## hamza_b23 Group Title how do we find eigenvectors of any matrix where detA =0? one year ago one year ago

1. hamza_b23 Group Title

i mean det (A -(lamda)I)= 0 say for example A = (1 2;-1 4)

2. hamza_b23 Group Title

i know the eigen values are lamda = 3 and lamda =2 but how to find eigenvectors

3. hamza_b23 Group Title

@KingGeorge

4. KingGeorge Group Title

If you have a matrix with a determinant of 0, I'm not sure what to do. However, for your example, let $A=\begin{bmatrix} 1&2\\-1&4\end{bmatrix}$Then we look at the determinant of $$A-\lambda I_2$$. So we have $\det\left(\begin{bmatrix} 1&2\\-1&4\end{bmatrix}-\begin{bmatrix} \lambda&0\\0&\lambda\end{bmatrix}\right) \\ =\det\left(\begin{bmatrix} 1-\lambda&2\\-1&4-\lambda\end{bmatrix}\right) \\ =(1-\lambda)(4-\lambda)-(2)(-1)$And set this equal to 0.

5. hamza_b23 Group Title

yeah i know how to find the eigenvalues completely like i am good at that but it asks what are the eigen vectors thats where i get confused

6. KingGeorge Group Title

To be honest, I'm not too great at eigenvectors either.

7. hamza_b23 Group Title

hmm true they are a pain. like i am reading the theorem you have to sub it back into (A - (lamda)I)x =0 and then solve for x using those lamdas but i reduce my matrix down for both the lamdas and i get x1 = x2 but wth does that mean

8. DanielxAK Group Title

You've pretty much solved one of the eigenvectors already. You just simplified it too much. x1 = x2 means x1 - x2 = 0. What values of x1 and x2 will solve that equation? Eigenvectors are not unique, so there's infinitely many that will work in this case. Then, your eigenvector is [x1 x2]. Note: Eigenvalues are unique, so your eigenvector only corresponds to that eigenvalue, not both. You'll have to plug in the next eigenvalue to get your other eigenvector.

9. KingGeorge Group Title

So if we have $$\lambda=2,3$$, then we have $$A=2I_2 x$$ or $$A=3I_2 x$$. Let's start with 3. That gives us$\begin{bmatrix} 1&2\\-1&4\end{bmatrix}\begin{bmatrix} x_1\\x_2\end{bmatrix}=3\begin{bmatrix} x_1\\x_2\end{bmatrix}$Solving for this, we get that $$x_1=x_2$$. So I guess that means that your eigenvector for $$\lambda=3$$ is just any vector where both $$x_1$$ and $$x_2$$ are the same.

10. KingGeorge Group Title

If you solve for $$\lambda=2$$ instead, you get that $$x_1=2x_2$$. So in this case, your eigenvector for $$\lambda=2$$ is any vector where $$x_2$$ is twice $$x_1$$.

11. hamza_b23 Group Title

hmm ya thats what it said but i wasnt getting the right answer do they use any x values like we can use any number right?

12. KingGeorge Group Title

You should be able to use any numbers you want, unless I'm grossly misunderstanding eigenvectors.

13. DanielxAK Group Title

Right. So, in the example where lamda is equal to 3, you have x1 - x2 = 0. So, [x1 x2] could be [1, 1] or [3,3] or [sqrt(2),sqrt(2)]. The simpler the better though, in my opinion.

14. hamza_b23 Group Title

true for some reason they have [1/sqrt2 1/sqrt2] thats where i was going crazy in how they determined that

15. KingGeorge Group Title

They wanted a normalized vector apparently (length of 1). Anyways, I've got to get going. Good luck.

16. hamza_b23 Group Title

ooo alright thanks a lot though

17. hamza_b23 Group Title

but @DanielxAK to get a normalized eigen vector we find the main eigenvector and normalize it?

18. DanielxAK Group Title

Yes, if x is your vector, then to normalize it, you divide it by the norm: x/||x||. The norm of x (||x||) is just the distance equation. (So, sqrt(x1^2 + x2^2) = ||x||) Also, note that the normalized eigenvector is unique. That's probably one reason why they're asking you to normalize it.

19. hamza_b23 Group Title

oo true probably thanks a lot man, just another question do you know anything about symmetric matrices?

20. DanielxAK Group Title

21. hamza_b23 Group Title

how do we show something is symmetric like i know A^T = A but its talking about some orthagonal stuff where A = CDC^T

22. DanielxAK Group Title

Can you state the whole question? Is it asking you to show A^T = A when A = CDC^T? I'm guessing C is an orthogonal matrix and D is a diagonal matrix?

23. hamza_b23 Group Title

yeahh it isnt a question my TA just said just know something about that because it may be on the test

24. hamza_b23 Group Title

is this like a spectral decomposition?

25. DanielxAK Group Title

Yes, sort of. It's a very specific type of decomposition. Matrices that have certain properties are easier to decompose. Decomposition is important in computing solutions to matrices. Some decompositions are more ideal than others. Anyway, If A = CDC*, then A* = (CDC*)* = (C*)*D*C*= CD*C* = CDC* = A. (D* = D since D is diagonal). Does that make sense?

26. hamza_b23 Group Title

hmm not quite

27. hamza_b23 Group Title

i didnt know (C*DC)* = (C*)*D*C* like i thought it was (AB)* = B*A* ?

28. DanielxAK Group Title

Correct. But you have three matrices instead of two. So, in the case of two matrices, you're reversing the order. The same is true for three.

29. hamza_b23 Group Title

hmm true do you mind showing me an example of spectral decomposition if you dont mind ?

30. hamza_b23 Group Title

i have A = ( 3 6 -1; 6 9 4; -1 4 3)

31. DanielxAK Group Title

I'm not going to work it out by hand. That'd be quite a bit of work for a 3x3 matrix. I did it on Matlab really quick though. So, you'd have it of the form: A = C*D*C' Where, roughly, C = [ -0.4551 0.5802 -0.6754 -0.8460 -0.0451 0.5313 -0.2778 -0.8132 -0.5114] D = [ 13.5417 0 0 0 3.9353 0 0 0 -2.4770]

32. hamza_b23 Group Title

ok sounds good and C= the normalized matrix for A?

33. DanielxAK Group Title

C is your unitary (or orthogonal) matrix. Spectral decomposition is also known as eigenvalue decomposition. If you worked out the eigenvalues of that matrix, you would see that diagonal of D makes up your eigenvalues and each column of C makes up your normalized eigenvectors (which correspond to its eigenvalue). For example, v = -0.4551 -0.8460 -0.2778 corresponds to the eigenvalue 13.5417.

34. hamza_b23 Group Title

ahh okayy i have to keep on practicing thenn i apparently have to do this by hand tommorow lol

35. DanielxAK Group Title

For a 3x3? I hope not. That's prone to error if done by hand. A 2x2 isn't too bad, but it gets much worse past that.

36. hamza_b23 Group Title

hmm truee just one final question, i have found the eigenvalues its just the vectors that throw me off

37. hamza_b23 Group Title

I have A = (3 1 1; 1 0 2; 1 2 0)

38. hamza_b23 Group Title

and i found the eigenvalues to be lamda1=1, lamda2=4, lamda3 = -2

39. hamza_b23 Group Title

and the question states that. Find the normalized eigenvectors and use them as columns in orthagonal matrix C

40. DanielxAK Group Title

Alright. First, build your matrix D using your eigenvalues since you know those. Then, compute the eigenvector for each eigenvalue. Normalize the eigenvectors. Then, to build your C, match the column of C with the diagonal of D. So, the eigenvalue of your first diagonal of D should match your eigenvector which is the first column of C.