Linear Algebra Check...

demon8991

demon8991

demon8991

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Just wanted to check ive worked out the eigenvectors and eigenvalues right for this.

A=
(0.75 0.55)
(0.35 0.55)

I got my eigenvalues as lambda 1 = 0.75 and lambda 2 = 0.55

and my eigenvectors for lambda 1 = (15,7) and lambda 2 = (11,3)

Is this right?

If so how on earth do i find a formula for A^n for all n belonging to the natural numbers?
 
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demon8991 said:
Just wanted to check ive worked out the eigenvectors and eigenvalues right for this.

A=
(0.75 0.55)
(0.35 0.55)

I got my eigenvalues as lambda 1 = 0.75 and lambda 2 = 0.55

and my eigenvectors for lambda 1 = (15,7) and lambda 2 = (11,3)

Is this right?
Click to expand...
I get different eigenvalues. Want to show your working?

If so how on earth do i find a formula for A^n for all n belonging to the natural numbers?
Click to expand...
With correct eigenvalues/eigenvectors, you can find a matrix P with inverse Q s.t. PAQ = D, where D is a diagonal matrix.
Then (PAQ)^n = PAQPAQ...PAQ = PA^nQ (the QPs in the middle all cancel out) so A^n = QD^nP. And D^n is easy to work out. This should all be in your notes...
 
DaveF said:
I get different eigenvalues. Want to show your working?

With correct eigenvalues/eigenvectors, you can find a matrix P with inverse Q s.t. PAQ = D, where D is a diagonal matrix.
Then (PAQ)^n = PAQPAQ...PAQ = PA^nQ (the QPs in the middle all cancel out) so A^n = QD^nP. And D^n is easy to work out. This should all be in your notes...
Click to expand...

Excuse my appauling hand writing

 
Check the line where you calculate the quadratic equation - you sure you haven't forgotten to subtract off .55 x .35?
 
demon8991 said:
damn should be 0.22 as the constant then?
Click to expand...
That's what I got, yes.

I'm a bit puzzled to how you managed to calculate eigenvectors for incorrect eigenvalues - at that stage you should have noticed something was up!
 
DaveF said:
That's what I got, yes.

I'm a bit puzzled to how you managed to calculate eigenvectors for incorrect eigenvalues - at that stage you should have noticed something was up!
Click to expand...

Ok right I got my eigenvalues as lambda 1 = 1.1 and lambda 2 = 0.2

Im struggling to grasp the conecpt of eigenvectors as my notes seem to just jump a stage out and i cant for the life of me see what the connection is.

Basically for my lambda 1 value i have the equations.

(0.75 - lambda 1)x + 0.55y = -0.35x + 0.55y = 0
0.35x + (0.55 - lambda 1)y = 0.35x + -0.55 = 0

lambda 2:

(0.75 - lambda 1)x + 0.55y = 0.55x + 0.55y = 0
0.35x + (0.55 - lambda 1)y = 0.35x + 0.35y = 0
 
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demon8991 said:
Basically for my lambda 1 value i have the equations.

(0.75 - lambda 1)x + 0.55y = -0.35x + 0.55y = 0
0.35x + (0.55 - lambda 1)y = 0.35x + -0.55 = 0

lambda 2:

(0.75 - lambda 1)x + 0.55y = 0.55x + 0.55y = 0
0.35x + (0.55 - lambda 1)y = 0.35x + 0.35 = 0
Click to expand...
OK, the first thing to get your head round is that if you've got the value of lambda right, you won't be able to solve the equations. Put like that, it sounds rather silly, but the point is that if v is an eigenvector, then so is 2v, or indeed any other multiple of v, so you won't get a unique solution.

In this case, with only 2 dimensions, you should notice that for a given value of lambda, your equations for finding x and y are essentially the same - one is just a multiple of the other. So you've only got one equation and 2 unknowns, which isn't enough to find x,y uniquely.

