## Empty one year ago Here's a fun question I found that has a simple fun answer, try it out. $$\bar x$$ is an eigenvector with corresponding eigenvalue$$\lambda$$ of the matrix product AB where A and B are both invertible matrices: $$AB \bar x = \lambda \bar x$$ Show that $$\lambda$$ is an eigenvalue of the matrix BA

1. UsukiDoll

both invertible matrices.. like non-singular...as in there's an inverse.

2. Miracrown

Essentially, we have to show that BAx = lambda x

3. UsukiDoll

O_O the Latex just exploded.

4. Miracrown

$(AB)^{-1} \space \space = B ^{-1} \space \space A ^{-1}$

5. Miracrown

I know it will be a trick with the order of multiplication. And I just realized that x doesn't need to be an eigenvector of BA, just the eigenvalue has to be the same...

6. Empty

:)

7. Miracrown

Take y = Bx so y is just an intermediate vector variable So let's rewrite the ABx = lam x with y

8. Miracrown

|dw:1434863967669:dw|

9. Miracrown

Now multiply both side by B. And this is the part that you have to know the commutative property well. Can you multiply these for me? :)

10. nincompoop

no.

11. UsukiDoll

BAy =(lambda)xB ? like that?

12. Miracrown

When you multiply matrices, you know that AB and BA will not always be equal.So that means it matters which side you multiply by. So if you multiply from the LEFT side... you have to keep B on the left.

13. Miracrown

$$\color{blue}{\text{Originally Posted by}}$$ @UsukiDoll BAy =(lambda)xB ? like that? $$\color{blue}{\text{End of Quote}}$$ When you first multiply by B, yes. But it has to stay left of 'x' So B lambda x And then because lambda is a constant, that how you can justify it moving it in front of B.

14. UsukiDoll

BAy=(lambda)Bx

15. UsukiDoll

D": BAy=B(lambda)x

16. Miracrown

So you get lambda B x but now, you cannot switch B and x they will stay as Bx

17. Miracrown

BAy = (lambda) Bx

18. UsukiDoll

oh.. ok so it is BAy=(lambda)Bx

19. Miracrown

yes

20. Miracrown

the lambda is moved in front because it's a constant, so the matrix laws don't bother it

21. Miracrown

Now, remember how this started with saying "y = Bx"? So now on the right side, replace the Bx with y

22. UsukiDoll

BAy=(lambda)y

23. Miracrown

Yes :)

24. Miracrown

So by definition, lambda is an eigenvector of BA with eigenvector y

25. UsukiDoll

:O neat... it's like we started from the left side of the equation and then use theorems and definitions to try and get to the right side of the equation.

26. Miracrown

It kinda relies on making that correct substitution at the beginning. A lot of these proofs are fairly simple once you get them

27. UsukiDoll

somehow I understand doing proofs more in small groups rather than by myself T_____T!!!!!

28. Miracrown

Yep. There is actually a different way to prove it using the definition 0=(A - lambda I) I can quickly show you if you'd like

29. UsukiDoll

except Elementary Number Theory... I can do the proofs on my own because the theorems are easy to understand

30. Miracrown

You're not wrong

31. Empty

Yeah the substitution Bx=y was exactly what I was looking for, fun. =P I wouldn't mind watching or playing around with whatever you two want to continue showing or expanding upon.

32. UsukiDoll

<--- still a noob at proofs, but at least tries X_x

33. Miracrown

LOL usuki, you're more clever than me, I assure you! (;

34. Empty

Idk, Mira you're pretty clever, also nice to see you back, it's been a while. =P

35. Miracrown

It sure has, bud! Anyhoo another way using the formula for solving for eigenvalue rather than the definition of the eigenvector is:|dw:1434865293339:dw| That's the equation you solve to get eigenvalues. So now multiple by (AB) inverse on the right

36. Miracrown

|dw:1434865385356:dw| So now with the property of inverses from the very beginning, if we open the bracket, they will switch places

37. Miracrown

and I can be ditched because it's just the identity matrix |dw:1434865599306:dw| So what we need at the end is 0 = BA - lambda I

38. Miracrown

So here, because we have the B and the A inverses written without any other vectors what you want to do is multiply by B on the LEFT and by A on the RIGHT |dw:1434865665195:dw|

39. Miracrown

|dw:1434865717135:dw|

40. Miracrown

This looks better when it's done in several steps, but writing on the board is a little annoying..=/. So yes, the inverses go away and you are left with the BA But the trick is to get the order right :)