Thread: Eigenvalues of a 3 x 3 Matrix

1. Eigenvalues of a 3 x 3 Matrix

I am trying to work out some Matrix questions in advance to my course start.

I am trying to find the eigenvalues for this 3 x 3 Matrix:

5 3 2
1 4 6
9 7 3

So far I have got to:

L=Lambda

5-L ... 3 ... 2
1 ... 4-L ... 6
9 ... 7 ... 3-L

I am not sure where to go next - Can anybody help?

Thanks,

Chris

2. Re: Eigenvalues of a 3 x 3 Matrix

Well, you evaluate that determinant now! That will, of course, depend on $\lambda$. In fact, it will be a cubic equation.

$\left|\begin{array}{ccc}5- \lambda & 3 & 2 \\ 1 & 4- \lambda & 6 \\ 9 & 7 & 3-\lambda \end{array}\right|= 0$

Expanding on the top row, that is equal to
$(5- \lambda)\left|\begin{array}{cc} 4-\lambda & 6 \\ 7 & 3- \lambda \end{array}\right|$ $- 3 \left|\begin{array}{cc}1 & 6 \\ 9 & 3-\lambda\end{array}\right|$ $+ 2\left|\begin{array}{cc}1 & 4-\lambda \\ 9 & 7 \end{array}\right|= 0$

$(5-\lambda)((4-\lambda)(3-\lambda)- 42)- 3(3-\lambda- 54)+ 2(7- 9(4- \lambda)= 0$
Multiply that out and solve for $\lambda$.

3. Re: Eigenvalues of a 3 x 3 Matrix

Thanks.

I have got:

–L³ + 12L² + 16L - 55

Whats the best way to solve this?

Thanks a lot,

Chris

4. Re: Eigenvalues of a 3 x 3 Matrix

Originally Posted by RealENG
Thanks.

I have got:

–L³ + 12L² + 16L - 55

Whats the best way to solve this?
You have the correct equation $-\lambda^3+12\lambda^2+16\lambda-55=0$, but it does not have convenient solutions. A graph of that function shows that the roots are approximately –2.6, 1.7 and 12.9. You will have to use a numerical method if you want better approximations than that.

5. Re: Eigenvalues of a 3 x 3 Matrix

Thanks.

I checked with MatLab and an online cubic calculator and got -2.568, 1.659, 12.909 as my eigenvalues, so that's all good.

How do I go about finding the Eigenvectors for these eigenvalues?

Any help is greatly appreciated.

Regards,

Chris

6. Re: Eigenvalues of a 3 x 3 Matrix

The eigenvector corresponding to a certain eigenvalue is the vector x such that $Ax=\lambda x$.
Lambda is one of the three eigenvalues.
There are thus 3 eigenvectors in this problem.

Using Matlab you can find the eigenvectors using: $[V,L] = eig(M)$ where V are the vectors (1 column is 1 vector), L are the eigenvalues (lambdas) and M is your input matrix.

The eigenvector corresponding to the eigenvalue, 12.9094 is $\begin{bmatrix}0.3936\\-0.5435\\-0.7414 \end{bmatrix}$.

I hope this helps...

7. Re: Eigenvalues of a 3 x 3 Matrix

you can use this site to calculate matrix stuff:

Calculator for Eigenvalues and Eigenvectors