characteristic polynomial

**piglet** · April 6th 2010, 02:46 AM

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

**Prove It** · April 6th 2010, 03:40 AM

Suppose that

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

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

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

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

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

It might make sense to move everything to one side.

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

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

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

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

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

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

For a nontrivial solution

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

Since this is a zero matrix, it is singular.

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

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

In your case:

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

You need to work out

$|A - \lambda I|$ .

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

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

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

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

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

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

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

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

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

Since this determinant is zero...

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

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

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

**piglet** · April 6th 2010, 04:01 AM

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

Thread: characteristic polynomial

LinkBack

Thread Tools

Display

characteristic polynomial

Similar Math Help Forum Discussions

Characteristic Polynomial

Characteristic Polynomial

Any polynomial is the characteristic polynomial of some linear operator?

characteristic polynomial

Characteristic polynomial

Search Tags