Finding Eigenvalues and questions

[flyingpig]

Homework Statement

http://img703.imageshack.us/img703/4489/unledzh.th.png

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The Attempt at a Solution

a)

Ax = λx

Ax = x

Ax - x = 0

(A - I)x = 0


I set up my matrix

[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP154819f61i5gh6d9fed500006aa20fbb6bgf59g1?MSPStoreType=image/gif&s=17&w=207&h=56

RowReduced it and I got

[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP155119f61i5gh6d9fed50000387i4eb69hc6g67e?MSPStoreType=image/gif&s=17&w=80&h=56

Because this matrix is linearly dependent (can I say the "set is linearly dependent?" I have trouble with these names like "set", "systems"), 1 is an eigenvalue.

b) I know I could just do it the long way, but is there a faster way? Because this is not in triangular form

 

[I like Serena]

I'm not sure what you mean with the long way, but afaik the method is as follows:

Ax=λx ⇒ det(A-λI)=0

Calculating this will give you a 3rd order polynomial to solve.
Luckily you already know one eigenvalue (which is 1), this makes it easier to solve and in passing by also verifies that 1 is an eigenvalue.

Do you know how to do this?

 

[flyingpig]

No... unfortunately

 

[I like Serena]

No... unfortunately

Well, here's a wiki page that explains how to take the determinant of a 3x3 matrix: http://en.wikipedia.org/wiki/Determinant#3-by-3_matrices

And your equation is:
$$\det \begin{pmatrix}-\lambda & -4 & -6 \\<br /> -1 & -\lambda & -3 \\<br /> 1 & 2 & 5-\lambda<br /> \end{pmatrix} = 0$$

Can you combine those to yield a 3rd order polynomial?

Btw, which tools are you allowed to use?
Are you allowed to use e.g. WolframAlpha?

 

[flyingpig]

No I know I could that, but this is a question from an exam paper, so no technologies. I am wondering if there is a faster way to get to a triangular matrix and just "read it off"

 

[I like Serena]

No I know I could that, but this is a question from an exam paper, so no technologies. I am wondering if there is a faster way to get to a triangular matrix and just "read it off"

Nope, I'm afraid not. This is the fastest way.
But you don't need to go to a triangular matrix to do this.
So though your answer to a) was correct (and yes, you can say that the set of equations is linearly dependent), doing it this way, would answer a) as well.