Thread: 3X3 Matrix, I can find the Characteristic polynomial but not the eigenvalues

1. 3X3 Matrix, I can find the Characteristic polynomial but not the eigenvalues

I need to learn to find eigenvalues and vectors.
I can find the Characteristic polynomial (and if i have the eigenvalues i can find the vectors)

However i cannot get from the Characteristic polynomial to the eigenvalues

for example if i have a 3x3 matrix

(1 0 4)
(0 2 0)
(3 1 -3)

The Characteristic polynomial is x^3 - 19x + 30 = 0 (x=lambda)

I know that the eigenvalues are Real eigenvalues:
{-5, 2, 3}

and then i can work out the eigenvectors.

But how do you get from this step (x^3 - 19x + 30 = 0) to the eigenvalues.

Thankyou!

2. u can use Horner's scheme (are you familiar with it) ? sorry I forgot do they acquainted you with that on your level of education?

or u can :

$x^3-19x+30=0$

$x^3-19x+30=x^3-9x -10x +30 = x(x^2-9) -10(x-3)= x(x^2-3^2) -10(x-3) =$

$= x(x-3)(x+3)-10(x-3) = (x-3) [x(x+3) -10]$

$= (x-3)(x^2+3x-10) = (x-3)(x-2)(x+5) =0$

so u have

$x-3=0 \Rightarrow x_1=3$

$x-2=0 \Rightarrow x_2=2$

$x+5=0 \Rightarrow x_3=-5$

3. Another thing you can do is use the "rational roots theorem": if $a_nx^2+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0$ is a polynomial equation with integer coefficients and $\frac{m}{n}$ is a rational root of the equation, then the denominator, n, must divide the leading coefficient, $a_n$, and the numerator, m must divide the constant term, $a_0$.

Here, your equation is $x^3- 19x+ 30= 0$, a polynomial equation with integer coefficients. The leading coefficient is 1 and the only integers that divide 1 are 1 and -1. Since the denominator of any rational root must be 1 or -1, the only rational roots must be integers. The constant term is 30 so any rational root must be a factor of 30. Since 30= 2(3)(5), the only possible rational roots are 1, -1, 2, -2, 3, -3, 5, -5, 6, -6, 10, -10, 15, -15, 30, and -30. Just put each of those into the equation to see whether or not they satisfy the equation and you will see that 2, 3, and -5 satisfy it.

Note- this does not guarentee that there are any rational roots to a polynomial equation but if there are, it will find them.

We should point out that, in general, finding eigenvalues is not at all an easy task!