# Finding Eigenvectors and Eigenvalues

1. Apr 29, 2014

### concon

1. The problem statement, all variables and given/known data
The Matrix A is as follows
A= [4 -4 0
2 -2 0
-2 5 3]

and has 3 distinct eigenvalues λ1<λ2<λ3
Let Vi be the unique eigenvector associated with λi with a 1 as its first nonzero component.
Let
D = [ λ1 0 0
0 λ2 0
0 0 λ3]

and P= [v1|v2|v3]

Find D and P.

2. Relevant equations
not going to right it out but I know you find eigenvectors by (λI-A)X = 0
and eigenvalues by det(λI-A).

3. The attempt at a solution

So I already found D with the eigenvalues as {0,2,3}

I tried to solve for P and got

[-1 -2 0
1 1 0
1 1 1]
but this is wrong.

I am kinds confused b/c the textbook says (λI-A)X=0 but my professor uses (A-λI)X=0 in class!

2. Apr 29, 2014

### jbunniii

The columns of $P = [v_1 | v_2 | v_3]$ should be eigenvectors of $A$. But calculating $Av_1$ and $Av_2$ shows that the first two columns are not eigenvectors.

Both are correct. One is simply the negative of the other. Since a vector is zero if and only if its negative is zero, the equations are equivalent.

3. Apr 29, 2014

### concon

What do you mean? That the eigenvectors that I calculated v1,v2,v3 are not correct? Or that they are not actual eigenvectors?

What should I do then?

4. Apr 29, 2014

### jbunniii

They are not eigenvectors:
$$Av_1 = \begin{bmatrix}4 & -4 & 0 \\ 2 & -2 & 0 \\ -2 & 5 & 3 \\ \end{bmatrix} \begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -8 \\ -4 \\ 10 \end{bmatrix}$$
which is not a multiple of
$$\begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix}$$
Similarly, $A v_2$ is not a multiple of $v_2$.

5. Apr 29, 2014

### concon

So then what is/how do i find P?

6. Apr 29, 2014

### jbunniii

The columns of $P$ should be eigenvectors of $A$. Your eigenvalues are correct. How did you calculate the corresponding eigenvectors? Maybe you just made an arithmetic mistake.

7. Apr 29, 2014

### concon

So I took each value 0,2,3 and did (A-λI)X = 0

For zero:
(4-0) -4 0
2 (-2-0) 0
-2 5 (3-0)

then I reduced and got
v1 = -1,-1,1
and did same process for the other values and got

v2=-2,-1,1
v3=0,0,1

Everytime I solve this problem I get the same values, but the signs change but the numbers are still the same. Have you tried solving it? Is this what you got? What could I be doing wrong?

8. Apr 29, 2014

### jbunniii

OK, that's fine. Using this $v_1$ we calculate $Av_1 = 0$, so $v_1$ is an eigenvalue associated with the eigenvalue $0$. In your original post you had v1 = [-1, 1, 1], so maybe you made a sign error last time you calculated it.
Nope, using this $v_2$ I calculate $A v_2 = [0, 0, -3]$, which is not a multiple of $[-2, -1, 1]$.
This is fine, assuming it goes with the eigenvalue 3.
I think you're making sign errors in some of your calculations. Can you show how you computed $v_2$?

9. Apr 29, 2014

### concon

Okay here is calculation again for v2:
λ=2

(4-2) -4 0
2 (-2-2) 0
-2 5 1

simplified:
2 -4 0
2 -4 0
-2 5 1

row 1 minus row 2
2 -4 0
0 0 0
-2 5 1

row 1 plus row 3

2 -4 0
0 0 0
0 1 1

divide row one by 2

1 -2 0
0 0 0
0 1 1

x= 2t
y= t
z= -t

thus v2 = 2,1,-1

10. Apr 29, 2014

### jbunniii

OK, that calculation is correct. You originally had $v_2 = [-2, 1, 1]$, so you probably made a sign error in your earlier attempt.

So let's summarize: you have eigenvalues $0,2,3$ with corresponding eigenvectors $v_1 = [-1, -1, 1]$, $v_2 = [2, 1, -1]$, and $v_3 = [0, 0, 1]$. So your $P$ matrix could be
$$P = [v_1 | v_2 | v_3] = \begin{bmatrix} -1 & 2 & 0 \\ -1 & 1 & 0 \\ 1 & -1 & 1 \end{bmatrix}$$
However, there is one more thing the original problem asked for:
Your $v_1$ has $-1$ as its first component, and your $v_2$ has $-2$. So you have to normalize your vectors to get $1$ as the first nonzero component. ($v_3$ is OK as is.)

11. Apr 29, 2014

### concon

Wait but doesn't normalizing a vector mean taking v1/||v1|| ?

How would that give me a 1 as the first nonzero component?
Also if I use v2 = 2,1,-1 do I also need to normalize it if the 2 isn't negative?

12. Apr 29, 2014

### jbunniii

Sorry, I meant you need to scale the vectors so that the first nonzero component is 1. So for example, if you had an eigenvector $v = [-3, 1, -5]$ you would multiply it by -1/3 to get $[1, -1/3, 5/3]$.

Note that any nonzero multiple of an eigenvector is still an eigenvector associated with the same eigenvalue, because if $Av = \lambda v$ then multiplying $v$ by a constant scalar $c$, we get $A(cv) = c(Av) = c(\lambda v) = \lambda(cv)$.

Generally there's no need to scale the eigenvectors in any particular way unless the problem requests it. Probably your instructor wants them scaled with a 1 in the first nonzero component so your answer is guaranteed to match his/hers if it is correct.

13. Apr 29, 2014

### concon

Okay so scaling would give me:
1 2 0
1 1 0
-1 -1 1
Is this correct?

14. Apr 29, 2014

### jbunniii

The first and third columns are OK. The second column's first component is 2, not 1, so it is not scaled as requested.

15. Apr 29, 2014

### concon

whoops forget to scale that one! Thanks for all your help I really appreciate it! For some reason the scaling thing didn't register with me on what the professor wanted and I was confused with the sign changing.