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

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KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0\[A = \left[\begin{matrix}6 & 2 \\ 2 & 3\end{matrix}\right]\]

phi
 2 years ago
Best ResponseYou've already chosen the best response.0Are you learning about SVD (singular value decomposition)?

phi
 2 years ago
Best ResponseYou've already chosen the best response.0but A is symmetric, so I think its eig vectors are orthogonal

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0basically we learned about diagonalization in chaptrer 5 and now it';s back in 7. I did the first part fidning the eigenvalues.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0but I'm not sure about this finding P business.....

phi
 2 years ago
Best ResponseYou've already chosen the best response.0so find its evecs and lambdas AS= S L S is the matrix of evecs (as cols) L is the diag matrix of lambdas and S^1 A S = L A is "diagonalized" to its eigen values

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0The way trhe book started out was finding the eigen values which I got to be 3 and 6.

phi
 2 years ago
Best ResponseYou've already chosen the best response.0are you sure about the eigenvalues??

phi
 2 years ago
Best ResponseYou've already chosen the best response.0because I do not get 3 and 6

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I guess I did that wrong lol now i'm konfused with that :P

phi
 2 years ago
Best ResponseYou've already chosen the best response.0Let me fix some typo's A x = λ x (Ax  λ x)= 0 (A  λ I) x =0 x (evec) 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  6λ 2   2 3λ  = 0

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0yeah that's what I thought... (6L) (3L)  (2)(2) = 18  9L +L^2 4

phi
 2 years ago
Best ResponseYou've already chosen the best response.0you don't know how to factor?

phi
 2 years ago
Best ResponseYou've already chosen the best response.0the + 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 (L2)(L7)=0 this means L= 2 or L=7

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry I had to do some holiday stuff.....

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0and that makes sense yeah.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0not completely, I kind of left mid way for something.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Save me please :). 2 Questions 1 hour... :(.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Eigenvalues 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
 2 years ago
Best ResponseYou've already chosen the best response.0How do we find the normal EV's again? and what trick?
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