# Find the characteristic polynomial

1. Jan 20, 2008

### Niles

1. The problem statement, all variables and given/known data
If I have a n x n matrix B, and I must show that a vector a is an eigenvector for the matrix B and I have to find the corresponding eigenvalue, what is the easiest way of doing this?

3. The attempt at a solution
I know I can find the characteristic polynomial, but I thought that there is perhaps an easier way?

2. Jan 20, 2008

### morphism

Look at Ba?

3. Jan 20, 2008

### Niles

Ahh yes, thanks :-)

4. Jan 20, 2008

### HallsofIvy

Staff Emeritus
Write out the defining equation for an eigenvalue, Bx= $\lambda$x, of course. You know x= a and you know B. Go ahead and do the multiplication on the left and compare it to the right side!

For example if you know that
$$A= \left[\begin{array}{cc} -3 & 15 \\ -2 & 8 \end{array}\right]$$
and
$$a= \left[\begin{array}{c} 5 \\ 2\end{array}\right]$$
is an eigenvector of B, and want to find the corresponding eigenvalue, $\lambda$

$$\left[\begin{array}{cc} -3 & 15 \\ -2 & 8 \end{array}\right]\left[\begin{array}{c} 5 \\ 2\end{array}\right]= \lambda\left[\begin{array}{c} 5 \\ 2\end{array}\right]$$

$$\left[\begin{array}{c} 15 \\ 6\end{array}\right]= \left[\begin{array}{c} \lambda 5 \\ \lambda 2\end{array}\right]$$
So we must have $15= 5\lambda$ and $6= 2\lambda$. Obviously, $\lambda= 3$ satisfies both. (If the same $\lambda$ had not satisfied both, then the given vector was not an eigenvector.)

(Edit: Or I could have just said "look at Ba"!)

5. Jan 20, 2008

### Niles

Nono, because in the example with "Ba" the eigenvalue was 0, so I thought the method was different from the one you showed in your example.

Thanks to both of you.