Eigenvalues, projection and symmetry. Help please.

[Jimmy84]

Homework Statement

In each case describe the eigenvalues of the linear operator and a base in R^3 that consist of eigenvectors of the given linear operator.

Write the matrix of the operator with respect to the given base.



The Orthogonal Projection on the plane 2x + y = 0

and the Symmetry with respect to the plane x - y +2z = 0

Homework Equations

The Attempt at a Solution

I have no idea where to start, with the projection problem my guess is starting by getting a base of 2x + y =0

Then I'm thinking about using Gramm Schmidt to get orthogonal bases of the given plane. but I don't have a clear idea of how to solve this problem I would appreciate any help and advice thanks a lot.

 

[darthfunnybot]

This problem is too confusing to explain in this thread without proper looking matrices and that is too tedious. On this link notice the different properties rotation and symmetry are given with respect to each plane's normal vector. Basically read in and around the red text.
Here are some nice looking examples

 

[HallsofIvy]

Do you know how to write a linear transformation as a matrix, in a given basis?

Apply the matrix to each basis vector in turn, writing the result as a linear combination of the basis vectors. The coefficients in that linear combination give a column of the matrix representation.

Here, the first transformation is projection of (x, y, z) onto the plane 2x+y= 0. In the xy-plane (since z doesn't change), that is a line with slope -2 and so a perpendicular has slope 1/2. The line through (1, 0), perpendicular to that line, is y= (1/2)x- 1/2. The line y= (1/2)x- 1/2 intersects y= -2x when -2x= (1/2)x- 1/2, (5/2)x= 1/2, x= 1/5, y= -2/5. That is, <1, 0, 0> is mapped to <1/5, -2/5, 0> which is parallel to <1, -2, 0>. The line through (0, 1), perpendicular to that line, is y= (1/2)x+ 1/2. The line y= (1/2)x+ 1/2 intersects y= -2x when -2x= (1/2)x+ 1/2, (5/2)x= -1/2, x= -1/5, y= 2/5. That is, <0, 1, 0> is mapped to <-1/5, 2/5, 0> which is parallel to <-1, 2, 0>. In the standard basis, this linear transformation is represented by the matrix
$$\begin{pmatrix}1 & -1 & 0 \\ -2 & 2 & 0 \\ 0 & 0 & 1\end{pmatrix}$$
and find the eigenvalues and eigenmatrices of that matrix. Of course, a projection is NOT "invertible", hence one row is multiple of another and it should be clear that one eigenvalue is 0.

More basically, you should be able to see that any vector perpendicular to the plane 2x+ y+ 0z= 0, that any multiple of <1, 1, 0> is mapped to the 0 vector so is an eigenvector with eigenvalue 0. And any vector already in that plane is mapped to itself so is an eigenvector with eigenvalue 1.

The second problem is very similar except that, as a reflection, any vector perpendicular to the plane is mapped to -1 times itself so is an eigenvector with eigenvalue -1 while any vector in the plane is mapped to the 0 vector.

 

[Jimmy84]

Do you know how to write a linear transformation as a matrix, in a given basis?

Apply the matrix to each basis vector in turn, writing the result as a linear combination of the basis vectors. The coefficients in that linear combination give a column of the matrix representation.

Here, the first transformation is projection of (x, y, z) onto the plane 2x+y= 0. In the xy-plane (since z doesn't change), that is a line with slope -2 and so a perpendicular has slope 1/2. The line through (1, 0), perpendicular to that line, is y= (1/2)x- 1/2. The line y= (1/2)x- 1/2 intersects y= -2x when -2x= (1/2)x- 1/2, (5/2)x= 1/2, x= 1/5, y= -2/5. That is, <1, 0, 0> is mapped to <1/5, -2/5, 0> which is parallel to <1, -2, 0>. The line through (0, 1), perpendicular to that line, is y= (1/2)x+ 1/2. The line y= (1/2)x+ 1/2 intersects y= -2x when -2x= (1/2)x+ 1/2, (5/2)x= -1/2, x= -1/5, y= 2/5. That is, <0, 1, 0> is mapped to <-1/5, 2/5, 0> which is parallel to <-1, 2, 0>. In the standard basis, this linear transformation is represented by the matrix
$$\begin{pmatrix}1 & -1 & 0 \\ -2 & 2 & 0 \\ 0 & 0 & 1\end{pmatrix}$$
and find the eigenvalues and eigenmatrices of that matrix. Of course, a projection is NOT "invertible", hence one row is multiple of another and it should be clear that one eigenvalue is 0.

More basically, you should be able to see that any vector perpendicular to the plane 2x+ y+ 0z= 0, that any multiple of <1, 1, 0> is mapped to the 0 vector so is an eigenvector with eigenvalue 0. And any vector already in that plane is mapped to itself so is an eigenvector with eigenvalue 1.

The second problem is very similar except that, as a reflection, any vector perpendicular to the plane is mapped to -1 times itself so is an eigenvector with eigenvalue -1 while any vector in the plane is mapped to the 0 vector.

Im still confused about how did you find the standard matrix. I know there is a formula for the standard matrix A = B B B^-1

if I'm not mistaken B is the matrix

1 0 0
0 1 0
0 0 0

and B is the a matrix composed by bases of the plane 2x - y = 0

(Im trying to follow the steps of what darthfunnybot just sent). I found two bases of the plane, then I found one base that is orthogonal to those two bases. then those vectors I think constitute the matrix B.

Can I find the standard matrix in this way, or do I need to change something in the process?? Thanks in advance.

 

[Jimmy84]

Do you know how to write a linear transformation as a matrix, in a given basis?

I'm learning that right now. but I'm still confused . I would appreciate some advice please.