KonradZuse
  • KonradZuse
In Exercises 2–9, find a matrix P that orthogonally diagonalizes A, and determine P^-1 AP.
Linear Algebra
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SOLVED
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chestercat
  • chestercat
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KonradZuse
  • KonradZuse
\[A = \left[\begin{matrix}6 & -2 \\ -2 & 3\end{matrix}\right]\]
KonradZuse
  • KonradZuse
@phi @kropot72
phi
  • phi
Are you learning about SVD (singular value decomposition)?

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KonradZuse
  • KonradZuse
nope....
phi
  • phi
but A is symmetric, so I think its eig vectors are orthogonal
KonradZuse
  • 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
  • KonradZuse
but I'm not sure about this finding P business.....
phi
  • 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
  • KonradZuse
The way trhe book started out was finding the eigen values which I got to be 3 and 6.
phi
  • phi
are you sure about the eigenvalues??
KonradZuse
  • KonradZuse
yeah
phi
  • phi
because I do not get 3 and 6
KonradZuse
  • KonradZuse
actually h/o.
KonradZuse
  • KonradZuse
I guess I did that wrong lol now i'm konfused with that :P
phi
  • 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
  • KonradZuse
yeah that's what I thought... (6-L) (3-L) - (-2)(-2) = 18 - 9L +L^2 -4
phi
  • phi
18 - 9L +L^2 -4=0
KonradZuse
  • KonradZuse
yeah :)
phi
  • phi
L^2 -9L +14=0
phi
  • phi
this factors
KonradZuse
  • KonradZuse
mhm
phi
  • phi
you don't know how to factor?
phi
  • 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
  • KonradZuse
Sorry I had to do some holiday stuff.....
KonradZuse
  • KonradZuse
and that makes sense yeah.
KonradZuse
  • KonradZuse
@ganeshie8 @Hero
KonradZuse
  • KonradZuse
not completely, I kind of left mid way for something.
KonradZuse
  • KonradZuse
@tkhunny
KonradZuse
  • KonradZuse
Save me please :). 2 Questions 1 hour... :(.
tkhunny
  • 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
  • KonradZuse
How do we find the normal EV's again? and what trick?

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