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Nox_M 62
Postad: 15 apr 22:06 Redigerad: 15 apr 23:06

Linjär Algebra

(a)När jag försöker lösa 'det(A-λI)=0' så får jag '-x³+7x²+ax-15x+10-2a=0'. Hur kan jag lösa den när jag har fler okända variabler?

Finns det något annat sätt att svara på (a) än att först hitta eigenvärde och eigenvektorer?

(Ps: I am not sure if my Swedish is correct so if my question is confusing, please ask me!

To the mods: is it allowed to write in English instead of Swedish?)

Dracaena Online 4937 – Moderator
Postad: 16 apr 01:12 Redigerad: 17 apr 13:53

Although you could attempt to calculate the eigenvalues/vectors and check the condition A=PDP^(-1), it is not at all necessary.

A matrix is diagonalizable if and only if for each eigenvalue the dimension of the eigenspace is equal to the multiplicity of the eigenvalue.

That is, suppose we have a 2x2 square matrix. It should be obvious that if n denotes rows, then n=2. If we can find two (n distinct) distinct eigenvalues for this given 2x2 matrix then we can conclude that it is indeed diagonalizable without having to show that A=PDP^(-1).

As far as I know, writing your thread in English is not per se breaking the rules but keep in mind you'll be far less likely to recieve help. 

Nox_M 62
Postad: 16 apr 10:33

My main issue is how to find the eigenvalues. If I use det(A-λI)=0 then I get an equation that has too many unknowns. Is there an alternative method I can use to find the eigenvalues?

oneplusone2 551
Postad: 16 apr 13:12 Redigerad: 16 apr 13:15

There is a lucky root to your characteristic polynomial, x=2. You can then factor the polynomial using polynomial division.

Once you have done that you will be able to see that there are 3 distinct eigenvalues in all cases but one. With 3 distinct eigenvalues, diagonalizability follows from gm <= am with a minimum of 1 for gm. In the special case you can simply substitute that value for a into your original matrix and manually evaluate the situation.

I have ignored complex solutions here.

Nox_M 62
Postad: 16 apr 14:11

So I factored the polynomial and got:

(λ-2)(-λ²+5λ-5+a)=0

I don't know where to go from here. How do I factor (-λ²+5λ-5+a)?

oneplusone2 551
Postad: 16 apr 14:21

complete the square

Nox_M 62
Postad: 16 apr 14:38

I did that and I got 

λ = a + 3,75

So does that mean that A is diagonalizable for all real number values of a? is -3,75 included?

oneplusone2 551
Postad: 16 apr 14:38

write out the polynomial with only linear factors

Nox_M 62
Postad: 16 apr 14:54 Redigerad: 16 apr 16:51

I completed the square but I made a wrong assumption along the way

-λ² + 5λ - 5 + a=0

λ² - 5λ = a - 5

λ² - 5λ + 6,25 = a - 5 + 6,25

(λ - 2,5)(λ - 2,5) = a + 1,25

(λ - 2,5)² = a + 1,25

λ - 2,5 = ±√(a + 1,25)

λ  = ±√(a + 1,25) + 2,5

Have I done it correctly? and if so can I assume that A is diagonalizable for all numbers when a ≥1,25?

 

oneplusone2 551
Postad: 16 apr 15:01

it looks correct,

now write the polynomial like this -(x-a1)(x-a2)(x-a3)

Nox_M 62
Postad: 16 apr 15:05 Redigerad: 16 apr 15:07

(λ  + √(a + 1,25) - 2,5)(λ - √(a + 1,25) - 2,5)(λ - 2)= 0

So, I see that for any choice of a≥1,25 there are 3 distinct eigenvalues except in the case when a=1,25, there are two eigenvalues, λ=2 and λ=2,5 (multiplicity 2)

Am I on the right track?

oneplusone2 551
Postad: 16 apr 15:10

yes now u must evaluate the case a=1.25 manually by inserting it into your original matrix

Nox_M 62
Postad: 16 apr 15:22

Why do I have the evaluate the special case of a=1,25 manually?

Can I not assume that since a=1,25 gives me two eigenvalues one of which has multiplicity 2, it automatically satisfies the condition that there are 3 eigenvalues for the 3 x 3 matrix?

oneplusone2 551
Postad: 16 apr 15:23 Redigerad: 16 apr 15:24

u dont know what the geometirc multiplicity of the a=1.25 is. you need to determine which vectors are associated with the a=1.25 case

Nox_M 62
Postad: 16 apr 15:32

I will put everything in writing and post it and if you have time later you can give me feedback.

Thank you very much for all the help so far! : ) 

Nox_M 62
Postad: 16 apr 16:49

I skipped the calculation steps (polynomial division, completing the square, row reduction) so that the information on the page is readable. Is my answer correct?

oneplusone2 551
Postad: 16 apr 21:03

I get the impression that you need to have good look in your textbook reguarding diagonalization and geometric and algebraic multiplicity. 

For the case of a=-1.25 we have established that there are 2 eigenvalues, x=2 and 2.5. You still have to investigate what vectors belong to the x=2 eigenvalue.

Nox_M 62
Postad: 16 apr 21:29

Yeah, my knowledge of linear algebra is very superficial. I need to do a whole lot of revision.

There is one eigenvector that belongs to λ=2

I automatically assumed that since λ=2 has multiplicity 1, it could only give me one eigenvector. Was my assumption wrong?

Now that I found that there are only 2 eigenvectors when a=-1.25, does that change my answer?

Thank you for being so patient with me : )

oneplusone2 551
Postad: 16 apr 21:46

No your assumption is correct.
However my idea was to give you the follow up question: Can you diagonalize A using the eigenvectors you have discovered for a=-1.25 ?

Nox_M 62
Postad: 16 apr 21:57

I need 3 linearly independent eigenvectors to diagonalize a 3 x 3 matrix. In this case, I only have 2 eigenvectors so I would assume they are not enough to diagonalize A. Is my reasoning correct?

oneplusone2 551
Postad: 16 apr 22:00

That is correct. I think you have a ok overview of the assignment now. Make sure that you read some in your textbook as well.

Nox_M 62
Postad: 16 apr 22:02

Thank you very much for all the help : ) 

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