Linear Algebra (eigenvectors, eigenvalues, and orthogonal projections)

[Wm_Davies]

Homework Statement

I am part way done with this problem... I don't know how to solve part e or part f. Any help or clues would be greatly appreciated. I have been trying to figure this out for a couple days now.

W={<x,y,z>, x+y+z=0} is a plane and T is the orthogonal projection on it.

a) Find the basis of {v1, v2} for this subspace.

b) These basis vectors are basis vectors for what eigenvalue?

c) Show that n=<1,1,1> is orthogonal to {v1 ,v2}

d) Show that n is orthogonal to all of W

e) n is an eigenvector for T for what eigenvalue

f) Using matrix with eigenvectors and one for eigenvalues, find the standard matrix of T.

Homework Equations

The Attempt at a Solution

I am part way done with this problem... I don't know how to solve part e or part f.

W={<x,y,z>, x+y+z=0} is a plane and T is the orthogonal projection on it.

a) Find the basis of {v1, v2} for this subspace.
I found the basis of W to be v1=<-1,1,0> and v2=<-1,0,1>

b) These basis vectors are basis vectors for what eigenvalue?
I found the eigenvalue to be zero after multiplying W with v1 and v2.

c) Show that n=<1,1,1> is orthogonal to {v1 ,v2}
I just took the dot product of n*v1 and n*v2 and both were zero

d) Show that n is orthogonal to all of W
My reasoning is that n is orthogonal to all of W because W is a linear combination of {v1 ,v2}

e) n is an eigenvector for T for what eigenvalue
?

f) Using matrix with eigenvectors and one for eigenvalues, find the standard matrix of T.
?

 

[lanedance]

so you need to find the matrix T, which projects any vector onto the plane W

 

[lanedance]

actually for e) you don't need to find T first, what is the projection of n onto the plane? once you consider that the eigenvalue should be obvious

 

[3wrldman]

I am interested to see the answer to e) and f) myself... Landedance (or anybody) can you finish this example??

 

[lanedance]

For b) the eigenvalue is not zero, as the vector is within the plane, the projection onto the plane will be the vector itself, so
T.v1=v1
From which u should be able to read off the eigenvalue

The reasoning behind a, c, d is ok

 

[3wrldman]

so why do you need a matrix with eigenvectors and one with eigenvalues to find the standard matrix T?

Is it a diagonalization problem where T=A and A=PDP^-1 where D is a matrix of the eigenvalues and P a matrix of the eigenvectors?

Is there a repeated eigenvalue? seems like T must be 3x3 since the basis, <-1,1,0> and <-1,0,1> are in R3 right? So there must be a 3rd vector? and a repeated eigenvalue?

Then you could use PDP^-1 to find A right?

 

[Wm_Davies]

Okay so if T is an othogonal projection onto W then the eigen values for {v1, v2} must be 1 then right?

And if n is orthogonal to W then T*(n) should be equal to zero right? That would make the eigenvalue for n equal to zero.

So to find the standard matrix of T I could set P=[n,v1,v2] which would then make the diagonal matrix D=[0,1,1]

I think that this is correct, but I am not sure.

(I am typing this on my iPod so I can't really put in the full matrices, but I am sure you get the idea.)

 

[lanedance]

Okay so if T is an othogonal projection onto W then the eigen values for {v1, v2} must be 1 then right?
And if n is orthogonal to W then T*(n) should be equal to zero right? That would make the eigenvalue for n equal to zero.
So to find the standard matrix of T I could set P=[n,v1,v2] which would then make the diagonal matrix D=[0,1,1]

I think that this is correct, but I am not sure.

(I am typing this on my iPod so I can't really put in the full matrices, but I am sure you get the idea.)

Yes, if I understand ur matrix correctly, then u have the right idea and the eigenvalues are correct

 

[Wm_Davies]

Okay, I am at my computer so just to make sure we both understand each other...

<br /> P = \begin{pmatrix} 1 &amp; -1 &amp; -1 \\ 1 &amp; 1 &amp; 0 \\ 1 &amp; 0 &amp; 1 \end{pmatrix}<br />

<br /> D = \begin{pmatrix} 0 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 1 \end{pmatrix}<br />

Since P^-1TP=D
So then PDP^-1=T

Then <br /> T = \begin{pmatrix} \frac{2}{3} &amp; \frac{-1}{3} &amp; \frac{-1}{3} \\ \frac{-1}{3} &amp; \frac{2}{3} &amp; \frac{-1}{3} \\ \frac{-1}{3} &amp; \frac{-1}{3} &amp; \frac{2}{3} \end{pmatrix}<br />

 

[lanedance]

it would be easier to normalise you eigenvectors first, as then P^{-1} = P^Tm but that's the idea

to check, see whether you your eigenvectors behave as you would expect

 

[Wm_Davies]

it would be easier to normalise you eigenvectors first, as then P^{-1} = P^Tm but that's the idea

to check, see whether you your eigenvectors behave as you would expect

What do you mean by P^Tm?

Is T in this sense a transpose or the matrix T? Also what is m?

Thank you for all your help so far it has really helped me clear up a lot of things!

 

[lanedance]

sorry for the poor notation, but yes i meant the transpose, m was a mistype