Linear algebra - Spectral decompositions: Eigenvectors of projections

[TorcidaS]

Homework Statement

Let P1 and P2 be the projections defined on R^3 by:

P1(x1, x2, x3) = (1/2(x1+x3), x2, 1/2(x1+x3))
P2(x1, x2, x3) = (1/2(x1-x3), 0, 1/2(-x1+x3))

a) Let T = 5P1 - 2P2 and determine if T is diagonalizable.
b) State the eigenvalues and associated eigenvectors of T.

Homework Equations

The Attempt at a Solution

For a), I believe it is diagonalizable because P1 + P2 gives us (x1, x2, x3). Although I'm could be wrong on that...

It's mainly b) that I'm concerned for. By the theorem, (T = c1P1 + c2P2...+.. where c are eigenvalues) 5 and -2 are the eigenvalues (although this sort of confuses me because I had thought the eigenvalues of projections are always 1 and 0).

How can we find the eigenvectors? Had this been a matrix it's simple, subtract the eigenvalue from the main diagonal, simplifiy, and find the nullspace.



Also, the solution to b) is for eigenvalue 5, the eigenvectors are <(1,0,1),(0,1,0)> and for eigenvalue -2, the eigenvector is <(-1,0,1)> (so my guess some matrix is formed?)


It feels like I'm missing something obvious here. Can anyone please help me out?

Thanks

 

[vela]

Homework Statement

Let P1 and P2 be the projections defined on R^3 by:

P1(x1, x2, x3) = (1/2(x1+x3), x2, 1/2(x1+x3))
P2(x1, x2, x3) = (1/2(x1-x3), 0, 1/2(-x1+x3))

a) Let T = 5P1 - 2P2 and determine if T is diagonalizable.
b) State the eigenvalues and associated eigenvectors of T.

Homework Equations

The Attempt at a Solution

For a), I believe it is diagonalizable because P1 + P2 gives us (x1, x2, x3). Although I'm could be wrong on that...

I don't see how that follows. Is there some theorem that you're using?

It's mainly b) that I'm concerned for. By the theorem, (T = c1P1 + c2P2...+.. where c are eigenvalues) 5 and -2 are the eigenvalues (although this sort of confuses me because I had thought the eigenvalues of projections are always 1 and 0).

T isn't a projection, so its eigenvalues don't have to be 0 or 1.

How can we find the eigenvectors? Had this been a matrix it's simple, subtract the eigenvalue from the main diagonal, simplifiy, and find the nullspace.

You should be able to write down the matrices for P1 and P2 by inspection, and then you can calculate the matrix for T. Or you can find the nth-column of the matrix for T by applying T to the nth basis vector.

Also, the solution to b) is for eigenvalue 5, the eigenvectors are <(1,0,1),(0,1,0)> and for eigenvalue -2, the eigenvector is <(-1,0,1)> (so my guess some matrix is formed?)


It feels like I'm missing something obvious here. Can anyone please help me out?

Thanks