Please find the transformation matrix A, for the linear transformation y=Ax. for:Orthogonal projection of R^3 on to the plane y=x. ALSO FIND THE EIGENVALUES OF A and SHOW THEIR SIGNIFICANCE?

Please help me, thank you.

Printable View

- Jan 12th 2013, 06:22 AMszak1592Find the transformation matrix???
Please find the transformation matrix A, for the linear transformation y=Ax. for:

**Orthogonal projection of R^3 on to the plane y=x. ALSO FIND THE EIGENVALUES OF A and SHOW THEIR SIGNIFICANCE?**

*Please help me, thank you.* - Jan 12th 2013, 07:38 AMa tutorRe: Find the transformation matrix???
Very nice lecture here:

Lecture 15: Projections onto subspaces | Video Lectures | Linear Algebra | Mathematics | MIT OpenCourseWare

Excellent, but rather more powerful than necessary here. - Jan 12th 2013, 07:54 AMa tutorRe: Find the transformation matrix???
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>? - Jan 12th 2013, 08:23 AMjakncokeRe: Find the transformation matrix???
------------------- I messed up

- Jan 12th 2013, 08:36 AMa tutorRe: Find the transformation matrix???
- Jan 12th 2013, 11:15 AMszak1592Re: Find the transformation matrix???
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.

- Jan 12th 2013, 11:35 AMHallsofIvyRe: Find the transformation matrix???
- Jan 12th 2013, 12:28 PMDevenoRe: Find the transformation matrix???
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:

$\displaystyle A = \begin{bmatrix} \frac{1}{2}&\frac{1}{2}&0\\ \frac{1}{2}&\frac{1}{2}&0\\0&0&1 \end{bmatrix}$

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}=

$\displaystyle \begin{bmatrix} \frac{1}{2}&\frac{1}{2}&0\\ \frac{1}{2}&\frac{1}{2}&0\\0&0&1 \end{bmatrix} \begin{bmatrix} \frac{1}{2}&\frac{1}{2}&0\\ \frac{1}{2}&\frac{1}{2}&0\\0&0&1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}&\frac{1}{2}&0\\ \frac{1}{2}&\frac{1}{2}&0\\0&0&1 \end{bmatrix} = A$

so A is a projection.

ok, so we found A. now let's find some eigenvalues.

$\displaystyle \begin{vmatrix} \frac{1}{2}-\lambda&\frac{1}{2}&0\\ \frac{1}{2}&\frac{1}{2}-\lambda&0\\0&0&1-\lambda \end{vmatrix} = (\dfrac{1}{2}-\lambda)^2(1-\lambda) - \dfrac{1}{4}(1-\lambda)$

$\displaystyle = (1 - \lambda)(\frac{1}{4} - \lambda + \lambda^2 - \frac{1}{4}) = (1 - \lambda)^2(-\lambda)$

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. - Jan 14th 2013, 06:38 PMLoidaWardRe: Find the transformation matrix???
I mean how did you find the orthogonal basis and what is it actually!!! Thanks. again.

____________________

Your DVD Zone, Take Your Choice without Any Doubt on The Newsroom Season 1 DVD - Jan 15th 2013, 10:40 AMHartlwRe: Find the transformation matrix???
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. - Jan 15th 2013, 12:06 PMHartlwRe: Find the transformation matrix???
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? - Jan 15th 2013, 02:00 PMHartlwRe: Find the transformation matrix???
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). - Jan 16th 2013, 03:57 AMHartlwRe: Find the transformation matrix???
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.