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Matrix diagonalisation

Hi, for the first part of this question I’m wondering what it wants, as it is 4 marks so I am assume it wants some sort of derivation, but my answer really isn’t looking 4 marks worthy, how would I derive this properly ? Because the first part (QE= ED) just seems obvious and the second part follows directly so I guess I would just have to show the first part?

BA786C76-7240-46A6-86CB-070F1D67247A.jpg.jpeg
Reply 1
And this is the question:
4C8A5DBA-08CC-4EB0-9BE6-70239CFD6876.jpg.jpeg
(edited 10 months ago)
Reply 2
For the first part you could have preceded with
Qe = le
where l is an evalue and e and evector, then put it in matrix form. Similarly you could mention the |E|!=0 part and which part that impacts on. But it is a couple of lines and pretty much standard textbook derivation.
Reply 3
Original post by mqb2766
For the first part you could have preceded with
Qe = le
where l is an evalue and e and evector, then put it in matrix form. Similarly you could mention the |E|!=0 part and which part that impacts on. But it is a couple of lines and pretty much standard textbook derivation.


Is it true that |E| =\ 0 for all ( not necessarily symmetric) matrices? I know for symmetric matrices they can be made all orthogonal but what about non symmetric matrices?
Reply 4
Original post by grhas98
Is it true that |E| =\ 0 for all ( not necessarily symmetric) matrices? I know for symmetric matrices they can be made all orthogonal but what about non symmetric matrices?

Theres an example towards the bottom for a nonsymmetric 2*2 matrix
https://www.statlect.com/matrix-algebra/linear-independence-of-eigenvectors
so really youre assuming the evectors are linearly independent or |E|!=0
Reply 5
Original post by mqb2766
Theres an example towards the bottom for a nonsymmetric 2*2 matrix
https://www.statlect.com/matrix-algebra/linear-independence-of-eigenvectors
so really youre assuming the evectors are linearly independent or |E|!=0

Ok, so is it important to mention that the equation only holds for linearly independent eigen vectors?
Reply 6
Original post by grhas98
Ok, so is it important to mention that the equation only holds for linearly independent eigen vectors?


Id have mentioned it when you assumed E^(-1) existed / that |E|!=0. You wouldnt probably have to relate it to linearly independent evectors, just that the matrix is invertible because the det is non-zero. However, theres no harm in a bit more understanding.
Reply 7
Original post by mqb2766
Id have mentioned it when you assumed E^(-1) existed / that |E|!=0. You wouldnt probably have to relate it to linearly independent evectors, just that the matrix is invertible because the det is non-zero. However, theres no harm in a bit more understanding.


I See, for part ii do you think this is enough working:
FBB97951-F80B-4CAB-886F-700976945F78.jpg.jpeg
Seeing as it is a 5 mark question, this working honestly looks worth 2 marks, is there more to it?

The top like is meant to have k on the outside of the brackets not as an exponent for D
(edited 10 months ago)
Reply 8
Original post by grhas98
I See, for part ii do you think this is enough working:
FBB97951-F80B-4CAB-886F-700976945F78.jpg.jpeg
Seeing as it is a 5 mark question, this working honestly looks worth 2 marks, is there more to it?

The top like is meant to have k on the outside of the brackets not as an exponent for D


Looks about right but the first line does look like youre jumping to the answer. Also, you could have written out the
EDE^(-1)EDE^(-1)E..... EDE^(-1)EDE^(-1)
and just paired all the interal E^(-1)E = I and pretty much got straight to the result.

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