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In Exercises 2–9, find a matrix P that orthogonally diagonalizes A, and determine P^1 AP.
 one year ago
 one year ago
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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KonradZuseBest ResponseYou've already chosen the best response.0
\[A = \left[\begin{matrix}6 & 2 \\ 2 & 3\end{matrix}\right]\]
 one year ago

phiBest ResponseYou've already chosen the best response.0
Are you learning about SVD (singular value decomposition)?
 one year ago

phiBest ResponseYou've already chosen the best response.0
but A is symmetric, so I think its eig vectors are orthogonal
 one year ago

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

KonradZuseBest ResponseYou've already chosen the best response.0
but I'm not sure about this finding P business.....
 one year ago

phiBest ResponseYou've already chosen the best response.0
so 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
 one year ago

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

phiBest ResponseYou've already chosen the best response.0
are you sure about the eigenvalues??
 one year ago

phiBest ResponseYou've already chosen the best response.0
because I do not get 3 and 6
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
I guess I did that wrong lol now i'm konfused with that :P
 one year ago

phiBest ResponseYou've already chosen the best response.0
Let 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
 one year ago

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

phiBest ResponseYou've already chosen the best response.0
you don't know how to factor?
 one year ago

phiBest ResponseYou've already chosen the best response.0
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 (L2)(L7)=0 this means L= 2 or L=7
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
Sorry I had to do some holiday stuff.....
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
and that makes sense yeah.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
not completely, I kind of left mid way for something.
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
Save me please :). 2 Questions 1 hour... :(.
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
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?
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
How do we find the normal EV's again? and what trick?
 one year ago
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