something is wrong

- Mimi_x3

something is wrong

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- Mimi_x3

|dw:1439098716867:dw|

- Mimi_x3

I'm trying to find the eigenvector

- Mimi_x3

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

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## More answers

- Mimi_x3

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

- UsukiDoll

did you compute the eigenvalues first?

- UsukiDoll

|dw:1439100292165:dw|

- UsukiDoll

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

- UsukiDoll

|dw:1439100402741:dw|

- UsukiDoll

\[(-\lambda)(-\lambda) -v_2(0) = 0 \]
\[\lambda^2-0=0\]
\[\lambda^2 = 0 \]
\[\lambda = \sqrt{0} \]
\[\lambda = 0 \]
both of our eigenvalues so be 0
\[\lambda_1 = 0, \lambda_2=0 \]

- UsukiDoll

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

- UsukiDoll

we should have at least a row of zeros after doing row operations to find eigenvectors
so when \[\lambda = 0 \]
|dw:1439100836674:dw| I drew the matrix in augmented format

- UsukiDoll

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

- UsukiDoll

eigenvectors are in the form of \[[v_1,v_2...v_n]\] except it's written vertically. |dw:1439101413091:dw|

- Mimi_x3

wait

- Mimi_x3

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

- UsukiDoll

huh? what was the original question?

- Mimi_x3

https://gyazo.com/3f46ed9dd85e81fc5b637b27dcfbe1b6

- Mimi_x3

I found the eigen value => 2

- Mimi_x3

but the eigen vector does not seem to be right

- UsukiDoll

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

- Mimi_x3

yeah

- UsukiDoll

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

- Mimi_x3

yeah so one eigen value

- UsukiDoll

I remember this.... if I remembered correctly, there's two eigenvector... one all 0's and one we just have to place \[\lambda = 2 \] back inside the matrix, do some row operations, and see if we can obtain a row of zeros.

- Mimi_x3

there's only one eigen vector..

- Mimi_x3

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

- UsukiDoll

there could be 1 repeat eigenvalue 1 eigenvector ( the phase portrait is an improper node) or 1 repeat eigenvalue 2 eigenvectors (the phase portrait is a proper node) .
we need to do a bit of matrix multiplication for this
|dw:1439102328212:dw|

- UsukiDoll

so if we have v_2 =0
v_1 must have been set at 0
(I'm starting to feel weak in real life... gotta eat dinner)

- Mimi_x3

ok
|dw:1439102632011:dw|

- Mimi_x3

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

- UsukiDoll

@Michele_Laino

- Michele_Laino

what is the starting matrix?

- UsukiDoll

https://gyazo.com/3f46ed9dd85e81fc5b637b27dcfbe1b6

- Michele_Laino

or better:
if the starting matrix is:
\[M = \left( {\begin{array}{*{20}{c}}
0&0 \\
0&1
\end{array}} \right)\]
then the eqiuation to eigenvalues is:
\[\left\{ {M - \lambda \left( {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right)} \right\}v = 0\]
being v a non null vector

- UsukiDoll

?! nugh. we know there is a repeated eigenvalue which is 2
however there could be 1 eigenvector or 2 eigenvectors. It's the eigenvectors that I got stuck on. We do know that v_2 = 0... but what is v_1?

- Michele_Laino

ok! if the starting matrix is:
\[N = \left( {\begin{array}{*{20}{c}}
2&1 \\
0&2
\end{array}} \right)\]
then we have to solve this vector equation:
\[\left\{ {N - \lambda \left( {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right)} \right\}v = 0\]
where, as usually, v can not be the zero vector

- UsukiDoll

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

- Michele_Laino

so, developing that formula, we get:
\[\left( {\begin{array}{*{20}{c}}
{2 - \lambda }&1 \\
0&{2 - \lambda }
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{{v_1}} \\
{{v_2}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
0 \\
0
\end{array}} \right)\]

- Michele_Laino

we have to request this condition:
\[\det \left( {\begin{array}{*{20}{c}}
{2 - \lambda }&1 \\
0&{2 - \lambda }
\end{array}} \right) = 0\]

- Michele_Laino

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

- UsukiDoll

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

- UsukiDoll

so now we plug 2 back into the matrix... we already have our row of zeros which is a good sign that we've done this correctly.. to get eigenvector

- