Finding eigenvectors of a 2x2 matrix

• Jan 6th 2010, 04:38 AM
matlabnoob
Finding eigenvectors of a 2x2 matrix
say you have the matrix:

A = (k, 1
0,2)

now i found the eigenvalues to be k and 2.

im trying to find the eigenvector for the eigenvalue k.

i thought its supposed to work when i do the following...
so if i use (A-lambda(I))v = 0, the matrix reduces to:

(k-k , 1
0, 2-k)
which is
(0, 1
0, 2-k)
so v2 = 0 and (2-k)v2 = 0

and we know v1 = 0...

but that means the eigenvector is (0,0) !! how is this possible?

because when i try working out the other way...

so back to our matrix A..

I do Av = lambda(v)

kv1 + v2 = kv1
2v2 = kv2

and i get eigenvector (1,0) which is correct..

so why does the first method not work?? i thought we can use any of those methods....(Doh)
• Jan 6th 2010, 05:46 AM
HallsofIvy
Quote:

Originally Posted by matlabnoob
say you have the matrix:

A = (k, 1
0,2)

now i found the eigenvalues to be k and 2.

im trying to find the eigenvector for the eigenvalue k.

i thought its supposed to work when i do the following...
so if i use (A-lambda(I))v = 0, the matrix reduces to:

(k-k , 1
0, 2-k)
which is
(0, 1
0, 2-k)
so v2 = 0 and (2-k)v2 = 0

and we know v1 = 0...

How do "we know v1= 0"? Why can't v1 be any number at all?

Quote:

but that means the eigenvector is (0,0) !! how is this possible?
No, it doesn't. It means an eigenvector is of the form (a, 0) for any number a.

Quote:

because when i try working out the other way...

so back to our matrix A..

I do Av = lambda(v)

kv1 + v2 = kv1
2v2 = kv2

and i get eigenvector (1,0) which is correct..
Yes, it is. It is your statement "and we know v1 = 0" that is incorrect. We have no reason to say v1= 0.

Quote:

so why does the first method not work?? i thought we can use any of those methods....(Doh)
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The first method works fine if you use it correctly!
• Jan 6th 2010, 06:21 AM
matlabnoob
aah thanks!
that was stupid of me assuming v1 =0
what on earth was i thinking?? (Doh)
• Feb 3rd 2010, 12:33 PM
matlabnoob
back to same problem =(
ok. so im trying to solve the same matrix..

(k , 1
0 , 2)

and im using the method as follows:

(k-k, 1
0, 2-k)

so i get

(0, 1
0, 2-k)

and then...

0v1 + v2 = 0
0v1 + (2-k)v2 = 0

which i get..

v2 = 0

so...

(0,0)

!!

how does that work out? that looks so wrong! (Crying)
• Feb 3rd 2010, 12:52 PM
Defunkt
Quote:

Originally Posted by matlabnoob
ok. so im trying to solve the same matrix..

(k , 1
0 , 2)

and im using the method as follows:

(k-k, 1
0, 2-k)

so i get

(0, 1
0, 2-k)

and then...

0v1 + v2 = 0
0v1 + (2-k)v2 = 0

which i get..

v2 = 0

so...

(0,0)

!!

how does that work out? that looks so wrong! (Crying)

Again, as is obviously evident from what you posted, there is no condition on $\displaystyle v_1$. You found, correctly, that $\displaystyle v_2=0$. This means that the eigenvector of eigenvalue k is any vector of the form $\displaystyle (x ~ 0)$ with $\displaystyle x$ a real number.
• Feb 3rd 2010, 02:44 PM
math2009
$\displaystyle f_A(\lambda)=\det(A-\lambda I_3)=\det \begin{bmatrix}k-\lambda &1 \\0 &2-\lambda \end{bmatrix} =(2-\lambda)(k-\lambda)=0~\rightarrow~\lambda=\{2,k \}$
• Feb 4th 2010, 05:03 AM
HallsofIvy
Quote:

Originally Posted by matlabnoob
aah thanks!
that was stupid of me assuming v1 =0
what on earth was i thinking?? (Doh)

Quote:

Originally Posted by matlabnoob
ok. so im trying to solve the same matrix..

(k , 1
0 , 2)

and im using the method as follows:

(k-k, 1
0, 2-k)

so i get

(0, 1
0, 2-k)

and then...

0v1 + v2 = 0
0v1 + (2-k)v2 = 0

which i get..

v2 = 0

so...

(0,0)

!!

how does that work out? that looks so wrong! (Crying)

Yes, and you made the same mistake you did a month ago! v1 is NOT 0. It can be any number.