alright that answers my question thank you for your help.

ah yes you are right, i did that and it works now thank you. but still, why is it that we have to say 'lambda=l'? cant we leave it as lambda?

ok so then I put this before the loop:

eigen:=subs(lambda=l,map(eval,eigen));

so this means that i have replaced lambda with l and fully evaluated the matrix and then i proceed with the loop.  I get all of the eigenvectors, but is this correct then? and also why do we have to replace lambda with 'l', cant we just proceed with lambda?

Thank you,

Hi, sorry again but i have another problem. See i am doing a similar thing to what you showed me, except its with a more complicated matrix. This time, the loop is not substituting lambda=l into the matrix, i attached the file so you can see.

View 11295_07feb.mw on MapleNet or Download 11295_07feb.mw
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If you could have a look that would be great, and thanks for your help before.

Oh i see thank you, i didnt think that i could use the map command within another map command. i was doing Eigenvectors(map(...)), obviously wrong.

Thank you so much

Yah I know how to do eigenvectors for loops but I am not sure how to do eigenvectors for this map command. I tried many times but I am not getting it. I tried to create a separate loop for the eigenvectors but maple tells me that the substitutions are not valid for the argument.

oh excellent i understand thank you very much, and how can i find the eigenvectors of these matrices?

Ok thanks that really helps, I understand what you have done. Now for my problem, actually I cant solve analytically for m in terms of lambda, however the equation does solve numerically. That is why I tried to run the loop for lambda before solving the determinant for m...

Now in this simpler case, cant we insert the loop before we solve the determinant?

Thanks again,

OK i understand what you have done and that is fine. But i do need lambda.

I want to find the solutions of m for a range of values of lambda, and then substitute them back in and display the result (exactly like you have done above). You have done this for one value of lambda, i.e 1, and thats fine, but i need to find it for many lambda. for example we could use lambda from 1 to 2. This is why i used lambda as a loop counter, but i am unable to display all of the resulting solutions and corresponding matrices.

If you can, could you please show me how to do that?

Thank you,

SORRY NOT M its eigeni

for j from 1 by 1 to 8 do  Eigenvectors(subs(m[i]=ans[j],eigeni)):  end do;

SORRY NOT M its eigeni

for j from 1 by 1 to 8 do  Eigenvectors(subs(m[i]=ans[j],eigeni)):  end do;

sorry one code is missing from the end of that,

for j from 1 by 1 to 8 do  Eigenvectors(subs(m[i]=ans[j],M)):  end do;

thanks

it can be done in matlab, i dont know how, but my advice is forget matlab

i am not sure to be honest im not too good with maple, somebody else can answer u

try again, cant do much with wat u sent. click on the green up arrow button to upload