# N term metrix question

1. Feb 5, 2008

2. Feb 5, 2008

### EnumaElish

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

3. Feb 5, 2008

### transgalactic

it makes no difference

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

??

4. Feb 5, 2008

5. Feb 5, 2008

### transgalactic

youll noticed that i used it
but in a different step
i dont think there is any difference
i have trouble to finish it

6. Feb 5, 2008

### HallsofIvy

Staff Emeritus
That's exactly what he did. Although he miswrote it as det A= ... when he has det A-$\lambda$I 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)

7. Feb 6, 2008

### transgalactic

for L=1
you got the expression of x2 in "n" terms
i cant 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
??

8. Feb 6, 2008

### HallsofIvy

Staff Emeritus
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.