Find the characteristic polynomial

[Niles]

Homework Statement

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?

The Attempt at a Solution

I know I can find the characteristic polynomial, but I thought that there is perhaps an easier way?

 

[morphism]

Look at Ba?

 

[Niles]

Ahh yes, thanks :-)

 

[HallsofIvy]

Write out the defining equation for an eigenvalue, Bx= \lambdax, 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"!)

 

[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.