1. ## characteristic polynomial

What's the easiest way to calculate the characteristic polynomial of a 3x3 matrix?

I'm aware this is the formula: det(A-lambda(I)), dont really know how to apply it though.

Say matrix was: 1 2 1
1 3 1
1 2 1

i dont want the solution, just the actual method.. or if someone could run through step by step it would be great

2. Suppose that

$\displaystyle A\mathbf{x} = \lambda \mathbf{x}$

for some matrix $\displaystyle A$, some vector $\displaystyle \mathbf{x}$ and some scalar $\displaystyle \lambda$.

Then $\displaystyle \mathbf{x}$ is called an eigenvector and $\displaystyle \lambda$ is the corresponding Eigenvalue.

The thing is, how do you find $\displaystyle \lambda$ and $\displaystyle \mathbf{x}$?

Notice if $\displaystyle A\mathbf{x} = \lambda \mathbf{x}$

It might make sense to move everything to one side.

Unfortunately $\displaystyle A$ is a matrix and $\displaystyle \lambda$ is a scalar. So you instead write

$\displaystyle A\mathbf{x} = \lambda I \mathbf{x}$.

Now that the matrices have the same dimensions, you can subtract them

$\displaystyle A\mathbf{x} - \lambda I \mathbf{x} = \mathbf{0}$

$\displaystyle (A - \lambda I)\mathbf{x} = \mathbf{0}$.

Clearly, this has a trivial solution if $\displaystyle \mathbf{x} = \left[\begin{matrix}0 \\ 0 \\ \vdots \\ 0\end{matrix}\right]$.

For a nontrivial solution

$\displaystyle A - \lambda I = \mathbf{0}$.

Since this is a zero matrix, it is singular.

So $\displaystyle |A - \lambda I| = 0$.

We can use this information to find $\displaystyle \lambda$.

$\displaystyle A = \left[\begin{matrix}1 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 1\end{matrix}\right]$.

You need to work out

$\displaystyle |A - \lambda I|$.

$\displaystyle A - \lambda I = \left[\begin{matrix}1 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 1\end{matrix}\right] - \left[\begin{matrix}\lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda\end{matrix}\right]$

$\displaystyle = \left[\begin{matrix}1 - \lambda & 2 & 1 \\ 1 & 3 - \lambda & 1 \\ 1 & 2 & 1 - \lambda\end{matrix}\right]$

Therefore $\displaystyle |A - \lambda I| = \left|\begin{matrix}1 - \lambda & 2 & 1 \\ 1 & 3 - \lambda & 1 \\ 1 & 2 & 1 - \lambda\end{matrix}\right|$

$\displaystyle = (1 - \lambda)\left|\begin{matrix}3 - \lambda & 1 \\ 2 & 1 - \lambda\end{matrix}\right| - 2\left|\begin{matrix}1 & 1 \\ 1 & 1 - \lambda\end{matrix}\right| + 1\left|\begin{matrix}1 & 3-\lambda \\ 1 & 2\end{matrix}\right|$

$\displaystyle = (1 - \lambda)[(3 - \lambda)(1 - \lambda) - 1\cdot 2] - 2[1(1 - \lambda) - 1\cdot 1] + 1[1 \cdot 2 - (3 - \lambda)1]$

$\displaystyle = (1 - \lambda)(3 - 3\lambda - \lambda + \lambda^2 - 2) - 2(1 - \lambda - 1) + 2 - 3 + \lambda$

$\displaystyle = (1 - \lambda)(1 - 4\lambda + \lambda^2) + 2\lambda -1 + \lambda$

$\displaystyle = 1 - 4\lambda + \lambda^2 - \lambda + 4\lambda^2 - \lambda^3 + 3\lambda - 1$

$\displaystyle = -\lambda^3 + 5\lambda^2 - 2\lambda$.

Since this determinant is zero...

$\displaystyle -\lambda^3 + 5\lambda^2 - 2\lambda = 0$

$\displaystyle -\lambda(\lambda^2 - 5\lambda + 2) = 0$

You should be able to solve for $\displaystyle \lambda$ now.

3. thanks so much, especially for the theory and worked through problem