Question

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

Answer 1

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

Answer 2

@phi @kropot72

Answer 3

Are you learning about SVD (singular value decomposition)?

Answer 4

nope....

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

are you sure about the eigenvalues??

Answer 11

yeah

Answer 12

because I do not get 3 and 6

Answer 13

actually h/o.

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

yeah :)

Answer 19

L^2 -9L +14=0

Answer 20

this factors

Answer 21

mhm

Answer 22

you don't know how to factor?

Answer 23

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

Answer 24

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

Answer 25

and that makes sense yeah.

Answer 26

@ganeshie8 @Hero

Answer 27

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

Answer 28

@tkhunny

Answer 29

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

Answer 30

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?

Answer 31

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