Question

After calculating, I got this rI -A is \[\left[\begin{matrix}3&-1&-2\\1&2&1\\2&1&3\end{matrix}\right]\]
I forgot how to find eigenvector from that, help me, please

Answer 1

@e.mccormick

Answer 2

I replace eigenvalue into the RI -A to get the matrix above. And ...I need you help, hehehe

Answer 3

RI? Mean \(\lambda I \)?

Answer 4

yup

Answer 5

OK. Did you get the determinant?

Answer 6

why? do we have to have it?

Answer 7

That gets you the eigenvalues.

Answer 8

ok, let me show you the original A \[\left[\begin{matrix}1&1&2\\1&2&1\\2&1&1\end{matrix}\right]\]

Answer 9

I have characteristic equation is \[\lambda^3-4\lambda^2-\lambda+4=0\\(\lambda+1)(\lambda -4)(\lambda -1)=0\]
so the eigenvalues are -1, 1 and 4
the matrix above is got from eigenvalue 4 and stuck

Answer 10

OK, so you did the value part already.

Answer 11

yes

Answer 12

For each eigenvalue, you solve \((A-\lambda I)=0\) by Gaussian Elimination.

Answer 13

rref?

Answer 14

missed an \(\vec{x}\) after the )....

Answer 15

ok, we know it, doesn't matter, go ahead

Answer 16

So for the first value, -1, we have:
\(\begin{bmatrix}
1-\lambda & 1 & 2\\
1 & 2-\lambda & 1\\
2 & 1 & 1-\lambda
\end{bmatrix}\vec{x}=0 \implies \)

\(\begin{bmatrix}
1+1 & 1 & 2\\
1 & 2+1 & 1\\
2 & 1 & 1+1
\end{bmatrix}\vec{x}=0 \implies \)

\(\left[ \begin{array}{ccc|c}
2 & 1 & 2&0\\
1 & 3 & 1&0\\
2 & 1 & 2&0\\
\end{array} \right]\)

Answer 17

why? I ask for 4 , not for -1 why do you force me to do the whole set??

Answer 18

And you put that in ref, not rref. Upper triangular form.

Answer 19

I was just picking it as an example. If you want to do another, do so.

Answer 20

I want 4. you know why? because it 's hardest.

Answer 21

If I can solve the hardest one. the easier one will be "no problem" , right?

Answer 22

I don't know that any of them are really any harder or easier. At that point, it is basically putting it into parametric form.

Answer 23

hey, what is parametric form?

Answer 24

Know parametic equations? Where you represent all variables as combinations of one of them?

Answer 25

Here is an explantion of what I mean:
http://www.scss.tcd.ie/Rozenn.Dahyot/CS1BA1/SolutionEigen.pdf

The lower part of this shows it.

Answer 26

yes, know it from cal3 but don't see the link to linear algebra

Answer 27

Ok, I think I am ok now. Thanks for the link @e.mccormick. I will show you the answer. hihi

Answer 28

one more thing, my prof wants me apply RI -A , not A - RI . ha!! each prof gives out his way to apply. I have to go back and forth between the 2

Answer 29

They are the same in the end. It is just if you start one way, you use that way all through.

Answer 30

hey, do we always have a row =0 from ref?

Answer 31

at least 1 , at most 2 if it is 3x3 matrix?

Answer 32

It is implied if you leave it off. The main thing to realize is that you are solving an Ax=b type equation where b is the zero vector and you want to get at x.

Answer 33

you confused me, because if it's so, x = A^-1 b =0

Answer 34

When the x is a series of variables, it distributes across the matrix.
\(\begin{align*}
ax_1 + bx_2 + cx_3 &= 0 \\
dx_1 + ex_2 + fx_3 &= 0 \\
gx_1 + hx_2 + ix_3 &= 0
\end{align*}\)

becomes:

\(\begin{bmatrix}
a & b& c \\
d&e&f\\
g&h&i
\end{bmatrix} \left \langle \begin{matrix} x_1\\x_2\\x_3 \end{matrix} \right \rangle=0\)

Answer 35

Or more accurately, \(=\vec{0}\) on the right of that....

Answer 36

but cannot pick \(\vec 0\)

Answer 37

The 0 vector on the right is the same thing as having the augmented matrix with zeros. All the zeros do is let you put say \(x_1\) in terms of \(x_2\) and \(x_3\).

Answer 38

got you. thanks e.mccormick