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## KonradZuse Group Title In Exercises 2–9, find a matrix P that orthogonally diagonalizes A, and determine P^-1 AP. one year ago one year ago

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1. KonradZuse

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

2. KonradZuse

@phi @kropot72

3. phi

Are you learning about SVD (singular value decomposition)?

4. KonradZuse

nope....

5. phi

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

6. KonradZuse

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

7. KonradZuse

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

8. 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

9. KonradZuse

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

10. phi

are you sure about the eigenvalues??

11. KonradZuse

yeah

12. phi

because I do not get 3 and 6

13. KonradZuse

actually h/o.

14. KonradZuse

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

15. 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

16. KonradZuse

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

17. phi

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

18. KonradZuse

yeah :)

19. phi

L^2 -9L +14=0

20. phi

this factors

21. KonradZuse

mhm

22. phi

you don't know how to factor?

23. 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

24. KonradZuse

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

25. KonradZuse

and that makes sense yeah.

26. KonradZuse

@ganeshie8 @Hero

27. KonradZuse

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

28. KonradZuse

@tkhunny

29. KonradZuse

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

30. 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?

31. KonradZuse

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