maths 1B last minute questions (1 Viewer)

He-Mann

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2) Hope this makes sense, I can't Engrish!



You could also start with A^T v = lambda v and find the characteristic equation (ah, this is the term I was looking for) of A^T and realise it's the same as the characteristic equation as A.
 

He-Mann

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Thanks for the solution, InteGrand! Do you know if there is another way to do it? I have a solution but it requires A to be diagonalizable. I had another idea that does this:



Ignore the above TeX, the following is what I meant:


Using factorization:



Not sure about it.
 
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InteGrand

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Reading He-Mann's solution more carefully for this now, I think this is what he was doing (wasn't sure on first reading).
 

He-Mann

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My mistake on the not being precise with where each zero-vector belonged to which vector space.
 

InteGrand

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4) let r_k be the rank of T_k and n_k the nullity of T_k (k = 1,2). Let d_j be the dimension of V_j (j = 1,2,3).

We have im(T_2) = T_2 (V_2) = V_3, so r_2 = d_3 (taking dimensions). By Rank-Nullity, we thus have n_2 = d_2 - d_3.

We also have im(T_1) = T_1 (V_1) = ker(T_2). Taking dimensions, we have r_1 = n_2. Hence by Rank-Nullity Theorem, d_1 = n_1 + r_1 = n_1 + n_2.

Now, d_1 - d_2 + d_3 = (n_1 + n_2) - (n_2) (using our established relationships above)

= n_1,

As required.
 

Drsoccerball

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Thanks guys :) ! Will be back for more soon, I pretty much skipped 70-80 % of my lectures so I need your help !
 

InteGrand

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Thanks for the solution, InteGrand! Do you know if there is another way to do it? I have a solution but it requires A to be diagonalizable. I had another idea that does this:



Ignore the above TeX, the following is what I meant:


Using factorization:



Not sure about it.
 

Drsoccerball

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If we are given the eigenvalues of the matrix with corresponding eigenvectors what is the simplest way to find the original matrix without doing A = MDM^(-1)?
 

InteGrand

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If we are given the eigenvalues of the matrix with corresponding eigenvectors what is the simplest way to find the original matrix without doing A = MDM^(-1)?
In some special cases there may be some shortcuts, but in general I don't think there's much we can do that's simpler than just doing A = MDM-1.
 

InteGrand

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By the way, a nice and easy check you should do after doing such a question (finding the matrix A when given its eigenvalues and eigenvectors): check that the sum of diagonal elements of the matrix A you found equals the sum of the given eigenvalues. (It should be the case, because for any square matrix, the sum of diagonal elements (called the trace of the matrix) is equal to the sum of its eigenvalues.)

Another test you could do (which is more tedious but would give further confidence in your answer) would be to multiply your matrix A with a given eigenvector v and check that it satisfies the relation Av = λv (where λ is the given eigenvalue associated with that eigenvector v).
 

Drsoccerball

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Okay... I have questions about convergence and divergence.

1) Is this a sufficient condition to prove convergence of b_k:



2) To prove that converges we can prove that

3) How do you prove something is conditionally convergent, yes I know that you have to prove the absolute is divergent and the normal is convergent but how do you prove it to be convergent? Do you just split up the negative terms and positive terms into two series and see if they both converge?
 

Drsoccerball

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Okay... I have questions about convergence and divergence.

1) Is this a sufficient condition to prove convergence of b_k:



2) To prove that converges we can prove that

3) How do you prove something is conditionally convergent, yes I know that you have to prove the absolute is divergent and the normal is convergent but how do you prove it to be convergent? Do you just split up the negative terms and positive terms into two series and see if they both converge?
From my knowledge 1 is completely wrong. If a_k diverges then that can prove that b_k diverges. Also from my knowledge of 2 if the ratio is 1 then we have to do integral test?

EDIT: and for 3 we just have to prove a_n > a_{n+1}, a_{n} -> 0 as n -> infty, a_n >= 0?
 
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leehuan

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By the way, a nice and easy check you should do after doing such a question (finding the matrix A when given its eigenvalues and eigenvectors): check that the sum of diagonal elements of the matrix A you found equals the sum of the given eigenvalues. (It should be the case, because for any square matrix, the sum of diagonal elements (called the trace of the matrix) is equal to the sum of its eigenvalues.)

Another test you could do (which is more tedious but would give further confidence in your answer) would be to multiply your matrix A with a given eigenvector v and check that it satisfies the relation Av = λv (where λ is the given eigenvalue associated with that eigenvector v).
Do you call it the trace or something? I don't quite remember.
Okay... I have questions about convergence and divergence.

1) Is this a sufficient condition to prove convergence of b_k:



2) To prove that converges we can prove that

3) How do you prove something is conditionally convergent, yes I know that you have to prove the absolute is divergent and the normal is convergent but how do you prove it to be convergent? Do you just split up the negative terms and positive terms into two series and see if they both converge?






For our purposes with 3) we must rely on the alternating series test.
 
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