# Finding a matrix P that orthogonally diagonalizes

1. Oct 16, 2009

### Iconate

1. The problem statement, all variables and given/known data
Find a matrix P that orthogonally diagonalizes I - vvT if

v = (1, 0, 1)

2. Relevant equations
Well, solving I - vvT will give me my A if I am correct, and the characteristic equation for A is

det($$\lambda$$I - A) = $$\lambda$$3 - 2$$\lambda$$2 = 0

Solving this gives me $$\lambda$$1 = 0 and $$\lambda$$2 = 2

Now when I substitute these values back into det($$\lambda$$I - A) I only get 2 basis vectors for my P. (1, 0, 1) and (-1, 0, 1). Since I dont 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?

Last edited: Oct 16, 2009
2. Oct 16, 2009

### 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...

Last edited: Oct 16, 2009
3. Oct 16, 2009

### lanedance

also i notice if
A = I - vvT

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

Last edited: Oct 16, 2009
4. Oct 16, 2009

### lanedance

and one more...

to clarify in this case, 0 is an eignvalue of vvT 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.

Last edited: Oct 16, 2009