I am trying to solve to get the eigenvectors, Can someone help me?

-1 1 2 | 0
1 1 0 | 0
0 -1 -1 |0

the answer is 1 But I dont know how to get to that
-1
1

4/18/2006 7:55:14 PM

first, you need to find the eigen values. you do that by setting the determinant of (A-c*I) to zero, and then solve for c. c are your eigen values.

for the eigenvectors, you find the kernel/null space of (A-c*I), for each eigenvalue. the vector that then spans the null space is your eigen vector.
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although it looks like you already have it in a (A-c*I) form since it's adjunct to [0;0;0], in which case you just find the null space of that matrix. you may want to row reduce it a little more to make things more clear.

4/18/2006 11:40:26 PM

ti-89

4/18/2006 11:47:47 PM

or hell, a t1-86 for that matter

or maple
or matlab

4/18/2006 11:56:03 PM

yeah, pretty much what virga said.

i went ahead and did it cause i need the practice for my diffy q test...so ill tell you what i did

you just need to reduce it to row-echelon form. doing that you get the first row in the matrix to be 1 0 -1 0, the second row to be 0 1 1 0 and the last row to be all zeros. So then, from that you use it like a set of equations. Interpret the first row as 1*U1+0*U2+-1*U3=0, then the second row is U2+U3=0. Make U3 arbitrary. What we were taught to do was say U3=S and then solve in the others in terms of it. So U1=S, U2=-s, U3=S, so your eigenvector is 1, -1, 1

4/19/2006 12:06:12 AM

[1; -1; 1] spans the null space.

and hi sarah!

4/19/2006 12:32:44 AM

hey adam! i have no idea what null space is. i was just trying to explain it in terms that non-math geniuses can understand! haha

4/19/2006 12:53:33 AM