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Oct 30 2014 09:51am
Hey I need some help with question 3 and 5



For question 3 I can show it for n = 2 but I am not sure how to prove it for all n. After getting the 2x2 case would I use induction to prove it for n? Not sure how though...
And question 5 all I have so far is P^TAP=B and so P^T = P^-1. I think the showing A^n is orthogonally similar to B^n is something trivial or easy but I can't really seem to prove it.. I was thinking it would be similar to how to show A^n = PD^nP^-1 for a diagonal matrix D.
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Oct 30 2014 01:45pm
For 3 you should be fine with just proving the case for symmetric 3x3's. You could even use the A matrix in question 4.
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Oct 30 2014 01:46pm
90% of linear algebra is just the same stuff over and over again, just with different terminology rofl
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Nov 1 2014 12:58pm
for question 3. to prove => you can use the definition of eva;s/evects which is Ax = λx

proving <= you can use the definition of a diagonal matrix A = PDP^-1

for question 5 just use the definition for being orthogonally similar. A = PDP^T
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Nov 1 2014 02:33pm
Quote (JDota72 @ Oct 30 2014 09:46pm)
90% of linear algebra is just the same stuff over and over again, just with different terminology rofl

you obviously don't know much

on topic:
for Q3 take X eigen vector of A associated to lambda, write (AX.X) = (B²X.X) = (since B is symetric) ||BX||² = lambda*||X||², which implies lambda >= 0 (where (AX.X) denotes the scalar product)

Q4 is simple diagonalization .. P is the matrix of the eigen vectors. Solve det(A-lambda*I) = 0 to find the eigen values, and AX = lambda*X for the eigen values you found.

for Q5 write A = QBQ', A^n = Q*B*(Q'*Q)*B*Q'* ....*Q*B*(Q'*Q)*B*Q' = Q*B^n*Q' (because Q'*Q = I and every term in between cancels out) which proves that A^n and B^n are orthogonally similar
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Nov 1 2014 04:30pm
Quote (HbSoe @ Nov 1 2014 03:33pm)
you obviously don't know much

on topic:
for Q3 take X eigen vector of A associated to lambda, write (AX.X) = (B²X.X) = (since B is symetric) ||BX||² = lambda*||X||², which implies lambda >= 0 (where (AX.X) denotes the scalar product)

Q4 is simple diagonalization .. P is the matrix of the eigen vectors. Solve det(A-lambda*I) = 0 to find the eigen values, and AX = lambda*X for the eigen values you found.

for Q5 write A = QBQ', A^n = Q*B*(Q'*Q)*B*Q'* ....*Q*B*(Q'*Q)*B*Q' = Q*B^n*Q' (because Q'*Q = I and every term in between cancels out) which proves that A^n and B^n are orthogonally similar


Wow thank you! That q3 solution is clever. I was trying to prove it by looking at each element of the matrix which is why I couldn't prove it for an n x n matrix but could for a 2x2

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