In Exercises 2–9, find a matrix P that orthogonally diagonalizes A, and determine P^-1 AP.

- KonradZuse

In Exercises 2–9, find a matrix P that orthogonally diagonalizes A, and determine P^-1 AP.

- chestercat

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- KonradZuse

\[A = \left[\begin{matrix}6 & -2 \\ -2 & 3\end{matrix}\right]\]

- phi

Are you learning about SVD (singular value decomposition)?

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- KonradZuse

nope....

- phi

but A is symmetric, so I think its eig vectors are orthogonal

- KonradZuse

basically we learned about diagonalization in chaptrer 5
and now it';s back in 7. I did the first part fidning the eigenvalues.

- KonradZuse

but I'm not sure about this finding P business.....

- phi

so find its e-vecs and lambdas
AS= S L
S is the matrix of e-vecs (as cols) L is the diag matrix of lambdas
and
S^-1 A S = L A is "diagonalized" to its eigen values

- KonradZuse

The way trhe book started out was finding the eigen values which I got to be 3 and 6.

- phi

are you sure about the eigenvalues??

- KonradZuse

yeah

- phi

because I do not get 3 and 6

- KonradZuse

actually h/o.

- KonradZuse

I guess I did that wrong lol now i'm konfused with that :P

- phi

Let me fix some typo's
A x = λ x
(Ax - λ x)= 0
(A - λ I) x =0
x (e-vec) is in the null space of (A - λ I)
reduce to rref to find the null space vector (x)
for x to exist (not be just the 0 vector) the matrix (A - λ I) cannot be full rank
that means its determinant is 0
to find the lambdas solve
det ( (A - λ I) ) =0
| 6-λ -2 |
| -2 3-λ | = 0

- KonradZuse

yeah that's what I thought... (6-L) (3-L) - (-2)(-2) = 18 - 9L +L^2 -4

- phi

18 - 9L +L^2 -4=0

- KonradZuse

yeah :)

- phi

L^2 -9L +14=0

- phi

this factors

- KonradZuse

mhm

- phi

you don't know how to factor?

- phi

the + in +14 means the values are the same
the - in -9 (coeff of L) means the largest is -. so both are negative
list the pairs of factors of 14:
1,14
2,7
which pair added together with the same sign (- in this case) give -9
yes -2 and -7
so
(L-2)(L-7)=0
this means L= 2 or L=7

- KonradZuse

Sorry I had to do some holiday stuff.....

- KonradZuse

and that makes sense yeah.

- KonradZuse

- KonradZuse

not completely, I kind of left mid way for something.

- KonradZuse

- KonradZuse

Save me please :). 2 Questions 1 hour... :(.

- tkhunny

Eigenvalues are 7 and 2.
Normal Eigenvectors are \(\left[-\dfrac{2}{\sqrt{5}},\dfrac{1}{\sqrt{5}}\right]^{T}\) and \(\left[-\dfrac{1}{\sqrt{5}},-\dfrac{2}{\sqrt{5}}\right]^{T}\)
What's the trick?

- KonradZuse

How do we find the normal EV's again?
and what trick?

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