Probably the simplest approach is to set x arbitrarily to 1. Then for lambda1, you have -.35x + .55y = 0, so y = .35/.55 = 7/11. So (1,7/11) is an eigenvector. As any multiple is also an eigenvector, (11,7) is a slightly "nicer" eigenvector to pick.

For lambda2 we have .55x+.55y = 0, so if x=1, y=-1 and (1,-1) is an eigenvector.

N.B. setting x=1 doesn't always work - you might be unlucky and have the eigenvector look like (0,1) which isn't going to have x=1 however you try to scale it. But if this happens it will be pretty obvious from the equations that x must be zero.
 
Ahhh i see, omg its soo simple when put like that!

Ive been racking my brains most of the day with this question!

Thanks!

So i end up with eigenvalues of 1.1 and 0.2

and eigenvectors of simplest form (11,7) and (1,-1)
 
demon8991 said:
Ahhh i see, omg its soo simple when put like that!

Ive been racking my brains most of the day with this question!

Thanks!

So i end up with eigenvalues of 1.1 and 0.2

and eigenvectors of simplest form (11,7) and (1,-1)
Click to expand...
I think so, yes.
 
DaveF said:
With correct eigenvalues/eigenvectors, you can find a matrix P with inverse Q s.t. PAQ = D, where D is a diagonal matrix.
Then (PAQ)^n = PAQPAQ...PAQ = PA^nQ (the QPs in the middle all cancel out) so A^n = QD^nP. And D^n is easy to work out. This should all be in your notes...
Click to expand...

These actually are my worst set of notes in linear algebra, i only have a few pages on it all.

I dont see what you set P equal to? how do you just find a random matrix?
 
demon8991 said:
These actually are my worst set of notes in linear algebra, i only have a few pages on it all.

I dont see what you set P equal to? how do you just find a random matrix?
Click to expand...
Sorry, I wasn't going into detail as I assumed (hoped) this would be in your notes and you just needed a little bit of guidance in the right direction.

You need to set Q to be the matrix formed out of your eigenvectors:

(11 1)
( 7 -1)

And P to be the inverse of Q, which I think is:

(1/18 1/18)
(7/18 -11/18)

And D is the diagonal matrix of eigenvalues:

(1.1 0 )
( 0 0.2)

Then assuming I haven't stuffed up anywhere, QDP should equal A.

This is all very hard to explain without proper maths symbols and diagrams, which is why I suggested you check your notes, which should do a better job than I can, but from your comments apparently not. :(

There's also a wiki page: http://en.wikipedia.org/wiki/Diagonalizable_matrix that may help, or if you want to google for more info, the keywords are "diagonalizable matrix".

Hope that's of some help.
 
Wicked thanks, this is really helping, ive gone off and done a a few different examples so im starting to get the hang of it.

The only thing that is bothering me is surely if A = B^-1 x D x B then A = D becuase the the B^-1 and B when multiplied together give the Identity matrix.
 
demon8991 said:
Wicked thanks, this is really helping, ive gone off and done a a few different examples so im starting to get the hang of it.

The only thing that is bothering me is surely if A = B^-1 x D x B then A = D becuase the the B^-1 and B when multiplied together give the Identity matrix.
Click to expand...
The reason B and B^-1 don't cancel each other out is because matrix multipllication isn't commutative (in other words AB doesn't always equal BA). So you can't rearrange B^-1 x D x B as D x B^-1 x B. (If you could, then you'd be right in your concern!).
 
Ahh course basic stuff, sorry about that.

Basically this is the question im trying to plough through with your help. It basically is the only one i could find that pulls on most of the stuff ive done with matrix algebra.



This is the equation ive got for A^n

-1/18 x ((-1 x 12.1^n) + (-7 x 0.2^n) (-1 x 12.1^n) + (11 x 0.2^n))
...........((-1 x 7.7^n) + (-7 x -0.2^n) (-1 x 7.7^n) + (11 x -0.2^n))

Im hoping that is right but my problem is for c) is actually finding out what the bacteria count is for day 10 as the original equation and this one are really confusing me, because i cant see where my x and y values are going to come from.

Thanks for all this mate really apreciate it.
 
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