Finding Eigenvalues: List of possible solutions for Lambda

[zak100]

Homework Statement

I got a solution for finding eigen values. It evaluates to:

(Lambda)^3 -12(lambda) -16=0, Then it that the list of possible integer solutions is:

+-1, +-2, +-4, _-8, +-16 (i.e. plus minus 1, plus minus 2 and so on).

I can't understand, why he says list of possible solution is this, when only lambda = 4 satisfies the eq..Zulfi.

Homework Equations

(Lambda)^3 -12(lambda) -16=0

The Attempt at a Solution

Only lamda =4 satisfies the eq while other values do not.
The solution provided by website isttached/.

Zulfi

 

[fresh_42]

If $\lambda_0$ is a solution to $\lambda^3-12\lambda-16=0$ then $(\lambda-\lambda_0)$ is a divisor of it. This means especially that $\lambda_0$ divides $16$. If there are integer solutions, those numbers are simply the list of possible divisors of $16$, which can be tested, in order to reduce the degree of the equation. How did you find that $\lambda_0=4$ is a solution?

 

[zak100]

Hi,
Thanks for your response.
4^3 -12*4 -16 =0
64-48 -16 =0

It evaluates to zero.

Other values do not.

Zulfi.

 

[Ray Vickson]

Homework Statement

I got a solution for finding eigen values. It evaluates to:

(Lambda)^3 -12(lambda) -16=0, Then it that the list of possible integer solutions is:

+-1, +-2, +-4, _-8, +-16 (i.e. plus minus 1, plus minus 2 and so on).

I can't understand, why he says list of possible solution is this, when only lambda = 4 satisfies the eq..Zulfi.

Homework Equations

(Lambda)^3 -12(lambda) -16=0

The Attempt at a Solution

Only lamda =4 satisfies the eq while other values do not.
The solution provided by website isttached/.

Zulfi

He is claiming that at least one root is to be found in the set {-1, 1, -2, 2, -4, 4, -8, 8, -16, 16}. That means that you can find a root by checking at most 10 values of $\lambda.$ Some of the values in the set are not roots of the equation, but at least one of them is a root, or so he says.

He got the set by looking for rational roots of the equation, using the so-called Rational Root Theorem; see, eg., https://en.wikipedia.org/wiki/Rational_root_theorem
or https://trans4mind.com/personal_development/mathematics/polynomials/rationalRootTheorem.htm

Note: this method finds rational roots if they exist. Therefore, if none of the numbers in the list are actually roots, it follows that the polynomial has only irrational roots.

 

[SammyS]

Homework Statement

I got a solution for finding eigen values. It evaluates to

(Lambda)^3 -12(lambda) -16=0, Then it that the list of possible integer solutions is:

+-1, +-2, +-4, _-8, +-16 (i.e. plus minus 1, plus minus 2 and so on).

I can't understand, why he says list of possible solution is this, when only lambda = 4 satisfies the eq..

Zulfi.

Homework Equations

(Lambda)^3 -12(lambda) -16=0

The Attempt at a Solution

Only lamda =4 satisfies the eq while other values do not.
The solution provided by website isttached/.

Zulfi

One other number satisfies this. It is in your list.

 

[zak100]

Hi,
Ray Vickson, thanks for providing the link. God bless you.

I would try.

Zulfi.

 

[zak100]

Hi,
Using p/q I am able to find possible solutions for λ
λ^3 -12λ -16 = 0

P|16 = 1, 2, 4, 8, 16

Q|1 = 1

Now P/q = +-1.2^4/1 or +- 4 * 4/1 or +- 1.2.8/1 or +- 1.16/1

So possible solutions for λ = +-1, +-2, +-4, +-8 and +-16 but only λ=4 satisfies the eq.Zulfi.

 

[fresh_42]

Hi,
Using p/q I am able to find possible solutions for λ
λ^3 -12λ -16 = 0

P|16 = 1, 2, 4, 8, 16

Q|1 = 1

Now P/q = +-1.2^4/1 or +- 4 * 4/1 or +- 1.2.8/1 or +- 1.16/1

So possible solutions for λ = +-1, +-2, +-4, +-8 and +-16 but only λ=4 satisfies the eq.Zulfi.

Nope.

 

[zak100]

Hi,
Can you please tell me what's wrong with this? I have followed the tutorial.

Zulfi.

 

[fresh_42]

Hi,
Can you please tell me what's wrong with this? I have followed the tutorial.

Zulfi.

No, you haven't.

One other number satisfies this. It is in your list.

... is still true. You could either try to find it by trying or you calculate $f(\lambda):=(\lambda^3-12\lambda -16):(\lambda -4)$ and solve the quadratic equation $f(\lambda)=0$.

 

[zak100]

Hi,
I got it. Its -2. instead of lambda I am using x.

X^3 -12x -16 =0

X =1: 1 -12 -16 = 0 (false)

X=-1: -1 -12 -16 =0 (false)

X = 2: 8-16-16=0 (false)

X= -2: -8 -+24 -16 =0 (true)

Zulfi.

 

[fresh_42]

Hi,
I got it. Its -2. instead of lambda I am using x.

X^3 -12x -16 =0

X =1: 1 -12 -16 = 0 (false)

X=-1: -1 -12 -16 =0 (false)

X = 2: 8-16-16=0 (false)

X= -2: -8 -+24 -16 =0 (true)

Zulfi.

