Thread: Eigenvalues

1. Eigenvalues

Find all the eigenvalues (real and complex) of the matrix:

$\left( \begin{array}{ccc} -6 & 22 & 11 \\ -2 & -7 & -2 \\ -1 & 26 & 10 \end{array} \right)$

I started off by using the formula for finding the characteristic polynomial, and reducing the determinant matrix to:

$det\left( \begin{array}{ccc} 0 & 26x+134 & -x^2+4x+49 \\ 0 & x+59 & -2x+22 \\ 1 & -26 & x-10 \end{array} \right)$

Which then got me to:

$det\left( \begin{array}{ccc} 26x+134 & -x^2+4x+49 \\ x+59 & -2x+22 \end{array} \right)$

When I take the determinant of that, I get a cubic polynomial which is quite difficult to solve.

Is there an easier way to do this, or am I just doing it completely wrong? Can I possibly reduce it more?

2. Originally Posted by Thomas
Find all the eigenvalues (real and complex) of the matrix:

$\left( \begin{array}{ccc} -6 & 22 & 11 \\ -2 & -7 & -2 \\ -1 & 26 & 10 \end{array} \right)$

I started off by using the formula for finding the characteristic polynomial, and reducing the determinant matrix to:

$det\left( \begin{array}{ccc} 0 & 26x+134 & -x^2+4x+49 \\ 0 & x+59 & -2x+22 \\ 1 & -26 & x-10 \end{array} \right)$

Which then got me to:

$det\left( \begin{array}{ccc} 26x+134 & -x^2+4x+49 \\ x+59 & -2x+22 \end{array} \right)$

When I take the determinant of that, I get a cubic polynomial which is quite difficult to solve.

Is there an easier way to do this, or am I just doing it completely wrong? Can I possibly reduce it more?
$P_x(A) = |xI - A|$
what is your characteristic polynomial again?

3. I found the characterisitic polynomial to be:

$x^3+3x^2+19x+57$

4. Originally Posted by Thomas
I found the characterisitic polynomial to be:

$x^3+3x^2+19x+57$
provided that is correct, i did not check it, we know that x = -3 is one root. thus, by the factor theorem (x + 3) is a factor. use long division to determine the others (the other two are complex)

5. I'm quite confident it's correct - I re-did all the steps about three times and have arrived at the same polynomial.

To find one root is x=-3, did you just use a graphing calculator?

6. Originally Posted by Thomas
I'm quite confident it's correct - I re-did all the steps about three times and have arrived at the same polynomial.

To find one root is x=-3, did you just use a graphing calculator?
no. you check the factors of the constant, that is, the factors of 57 (consider the negative and positive factors) and plug them in to see if you get zero, this method is by the rational roots theorem

7. Originally Posted by Thomas
I found the characterisitic polynomial to be:

$x^3+3x^2+19x+57$
did u notice that x=-3 is a factor?? Ü

8. Ahh, I didn't remember that. Thank you. I'll post what I get for my values once I'm done!

9. So I get $x^2+19=0$

Can anyone refresh my memory on complex numbers?

Would the answer be $x=+/-\sqrt{19}\:i$?

10. Originally Posted by Thomas
So I get $x^2+19=0$

Can anyone refresh my memory on complex numbers?

Would the answer be $x=+/-\sqrt{19}\:i$?
yes, it is $\pm \sqrt{19}~i$

11. Awesome, now wish me luck trying to enter this into my online assignment.

12. Originally Posted by Thomas
Find all the eigenvalues (real and complex) of the matrix:

. . .

When I take the determinant of that, I get a cubic polynomial which is quite difficult to solve.

Is there an easier way to do this, or am I just doing it completely wrong? Can I possibly reduce it more?
Hi, Thomas.

Your answers are correct (-3, +/- 4.35889894354067i); however, I am wondering if you needed algebraic answers. (Just curious why you didn't use MATLAB or an online tool for calculating numeric results.)

David

13. I'm not sure why I didn't... I just did it the long way I guess.

As for algebraic answers... the online assignment program is set to allow answers within 1%, so it's fine handing in answers with decimals.

14. Originally Posted by Thomas
Find all the eigenvalues (real and complex) of the matrix:

$\left( \begin{array}{ccc} -6 & 22 & 11 \\ -2 & -7 & -2 \\ -1 & 26 & 10 \end{array} \right)$
Just for future reference, these questions are often rigged so that you can take factors out of the determinant (by doing elementary row and column operations) without having to multiply it out. In this case, the 22 and the 11 in the top row of the matrix suggest that you might try subtracting twice the right column from the middle column. Then you get

$\begin{vmatrix}-6-x & 22 & 11 \\ -2 & -7-x & -2 \\ -1 & 26 & 10-x \end{vmatrix} = \begin{vmatrix}-6-x & 0 & 11 \\ -2 & -3-x & -2 \\ -1 & 6+2x & 10-x \end{vmatrix}$.

When you see that, you know that you have struck lucky, because there is a factor 3+x going down the middle column. Take this factor out, and you are left with $(3+x)\begin{vmatrix}-6-x & 0 & 11 \\ -2 & -1 & -2 \\ -1 & 2 & 10-x \end{vmatrix}$. Now add twice the middle row to the bottom row, and expand down the middle column, to get $(3+x)\begin{vmatrix}-6-x & 11 \\ -5 & 6-x \end{vmatrix} = -(3+x)(x^2 - 36 + 55) = -(3+x)(x^2 + 19)$. There's the eigenvalue equation, neatly factorised, without having to do any heavy calculations at all!