# Eigenvalue(s) for a matrix

1. Aug 1, 2012

### Lord_Sidious

1. The problem statement, all variables and given/known data

Find the eigenvalues of A, when A =[3,1,1; 0,5,0; -2,0,1]

2. Relevant equations

3. The attempt at a solution

I took the det(A-$\lambda$I)=0 and found the characteristic polynomial to be:

-$\lambda$$^{3}$+15$\lambda$$^{2}$-73$\lambda$+115

I couldn't figure out whole roots to the equation and wolframalpha gave me
$\lambda$=5, 5-√2, 5+√2

Does this seem like the right answer? I have checked and double checked the determinant. Would I still be able to find a basis for the eigenspace?

2. Aug 1, 2012

### sharks

For matrix:
$$A= \begin{bmatrix}3 & 1 & 1 \\ 0 & 5 & 0 \\ -2 & 0 & 1\end{bmatrix}$$
Here is the characteristic equation that i get:
$-λ^3+9λ^2-25λ+25=0$

Last edited: Aug 1, 2012
3. Aug 1, 2012

### voko

Because the second column is [0, 5, 0], the polynomial should look like (5 - x)Q(x), where Q(x) is a second-degree polynomial. So it should be very easy to get its roots. Do it.

4. Aug 1, 2012

### Lord_Sidious

Sorry the matrix was $$A= \begin{bmatrix}3 & 1 & 1 \\ 0 & 5 & 0 \\ -2 & 0 & 7\end{bmatrix}$$
not 2,0,1 in the last row. That was my mistake.

5. Aug 1, 2012

### MednataMiza

Yes, those are the eigenvalues.

6. Aug 1, 2012

### Staff: Mentor

The second row is [0, 5, 0].

7. Aug 1, 2012

### voko

This is not clear from the original notation. Anyway, that does not matter for determinants so my argument holds.

8. Aug 2, 2012

### vela

Staff Emeritus
Expand the determinant along the second row.

9. Aug 2, 2012

### sharks

Well, the original notation used is actually very common and can be recognized as a standard, since it is also used in popular software like MATLAB.

10. Aug 2, 2012

### voko

Let's put it this way: not clear for me, I always seem to forget the conventions for rows vs columns in a linear transcription. This may stem from the fact that even when working with paper and pencil, I am about equally likely to write down a matrix column after column and row after row. Let's hope this thread will leave a dent in my state of confusion :)

11. Aug 2, 2012

### sharks

Your characteristic equation from post #1 is correct. Now, you just need to solve it. Usually, you would start by trial and error to find one factor of the polynomial (you already know the easy one is (λ-5)) and then do long division to obtain a quadratic in the quotient, which you would then solve using the quadratic formula.

12. Aug 2, 2012

### voko

The point I made in #3 is that none of this is required. By tackling the determinant along [0, 5, 0], one gets the polynomial directly in the factored form.