# Linear Algebra: Characteristic Equation Factoring

• henry3369
In summary, the eigenvalues of the homework equation are complex and it factors into three linear terms.
henry3369

## Homework Statement

If I have the characteristic equation:
3 + 3λ2 + 9λ + 5
And I'm told that one of its eigenvalues is -1.
How do I find the rest of the eigenvalues?

3 + 3λ2 + 9λ + 5

## The Attempt at a Solution

The furthest I can get is:
3 + 3λ2 + 9λ + 5 = (λ + 1) x (some factor of the original characteristic equation)

I can't seem to figure out what "some factor of the original characteristic equation" is. I tried it by inspection and I don't see how (λ + 1) helps. Is there some method that I'm missing to find that part?

henry3369 said:

## Homework Statement

If I have the characteristic equation:
3 + 3λ2 + 9λ + 5
And I'm told that one of its eigenvalues is -1.
How do I find the rest of the eigenvalues?

3 + 3λ2 + 9λ + 5

## The Attempt at a Solution

The furthest I can get is:
3 + 3λ2 + 9λ + 5 = (λ + 1) x (some factor of the original characteristic equation)

I can't seem to figure out what "some factor of the original characteristic equation" is. I tried it by inspection and I don't see how (λ + 1) helps. Is there some method that I'm missing to find that part?

Apply polynomial long division http://en.wikipedia.org/wiki/Polynomial_long_division.

STEMucator
First, you have not written an equation, but a cubic expression. Probably an oversight on your part.

Your expression contains $- \lambda^3$. So it must arise from an expression of the form $- (\lambda -a)(\lambda -b)(\lambda -c)$. You can get rid of the negative in the characteristic equation by simple algebra. You say correctly that one factor of the remaining cubic must be $(\lambda - (-1) )$ or $(\lambda + 1)$. A simple long division of the given expression (with the signs changed) will leave you with a quadratic expression. You can then use the standard quadratic formula to factorise the quadratic.

MarcusAgrippa said:
Your expression contains $- \lambda^3$. So it must arise from an expression of the form $- (\lambda -a)(\lambda -b)(\lambda -c)$.

It's not always the case a cubic polynomial will factor into that form. In this problem you have three linear factors, which is nice. In general though, there could be an irreducible quadratic generated by the long division. A more general form would be:

$$ax^3 + bx^2 + cx + d = (x \pm e)(fx^2 \pm gx \pm h^2)$$

Where the cubic is factored into the product of a linear factor and a quadratic factor. The quadratic factor may or may not be reducible depending on where you are.

Also see: http://mathworld.wolfram.com/CubicFormula.html

Last edited:
Zondrina said:
It's not always the case a cubic polynomial will factor into that form. In this problem you have three linear factors, which is nice. In general though, there could be an irreducible quadratic generated by the long division.
The characteristic equation can have complex roots: the eigenvalues can be complex. And with them, it factors into three linear factors.

Last edited:
ehild said:
The characteristic equation can have complex roots: the eigenvalues can be complex. And with them, it factors into three linear factors.

I completely ignored ##\mathbb{C}## there. My bad.

While it is true that any cubic polynomial can be factored to three linear terms, I think I would first start looking for a quadratic factor- call it $a\lambda^2+ b\lambda+ c$. Then we must have $(\lambda+ 1)(a\lambda^2+ b\lambda+ c)= (a\lambda^3+ b\lambda^2+ c\lambda)+ (a\lambda^2+ b\lambda+ c)= a\lambda^3+ (a+ b)\lambda^2+ (b+ c)\lambda+ c= -\lambda^3+ 3\lambda^2+ 9\lambda+ 5$
so that we must have $a= -1$, $a+ b= 3$, $b+ c= 9$, and $c= 5$

That is, of course, 4 equations for 3 unknowns. It is because we know that x+ 1 is a factor that we known these equations are consistent.

## 1. What is a characteristic equation in linear algebra?

A characteristic equation in linear algebra is an equation that describes the behavior of a given linear transformation on a vector space. It is used to find the eigenvalues and eigenvectors of a matrix, which are important in understanding the behavior of a system.

## 2. How is the characteristic equation factored?

The characteristic equation is factored by first finding the roots of the equation. These roots are the eigenvalues of the matrix. The factors of the characteristic equation will then be the corresponding eigenvectors.

## 3. Why is factoring the characteristic equation important?

Factoring the characteristic equation allows us to find the eigenvalues and eigenvectors of a matrix, which are crucial for understanding the behavior of a system. They also provide important information about the stability and dynamics of a system.

## 4. How do you use the characteristic equation to solve a system of linear equations?

The characteristic equation can be used to find the eigenvalues and eigenvectors of a matrix. These eigenvectors can then be used as a basis for the solutions to a system of linear equations. The corresponding eigenvalues can also provide information about the stability and behavior of the system.

## 5. Can the characteristic equation be factored for any matrix?

Yes, the characteristic equation can be factored for any square matrix. However, the difficulty of factoring the equation may vary depending on the complexity of the matrix.

Replies
2
Views
581
Replies
5
Views
13K
Replies
4
Views
1K
Replies
5
Views
1K
Replies
19
Views
3K
Replies
6
Views
2K
Replies
14
Views
771
Replies
12
Views
2K
Replies
3
Views
2K
Replies
14
Views
2K