Eigenvalue(s) for a matrix

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-[itex]\lambda[/itex]I)=0 and found the characteristic polynomial to be:

-[itex]\lambda[/itex][itex]^{3}[/itex]+15[itex]\lambda[/itex][itex]^{2}[/itex]-73[itex]\lambda[/itex]+115

I couldn't figure out whole roots to the equation and wolframalpha gave me
[itex]\lambda[/itex]=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?

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##

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.

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.

MednataMiza

Yes, those are the eigenvalues.

Mark44

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

voko

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

vela

Expand the determinant along the second row.

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.

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 :)

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.

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.