Very nice lecture here:
Lecture 15: Projections onto subspaces | Video Lectures | Linear Algebra | Mathematics | MIT OpenCourseWare
Excellent, but rather more powerful than necessary here.
Very nice lecture here:
Lecture 15: Projections onto subspaces | Video Lectures | Linear Algebra | Mathematics | MIT OpenCourseWare
Excellent, but rather more powerful than necessary here.
In this case however there is a simple solution.
<x,y,z> -> <a,a,z>
The vector representing this being <a-x,a-y,0> and this vector must be orthogonal to any vectors in the xy plane, eg <1,1,0> and <0,0,1>.
So <a-x,a-y,0> . <1,1,0> = 0 so a-x + a-y=0 and a=(x+y)/2
What matrix sends <x,y,z> to <(x+y)/2,(x+y)/2,z>?
Thanks all. I do understand how you got the transformation matrix, but I don't get it, how did you get the eigenvectors, I mean what we are taught is that using the eigen values, we find eigenvectors using (A-lambda*I)x=0. so please can you do it by this method or explain to me how did you do it? I mean how did you find the orthogonal basis and what is it actually!!! Thanks. again.
by definition a projection (in R^{3}) onto a plane is a mapping P:R^{3}-->R^{3} of rank 2 such that P^{2} = P.
an orthogonal projection means that R^{3} = ker(P) ⊕ im(P), or equivalently: P^{T}P = P (which combines the orthogonality condition P = P^{T}, and the projection condition P^{2} = P).
let's see how we can leverage these definitions to get an explicit matrix for A in the standard basis B = {(1,0,0),(0,1,0),(0,0,1)}.
let's call our plane U, just to give it a name. note that a plane has dimension 2, so to find a basis for U, it suffices to find 2 LI vectors in U. clearly, (1,1,0) and (0,0,1) are 2 such vectors, that is:
U = im(A) = span({(1,1,0),(0,0,1)}.
we get a bit of a lucky break, here: (0,0,1) is already in U, which tells us that A(0,0,1) = (0,0,1), which gives us the 3rd column of A "for free".
we also know, since A is linear, and (1,1,0) is in U: (1,1,0) = A(1,1,0) = A(1,0,0) + A(0,1,0); which tells us that if we find A(1,0,0), to get A(0,1,0) we just compute (1,1,0) - A(1,0,0).
so suppose A(1,0,0) = (a,b,c). since A(1,0,0) is in U: a = b. not much progress, but it's a start.
we're going to have to use the orthogonality, somewhere. what does this mean for A? it means that if u.v = 0 in R^{3}, then Au.Av = 0 as well (A preserves orthogonality).
this means that if (x,y,z).(1,1,0) = 0, then (A(x,y,z)).(A(1,1,0)) = (A(x,y,z)).(1,1,0) = 0, and if (x,y,z).(0,0,1) = 0, (A(x,y,z)).(0,0,1) = 0.
well which vectors in R^{3} are orthogonal to all of U?
if (x,y,z).(1,1,0) = 0, then x+y = 0, so (x,y,z) is of the form (x,-x,z).
if (x,y,z).(0,0,1) = 0, then z = 0. together, we get the the vectors perpendicular to all of U are all in span{(1,-1,0)}.
suppose (x,y,z) is such a vector (perpendicular to U). now A(x,y,z) is in U, so is of the form (a,a,b). but since A(x,y,z) is also perpendicular to U, (a,a,b) = t(1,-1,0) = (t,-t,0). so:
a = t
a = -t <---so 2a = 0, so a = 0.
b = 0
this means A(1,-1,0) = (0,0,0). how does this help us?
remember, we were trying to find A(1,0,0). now we have THREE vectors we know A of:
A(1,1,0) = (1,1,0)
A(0,0,1) = (0,0,1)
A(1,-1,0) = (0,0,0).
note that (1,0,0) = (1/2)[(1,1,0) + (1,-1,0)].
