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|dw:1439098716867:dw|

I'm trying to find the eigenvector

so 0v1 + 1v2 = 0
v2 = 0
v1 = 0

https://www.wolframalpha.com/input/?i=eigenvectors+{{2%2C1}%2C{0%2C2}}
BUT LOOK HERE

did you compute the eigenvalues first?

|dw:1439100292165:dw|

then using
\[\det(A- \lambda I) = 0\]
|dw:1439100348042:dw|

|dw:1439100402741:dw|

we will have 2 eigenvectors... and one of them all 0's.

we have a row of zeros, so we're on the right track. We have a single equation which is
v_2 = 0

wait

my eigen values is 2 from
|dw:1439101590357:dw|

huh? what was the original question?

https://gyazo.com/3f46ed9dd85e81fc5b637b27dcfbe1b6

I found the eigen value => 2

but the eigen vector does not seem to be right

ok ... so we have to find eigenvalues and eigenvectors of
2 1
0 2
is this the right 2 x 2 matrix?

yeah

ok gonna take a shortcut
|dw:1439101734333:dw|
ah! we have a repeated eigenvalue
\[\lambda = 2 \]

yeah so one eigen value

there's only one eigen vector..

so i get this at the end
|dw:1439101932129:dw|

ok
|dw:1439102632011:dw|

0v1 + v2 = 0
0v1 + 0v2 = 0

what is the starting matrix?

https://gyazo.com/3f46ed9dd85e81fc5b637b27dcfbe1b6

isn't v is the right eigenvector and w the left eigenvector?

namely:
\[{\left( {2 - \lambda } \right)^2} = 0\]

yeah where there is a repeated eigenvalue and that's 2.

the value \lambda =2 is double

yeah

right! v_2=0, and v_1 can be any real number, so we have a line of eigenvector

I did matrix multiplication way earlier
got
|dw:1439107968867:dw|

can v_1 be 1? but how?!

where k belong in a set of reals.

yes! that's right!

so k belongs in a set of reals
k must be a real number

like a 1?

k is varying in the real line, namely k is not fixed

:/ stuck! k can be anything 1 2 3 4 5 6 7 8 9

k can be any real number

so it can be rational or irrational.

both rational and irrational

so what are we doing ? just leaving the other vector as
|dw:1439108587318:dw| ?

yes!

ok but does it count as another eigenvector?