Question

Find an othogonal matrix P such that (P-transpose*A*P) diagonalizes matrix A. I use the usual eigenvalue-vector method but the eigenvecoters i get are dodge. could someone please go through how they get this!¡ Thanks in advance!!

Answer 1

A=(0 3 0
3 0 4
0 4 0)

this might help -.- hahah

Answer 2

did you find the eigenvalues?
find the characteristic equation
\[\left| \left[\begin{matrix}-\lambda & 3 & 0 \\ 3 & -\lambda & 4 \\ 0& 4 & -\lambda \end{matrix}\right]\right|=0\]
you get
\[−λ(λ^2 -16) -3(−3λ) = 0 \\ λ( -λ^2 +16+9)=0 \\ λ( -λ^2 +25)=0 \]
solve for lambda.

what do you get ?

Answer 3

got plus and minus 5, then i did the whole eigenvector thing but thats where i com across the problem. Plug in 5 and X1, X2, X3 are all equal zero, same thing for -5 and 0 isn't an option... Probably solving the equations incorrectly!

Answer 4

doing the rref i get a value in each pivot position with 0's in the rest, do i take those values as the eigenvectors? go through it for me cause I'm a tad at a loss! thanks

Answer 5

yes, the eigenvalues are 5, -5, and 0
notice the equation that defines the eigenvalue/vector is
\[ A x = \lambda x\\ A x - \lambda x=0 \\ \left(A -λ I\right) x = 0 \]
which tells you that the eigenvector is in the null space of the matrix
(A - λI)
to find the eigenvectors, find the null space of this matrix.

for each eigenvalue, create the matrix
\[ \left[\begin{matrix}-\lambda & 3 & 0 \\ 3 & -\lambda & 4 \\ 0& 4 & -\lambda \end{matrix} \right]\]
and use Gauss-Jordan elimination
https://en.wikipedia.org/wiki/Kernel_(matrix)

Answer 6

Yep, I do this part and get two vectors (each) for -5 and 5, 0 is not an option. I pick the three which are L.I. and come out with this
B=(3 0 0
5 -5 4
0 4 5)

Could you check if this is correct?

Once I've got that I have to orthonormalize B in order to diagonlize A right?

Answer 7

You should get 3 eigenvectors, including an eigenvector for eigenvalue 0

for example, if you start with
0 3 0
3 0 4
0 4 0

which is for lambda=0. Perform gauss-jordan elimination.
we can swap rows
3 0 4
0 3 0
0 4 0
divide the first and second rows by 3

1 0 4/3
0 1 0
0 4 0

add -4 * 2nd row to the 3rd row
1 0 4/3
0 1 0
0 0 0

if this is multiplies the vector x y z, with z=1. you find y=0, x= -4/3

the eigenvector is [-4/3 0 1]
multiply by 3 to get
v= [- 4 0 3]^T
as a test multiply A v to get [0 0 0]
so Av = 0 v works.

Answer 8

notice that matrix A is symmetric. that means its eigenvectors are orthogonal.

but we do need to make them unit length.

v = v= [-4 0 3]/sqrt(25)= [-4/5 0 3/5]

you must do the same thing for eigenvalues 5 and -5
the corresponding eigenvectors will be the columns of matrix P and
P^T A P = a diagonal matrix

Answer 9

Awesome dude! been going round in circles for a couple days. thanks¡¡¡!!!