Eigenvalues of a 3x3 matrix

**simplysparklers** · May 4th 2009, 10:06 AM

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

**vemrygh** · May 4th 2009, 12:30 PM

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

**HallsofIvy** · May 4th 2009, 12:47 PM

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 $\lambda= 7$ .

**simplysparklers** · May 4th 2009, 05:00 PM

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

**roshanhero** · May 4th 2009, 09:04 PM

How to find the eigen vector of the same question?

**Gamma** · May 4th 2009, 09:44 PM

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.

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