Finding the characteristic equation/eigenvalues

[012983]

Homework Statement

Given the matrix

3 2 1
0 0 2
0 2 0

Find the characteristic equation and the eigenvalues.

Homework Equations

|$$\lambda$$I-A|

The Attempt at a Solution

If the characteristic polynomial is

($$\lambda$$-3) 2 1
0 ($$\lambda$$-0) 2
0 2 ($$\lambda$$-0)

then doesn't that mean the equation is $$\lambda$$$$^{2}$$($$\lambda$$-3) with eigenvalues 3 and 0? Why do the calculators give me eigenvalues of 3, 2, and -2? I know the determinant of the matrix is -12 (3*2*-2?) but I'm not sure how to obtain the eigenvalues. Thanks.

 

[lineintegral1]

The eigenvalues are found by taking the determinant of the matrix that has lambda times identity subtracted from it (so, just new matrix with the lambdas in the entries). This gets your a characteristic polynomial. Then solve the quadratic equation.

 

[012983]

Ah, nevermind. I figured it out now, thanks.

 

[flyingpig]

This is not a triangular matrix...

 

[Mark44]

The eigenvalues are found by taking the determinant of the matrix that has lambda times identity subtracted from it (so, just new matrix with the lambdas in the entries). This gets your a characteristic polynomial. Then solve the quadratic equation.

In this case, it will be a cubic equation.

 

[wisvuze]

you can always triangularize it yourself if that makes things easier for you

 

[012983]

Yeah, I figured it out... was pretty simple to me once I looked at it differently.

Meanwhile, though, I'm stuck on a 3x3... I've redone it four times to no avail.

[3 2 -3]
[-3 -4 9]
[-1 -2 5]

[($$\lambda$$-3) 2 -3]
[-3 ($$\lambda$$+4) 9]
[-1 -2 ($$\lambda$$-5)]

Using row expansion I calculate the following:

($$\lambda$$-3)($$\lambda$$^2+4$$\lambda$$-5$$\lambda$$-20+18)-2(-3$$\lambda$$+15+9)+((-3)(6-(-$$\lambda$$-4)))
= ($$\lambda$$-3)($$\lambda$$^2-$$\lambda$$-2)+6$$\lambda$$-30-18+((-3)(10+$$\lambda$$))
= ($$\lambda$$^3-4$$\lambda$$^2+$$\lambda$$+6)+6$$\lambda$$-48-30-3$$\lambda$$
= ($$\lambda$$^3-4$$\lambda$$^2+$$\lambda$$+6)+6$$\lambda$$-78-3$$\lambda$$

Calculator says 2 and 0 are the eigenvalues, but I probably won't pull that out of that equation. Now what?

Thanks!
= $$\lambda$$^3-4$$\lambda$$^2+4$$\lambda$$-72

 

[lineintegral1]

In this case, it will be a cubic equation.

Absolutely right, typo on my part.

 

[Mark44]

Yeah, I figured it out... was pretty simple to me once I looked at it differently.

Meanwhile, though, I'm stuck on a 3x3... I've redone it four times to no avail.

[3 2 -3]
[-3 -4 9]
[-1 -2 5]

[($$\lambda$$-3) 2 -3]
[-3 ($$\lambda$$+4) 9]
[-1 -2 ($$\lambda$$-5)]

You have an error right off the bat. Following your style, the row 2, column 2 entry should be $\lambda$ - 4, and most other signs are incorrect.

In your approach, it looks like you are evaluating det($\lambda$I - A). Most textbooks I've seen do it the other way around, det(A - $\lambda$I).

Following this approach, your determinant should be
$$\left| \begin{array} {c c c} 3 - \lambda & 2 & -3\\ -3 &-4 - \lambda & 9\\ -1&-2& 5- \lambda \end{array}\right|$$

Using row expansion I calculate the following:

($$\lambda$$-3)($$\lambda$$^2+4$$\lambda$$-5$$\lambda$$-20+18)-2(-3$$\lambda$$+15+9)+((-3)(6-(-$$\lambda$$-4)))
= ($$\lambda$$-3)($$\lambda$$^2-$$\lambda$$-2)+6$$\lambda$$-30-18+((-3)(10+$$\lambda$$))
= ($$\lambda$$^3-4$$\lambda$$^2+$$\lambda$$+6)+6$$\lambda$$-48-30-3$$\lambda$$
= ($$\lambda$$^3-4$$\lambda$$^2+$$\lambda$$+6)+6$$\lambda$$-78-3$$\lambda$$

Calculator says 2 and 0 are the eigenvalues, but I probably won't pull that out of that equation. Now what?

Thanks!
= $$\lambda$$^3-4$$\lambda$$^2+4$$\lambda$$-72

 

[012983]

You have an error right off the bat. Following your style, the row 2, column 2 entry should be $\lambda$ - 4, and most other signs are incorrect.

In your approach, it looks like you are evaluating det($\lambda$I - A). Most textbooks I've seen do it the other way around, det(A - $\lambda$I).

Following this approach, your determinant should be
$$\left| \begin{array} {c c c} 3 - \lambda & 2 & -3\\ -3 &-4 - \lambda & 9\\ -1&-2& 5- \lambda \end{array}\right|$$

Yeah, I wasn't too sure about what I was reading in my textbook versus what I was reading on here. My book does indeed do it det($\lambda$I - A)... I assumed it didn't matter if it was det($\lambda$I - A) or et(A - $\lambda$I). My mistake.

Thank you for your help!

 

[wisvuze]

It doesn't matter, some books ( like this one: https://www.amazon.com/dp/1441924981/?tag=pfamazon01-20, and I think the Hoffman book as well (but I might be wrong about that) ) uses the det( cI - A) convention. He was saying that you messed up a sign in the juxaposition of the two matrices, not that you can't use that convention at all