Solving the N-Term Matrix Question: My Approach and Results | IMG100 Imageshack

[transgalactic]

i added my question and how i tried to solve it in the link

http://img100.imageshack.us/my.php?image=img8257ge1.jpg

 

[EnumaElish]

Why not set it up as |A − λI| = 0 (where I is the identity matrix), then solve for λ?

 

[transgalactic]

it makes no difference

my gowl was to find the determinat
and that what i have done
now what next??
i encountered some really tough problems

??

 

[EnumaElish]

Just start with the formula I stated, then proceed from there.

 

[transgalactic]

i know your formula
youll noticed that i used it
but in a different step
i don't think there is any difference
i have trouble to finish it

 

[HallsofIvy]

Why not set it up as |A − λI| = 0 (where I is the identity matrix), then solve for λ?

That's exactly what he did. Although he miswrote it as det A= ... when he has det A-\lambdaI after the "=".


transgalactic, for \lambda= 1 you have x_2+ x_3+ \cdot\cdot\cdot+ x_n= 0, -x_2= x_2, ..., -x_n= x_n. What does that give you? (It should give you one simple vector.)

For \lamba= 0, you have x_1+ x_2+ \cdot\cdot\dot+ x_n= 0. What does that give you? (n- 1 vectors)

 

[transgalactic]

for L=1
you got the expression of x2 in "n" terms
i can't do it indefinetly

for L=0
i know that i have n-1 vectors

what do i do now
what do i right as the answer of the question
??

 

[HallsofIvy]

For \lambda= 1 you have, as I said before, -x_2= x_2,-x_3= x_3, etc. That tells you that x_2= x_3= \cdot\cdot\cdot= 0! Then, of course, the first equation, x_2+ x_1+ \cdot\cdot\cdot+ x_n= 0 is automatically satisfied. Every number except x_1 must be 0. Since x_1 does not appear in any equation, it is arbitrary. All eigenvectors corresponding to \lambda= 1 are of the form <a, 0, 0, ..., 0> which is spanned, of course, by <1, 0, 0, ..., 0>.

If \lambda= 0 then you have the single equation x_1+ x_2+ \cdot\cdot\cdot+ \x_n= 0 which is the same as x_n= -(x_1+ x_2+ x_3+ \cdot\cdot\cdot+ x_{n-1}).
Now do as I have suggested before: take each x_i equal to 0 in turn, the others 0, and solve for x_n. That will give you the n-1 vectors you need.