# Thread: Finding eigenvectors of a 2x2 matrix

1. ## 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....

2. 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?

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.

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.

so why does the first method not work?? i thought we can use any of those methods....
The first method works fine if you use it correctly!

3. aah thanks!
that was stupid of me assuming v1 =0
what on earth was i thinking??

4. ## 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!

5. 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!
Again, as is obviously evident from what you posted, there is no condition on $v_1$. You found, correctly, that $v_2=0$. This means that the eigenvector of eigenvalue k is any vector of the form $(x ~ 0)$ with $x$ a real number.

6. $f_A(\lambda)=\det(A-\lambda I_3)=\det \begin{bmatrix}k-\lambda &1 \\0 &2-\lambda \end{bmatrix} =(2-\lambda)(k-\lambda)=0~\rightarrow~\lambda=\{2,k \}$

7. Originally Posted by matlabnoob
aah thanks!
that was stupid of me assuming v1 =0
what on earth was i thinking??
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!
Yes, and you made the same mistake you did a month ago! v1 is NOT 0. It can be any number.