# Find the eigenvalues of a matrix

1. Jun 16, 2007

1. The problem statement, all variables and given/known data
i'm trying to find the eigenvalues of a matrix and i have the solution but i don't understand how it gets from the step 1 to step 2??? could someone please explain.

let # = lambda

Step 1: (1-#)[(2-#)(-1-#)+1]+[3(-1-#)+2]+4[3-2(2-#)] = 0
Step 2: (1-#)(#+2)(#-3) = 0

2. Relevant equations

3. The attempt at a solution
i jst kept expanding the other half of the equation with my final result bein
useless

(1-#)(#^2 -#-1)+(5#-5) = 0

2. Jun 16, 2007

### Dick

If you expand the first equation completely you get -L^3+2*L^2+5*L-6=0 (L=lambda). Now you just want to factor this polynomial. Probably the easiest way to do this is to look for zeros of the polynomial. Eg L=1 is a zero so L-1 is a root (we also use the clue that if p/q is a rational root the p divides the constant polynomial coefficient and q divides the highest power coefficient). Once you found one root divide it out and try to factor what's left. It's pretty standard stuff.

3. Jun 29, 2007

### daniel_i_l

If the coefficient of the variable with the highest power is one:
$$x^n +a_{n-1} x^{n-1} + ... + a_1 x + a_0$$
then if the polynomial has a integer root then the last coefficient (a_0) is a multiple of the root. For example, given the polynomial:
$$x^2 + 5x + 6$$
the possible integer roots are +/-1 , +/-2, +/-3 and +/-6
a quick check shows that 2 and 3 are roots so the polynomial is equal to
$$(x-3)(x-2)$$
This is especially useful for finding e-values from the e-polynomial since in most problems the e-values are integers and the coefficient of x^n is always 1.