Finding eigenvalues of a 3x3 matrix

[hahaha158]

Homework Statement

Find the eigenvalues

| 1 2 -1|
| -5 7 -5 |
| -9 8 -7|

Homework Equations

The Attempt at a Solution

I know that i need to add a -λ to every term in the trace so my matrix becomes

| 1-λ 2 -1|
| -5 7-λ -5|
| -9 8 -7-λ|

Then i need to take the determinant and find the values of λ when det=0

However i tried to do this without doing any shortcuts and ended up with an equation in terms of λ^3,λ^2,λ, and constants.

I am not sure how i can solve this, and i feel like there should be a shorter way to simplify the matrix at the start to make taking the determinant easier, but I'm not sure since there are no 0's i can use to simplify the determinant. Can anyone please help?

Thanks

 

[jbunniii]

The Attempt at a Solution

I know that i need to add a -λ to every term in the trace so my matrix becomes

| 1-λ 2 -1|
| -5 7-λ -5|
| -9 8 -7-λ|

Then i need to take the determinant and find the values of λ when det=0

However i tried to do this without doing any shortcuts and ended up with an equation in terms of λ^3,λ^2,λ, and constants.

I am not sure how i can solve this, and i feel like there should be a shorter way to simplify the matrix at the start to make taking the determinant easier, but I'm not sure since there are no 0's i can use to simplify the determinant. Can anyone please help?

Unfortunately there's no shortcut in general. The characteristic polynomial of an $n \times n$ matrix is $n$th order, so for $n = 3$, you have to solve a cubic equation. There are formulas for the roots of a cubic or quartic ($n = 4$) equation, but they are nasty. (It's even worse if $n \geq 5$: such equations are generally not solvable by radicals!) However, if some of the eigenvalues happen to be rational, you are in luck. Try using the rational root theorem to check whether this is the case.

 

[hahaha158]

Unfortunately there's no shortcut in general. The characteristic polynomial of an $n \times n$ matrix is $n$th order, so for $n = 3$, you have to solve a cubic equation. There are formulas for the roots of a cubic or quartic ($n = 4$) equation, but they are nasty. (It's even worse if $n \geq 5$: such equations are generally not solvable by radicals!) However, if some of the eigenvalues happen to be rational, you are in luck. Try using the rational root theorem to check whether this is the case.

i tried doing it the long way and ended up with the equation

-λ^3+λ^2+8λ-12=0

I tried with wolfram and it told me that this gives me the roots -3 and 2, which are correct. However i am not sure how to solve this equation, do you think you could explain?

 

[jbunniii]

i tried doing it the long way and ended up with the equation

-λ^3+λ^2+8λ-12=0

I tried with wolfram and it told me that this gives me the roots -3 and 2, which are correct. However i am not sure how to solve this equation, do you think you could explain?

Yes, try the rational root theorem as I suggested. Your equation is
$$-\lambda^3 + \lambda^2 + 8\lambda - 12 = 0$$
The coefficients are all integers, so the rational root theorem applies. If this polynomial has a rational root, i.e., the equation has a solution of the form $m/n$ where $m$ and $n$ are integers, then $m$ must be a divisor of $-12$, and $n$ must be a divisor of $-1$. Therefore, the following are the only possibilities: $n = \pm 1$, $m = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$. In other words, the only possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$. Now just plug in each of these and see if it is a root.

In this case we get lucky and find that $-3$ and $2$ are roots. But every cubic polynomial has three roots (counting repetitions), so we need to make sure there are no others. To do this, note that $\lambda + 3$ and $\lambda - 2$ must be factors of $-\lambda^3 + \lambda^2 + 8\lambda - 12$, so using polynomial division if necessary, we can find the third factor:
$$-\lambda^3 + \lambda^2 + 8\lambda - 12 = -(\lambda+3)(\lambda-2)(\lambda-2)$$
which shows that $\lambda = -3$ and $\lambda = 2$ are the only solutions, with the latter being a repeated root.