# 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?

And this is the question:
(edited 1 year ago)
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.
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?
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
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?
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.
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:

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 1 year ago)
Original post by grhas98
I See, for part ii do you think this is enough working:

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.