# Math Help - Eigenvalues of a 3x3 matrix

1. ## Eigenvalues of a 3x3 matrix

Hiya,
Algebra exam coming up on Wednesday, and I'm stumped on how to get the egenvalues of a 3x3 matrix. I think I did them about 2 years ago, and they're gone completely out of my head.

Given: $A={\left(\begin{array}{ccc}2&4&0\\4&2&3\\0&3&2\end {array}\right)}$

I know the eigenvalues comes out as $\lambda={2},{-3},{7}$ .
I know you subtract ${\lambda}$ from the diagonal line at first, but then where do you go with it? I would really appreciate it if you could show step-by-step work. Thanks
Jo

2. To find the eigenvalues, find the roots of the polynomial $\text{det}(A-\lambda I)$.

3. So you are asking, basically, how to find those eigenvalues?

The number $\lambda$ is an eigenvalue of A if and only if $Av= \lambda v$, which is equivalent to $Av- \lambda v= (A- \lambda I)v= 0$ has "non-trivial solutions"- that is, a solution other than the obvious v= 0. If $A- \lambda I$ is invertible, then we could multiply on both sides by its inverse to get the unique solution. But on the right we have that matrix times the 0 vector which must be 0. That is, there exist other, non-zero solutions, if and only if $A- \lambda I$ is not invertible which means that its determinant must be 0:
$\left|\begin{array}{ccc}2-\lambda & 4 & 0 \\ 4 & 2-\lambda & 3 \\ 0 & 3 & 2-\lambda\end{array}\right|= 0$
You can expand that by the first row:
$(2-\lambda)\left|\begin{array}{cc}2-\lambda & 3 \\ 3 & 2-\lambda\end{array}\right|- 4\left|\begin{array}{cc}4 & 3 \\ 0 & 2-\lambda\end{array}\right|$
$=(2-\lambda)((2-\lambda)^2- 9)- 4(4(2-\lambda))= (2-\lambda)(\lambda^2- 4\lambda- 5- 16)$
$=(2- \lambda)(\lambda^2- 4\lambda-21)= 0$
$=(2- \lambda)(\lambda+ 3)(\lambda- 7)= 0$
which has solutions $\lambda= 2$, $\lambda= -3$, and $\lambda= 7$.

4. Oh yes! It's all coming back to me me now! Thank you so much for your explaination HallsofIvy!!

5. How to find the eigen vector of the same question?

6. To find the eigenvector associated with the eigenvalue $\lambda$, you simply solve $(A-I\lambda)\vec{v}=\vec{0}$

That is subtract the eigenvalue from the diagonal and find a basis for the null space of this transformation.