[\begin{matrix}1/2 \\ 1/2 \\ 0\end{matrix}\right] + C_3\left[\begin{matrix}0 \\0 \\ 1\end{matrix}\right]\]which, by factoring out 1/2 and combining the first 2 terms, can be simplified to\[C_1\left[\begin{matrix}1 \\1 \\ 0 \end{matrix}\right] + C_3\left[\begin{matrix}0 \\0 \\ 1\end{matrix}\right]\]And the null space is going to be vectors that, when multiplied with P, give us 0, which will occur at \[C_1\left[\begin{matrix}1 \\-1 \\ 0\end{matrix}\right]\]We can check this looking at whether a vector from the range dotted with a vector from the kernel gives us 0 (it does.)

Answer 17

Joe, the vector will be of the form (a,a,b), but if our initial vector is (x,y,z), the final vector will NOT be (x,x,z)

Answer 18

when i say (x,x,z), i dont mean that:
\[(x,y,z) \rightarrow (x,x,z)\]
i just mean that all vectors on the plane will have the same x and y coordinate. This is shown by your projection matrix, since the first and second rows are equal.

Answer 19

Yes, you are quite correct. What you put, that the range would be x(1,1,0) + z(0,0,1), is right (that's what I wrote in my correct solution as well.) I simply misconstructed the projection matrix from that.

Answer 20

i said i was going to just spectate, but where are you getting these numbers from

Answer 21

(removed a couple of extraneous posts)

Answer 22

myininaya, we're projecting a vector in 3 dimensions onto the plane defined by x = y. This is the plane containing all vectors with the x and y coordinates equal. x(1,1,) + z(0,0,1) is a linear combination expressing all such vectors.

Answer 23

im gonna have to read up on constructing projection matrices, i cant see it naturally yet >.<

Answer 24

So... basically, joe had the right answer; I actually constructed the projections (which isn't needed to solve the problem) to illustrate where the spaces were coming from.

Answer 25

i remember reading something about two vectors a and b are orthogonal if a dot b=0

Answer 26

Yup, that is the definition of orthogonality in vector spaces.

Answer 27

Anyway, the easiest way to solve this problem is how joe did it - think about what kinds of vectors will exist in the plane you're projecting onto and then write those vectors as a linear combination of independent basis vectors to find the range and then figure out what sorts of vectors will be orthogonal to a vector in the plane and write a linear combination of basis for those to find the null space.

Answer 28

*of basis vectors

Answer 29

gj you guys!

Answer 30

im drawing a picture (or trying to) of whats in my head >.> one sec lol

Answer 31

dont judge me <.< lol

Answer 32

pretty

Answer 33

Yeah, that's exactly right. When I said the solution was wrong, I meant my projection matrix, not your spaces.