so A(1,0,0) = A((1/2)[(1,1,0) + (1,-1,0)] = (1/2)A(1,1,0) + (1/2)A(1,-1,0) = (1/2)(1,1,0) + (1/2)(0,0,0) = (1/2,1/2,0) <---first column of A! woohoo!!
and remember that A(0,1,0) = (1,1,0) - A(1,0,0) = (1,1,0) - (1/2,1/2,0) = (1/2,1/2,0) <---second column of A!
so:
does this work?
well, clearly ranK(A) = 2. and just as clearly:
A(x,y,z) = (x/2+y/2,x/2+y/2,z), which is in U, since x/2+y/2 = x/2+y/2, so im(A) is a 2-dimensional subspace of U, so must BE U.
A is a symmetric matrix (A = A^{T}).
finally A^{2} =
so A is a projection.
ok, so we found A. now let's find some eigenvalues.
so the eigenvalues of A are 0 and 1. what does this mean?
well suppose v is an eigenvector belonging to 0. this means that Av = 0v = 0, that is, v is in the nullspace of A, which as we saw are precisely the vectors perpendicular to U.
on the other hand, suppose v is an eigenvector belonging to 1. this means that Av = 1v = v, that is, v is in U.
EDIT: there is an easier way to find the eigenvalues, by the way:
from A^{2} = A, we get A^{2} - A = 0, that is, A(A - I) = 0, so A satisfies the polynomial x(x - 1).
since rank(A) = 2 > 0, A is not the 0-matrix, nor the identity matrix (which has rank 3), so this must be the minimial polynomial of A, in which case the only eigenvalues of A are 0 and 1.
I mean how did you find the orthogonal basis and what is it actually!!! Thanks. again.
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Pt in a plane: x = (a,a,b)
Any other pt in plane: y = (ea,ea,fb) e,f arbitrarary (Eq of the plane)
y = Ax → A = (e,0,0);(e,0,0);(0,0,f)
det (A – λ) = 0 → λ = 0,e,f
eigenvectors: (A – λ)n = 0
λ = 0 → n1 = (0,0,0)
λ = e → n2 = (1,1,0)
λ = f → n3 = (0,0,1)
EDIT: I have to ck Algebra. I tend to get careless once I see the method.
EDIT: OK, that makes sense, n2 perp n3 and any point in plane is LC of n2 & n3.
whoops, previous post correct answer to wrong question.
pair of orthogonal vectors in the plane x=y: n1= (1/sqr2, 1/sqr2, 0), n2 = (0,0,1) by inspection
Any point in R3: p = (x,y,z)
projection of p on plane = (p.n1)n1 + (p.n2)n2 = ((x+y)/2, (x+y)/2, z)
pproj = A (x,y,z)
((x+y)/2, (x+y)/2, z)^{T} = A(x,y,z)^{T}
Solve for A and determine eigen values and eigenvectors by the std procedure. Does anyone want to see the algebra?
Reference previous post:
Come to think of it, you don’t have to calculate eigenvalues and eigenvectors.
The projection matrix A sends (0,0,1) in the (0,0,1) dir, (1/sqr2,1/sqr2,0) into the (1/sqr2,1/sqr2,0) direction, and a vector perpendicular to these two to (0,0,0) (so λ = 0 for this eigenvector, its significance).
So n1=(0,0,1), n2= (1/sqr2,1/sqr2,0), and n3 = (1/sqr2, -1/sqr2, 0).
Sorry to hog the posts. My memory is not that good and things don't occurr to me right away.
Since the projection of any vector in the x=y plane on the plane is the vector itself, Av=v for any vector in the plane so λ = 1, which is a repeated root because there are three roots (one is 0) and all eigenvectors are in the plane with λ = 1.
Summary: The roots are 0,1,1 where 0 corresponds to the plane normal and 1 corresponds to any vector in the plane.
EDIT: Actually, you can deduce that λ = 0 is a root corresponding to the vector n normal to the plane because its projection on the plane is zero so that An = 0.
EDIT: Good Grief! HallsofIvy did this already! My apologies.