Finding a matrix P that orthogonally diagonalizes

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Homework Statement

Find a matrix P that orthogonally diagonalizes I - vv^T if

v = (1, 0, 1)

Homework Equations

Well, solving I - vv^T will give me my A if I am correct, and the characteristic equation for A is

det(\lambdaI - A) = \lambda³ - 2\lambda² = 0

Solving this gives me \lambda₁ = 0 and \lambda₂ = 2

Now when I substitute these values back into det(\lambdaI - A) I only get 2 basis vectors for my P. (1, 0, 1) and (-1, 0, 1). Since I don't have 3 vectors I'm not going to yet normalize them, which has to be done before they become the column vectors of P.

Any help for finding this P would be great

EDIT : I guess technically I also have a 3rd basis vector (0, 0, 0) with \lambda=0 am I able to use the zero vector?

 

[lanedance]

are you sure about you determinant eqaution? maybe show how you got there... and your matrix

hint: if I've done it right & i multiply your first eignevector by the matrix A, I get 1x the eigenvector...

 

[lanedance]

also i notice if
A = I - vv^T

and u is an eignevector of A, with value m, then you should be able to show that u is also an eignevector of vv^T with related eigenvalues, and it should be reasonably easy to guess the eigenvectors of vv^T ro maybe even A

 

[lanedance]

and one more...

to clarify in this case, 0 is an eignvalue of vv^T but not A

When 0 is an eigenvector, it means the matrix is singular, with the eigenvectors corresponding to the 0 eigenvalue, belonging to the null space of the matrix

note that the zero vector, 0 is never included in the solutions to eigenvalue equation
Au = mu
as it it always a solution reagrdless of eigenvalue (considered the trvial solution).

So the zero eigenvalue will have corresponding non-zero eigenvector/s.