Now you have two zeroes, but three are possible. What can you say about the third?

 

[zak100]

Hi,
I can't see the 3rd one.

X = 4: 64 -48 -16 =0(true)

X= -4: -64 +48 -16 =0 (false)

X=8: Not possible

X=-8: Not possible

X=16: Not possible

X=-16: not possible

Zulf.

 

[fresh_42]

But couldn't the third be a zero which is not rational? That's where the long division is more useful than testing zeroes. Once you have a zero, $x=4$, we can divide the polynomial by $x-4$ and solve the rest. It is done the same way as a long division by numbers. Here's an example how it's done: https://www.physicsforums.com/threa...r-a-polynomial-over-z-z3.889140/#post-5595083

The reason is, that every polynomial $p(x)=x^n+a_1x^{n-1}+\ldots +a_1x+a_0$ can be written as $p(x)=(x-z_1)\cdot \ldots \cdot (x-z_n)$. The $z_i$ are usually complex numbers, but not in our case.

 

[SammyS]

Hi,
I can't see the 3rd one.
...

Zulf.

In the following thread of yours, you have the same polynomial,

Find the factors of an equation

I wrote the following in Post #4 of that thread:

This looks to be basically same problem as in your other recent post.

You can divide $\ x^3 -12x -16 \$ by $\ x−4 \$ using polynomial long division or its short cut, synthetic division.​

.

 

[zak100]

Yes. You are right SammyS. Ray Vickson provided me the link for long division again discussed by fresh_42. Thanks everybody. God bless you people.If I put zero, I would get:
-16 = 0

How it can be the root of cubic eq?

Any way you are right. I would explore the long division aspect tomorrow.

Zulfi.

 

[SammyS]

Yes. You are right SammyS. Ray Vickson provided me the link for long division again discussed by fresh_42. Thanks everybody. God bless you people.If I put zero, I would get:
-16 = 0

How it can be the root of cubic eq?

Any way you are right. I would explore the long division aspect tomorrow.

Zulfi.

What was your result for dividing $\ x^3 -12x -16 \$ by $\ x-4\$ ?

 

[Mark44]

X= -2: -8 -+24 -16 =0 (true)

This is actually incorrect, but likely is a typo.
If x = -2, the value of the expression $x^3 - 12x - 16$ is ##(-2)^3 - 12(-2) - 16 = -8 + 24 - 16 = 0.
You wrote -+24, which is the same as -24.

 

[zak100]

Hi Sammys, fresh_42,and Ray Vickson

I have the eq :

X^3 -12x -16

X^2 term is missing. So I would add 0x^2 in the above:

X^3 +0x^2-12x -16 divided by the obtained factor x-4

x-4)x^3 +0x^2-12x -16(x^2+4x +4
x^3-4x^2
- +
------------------
4x^2-12x
4x^2-16x
- +
------------------------
4x -16
4x-16
- +
____________________
zero
Ans: x^2+4x +4

root = -2

Thanks.
Zulfi.

 

[ehild]

Hi Sammys, fresh_42,and Ray Vickson

I have the eq :

X^3 -12x -16

X^2 term is missing. So I would add 0x^2 in the above:

X^3 +0x^2-12x -16 divided by the obtained factor x-4

x-4)x^3 +0x^2-12x -16(x^2+4x +4
x^3-4x^2
- +
------------------
4x^2-12x
4x^2-16x
- +
------------------------
4x -16
4x-16
- +
____________________
zero
Ans: x^2+4x +4

root = -2

x=-2 is double root of the equation x^2+4x +4=0. It can happen that some of the eigenvalues are equal.
It means that the original expression can be factorized as x^3 -12x -16=(x+2)[x+2)(x-4).

 

[zak100]

Hi Mark,
Now I have to find out the eigen vectors. It says:

Once the eigenvalues of a matrix (A) have been found, we can find the eigenvectors
by Gaussian Elimination.
• STEP 1: For each eigenvalue λ, we have
(A − λI)x = 0,
where x is the eigenvector associated with eigenvalue .
• STEP 2: Find x by Gaussian elimination. That is, convert the augmented matrix
(A − λI)X = 0
to row echelon form, and solve the resulting linear system by back substitution.

Right side is still zero. Do we have to convert into echelon form by using augmented matrix.

Zufi.

 

[zak100]

Hi Mark,
Case 1: _ = 4

We must find vectors x which satisfy (A − λI)x = 0.
Now they are using the augmented matrix (attached). Is there any way to avoid it

Zulfi.

 

[Mark44]

Now they are using the augmented matrix (attached). Is there any way to avoid it

Augmented matrices are not needed. Notice that after it says "Construct the augmented matrix ..." every augmented matrix has a fourth column of all zeroes. None of the row operations can possibly change this column, so what's the sense in keeping it? If you start with the un-augmented matrix in this section, doing the same row operations as shown in the screen shot will produce a 3 x 3 matrix that is the same as the first three columns in the last matrix they show.
The augmented matrix they show represents the matrix equation $(A - 4I)\vec x = \vec 0$. Since the right side is the zero vector, you don't need to use an augmented matrix.

 

[Mark44]

@zak100, if you have more questions about eigenvalues and eigenvectors, please post them in the Calculus & Beyond section, not in the Precalculus section. Eigenvectors and eigenvalues are typically well beyond precalculus topics.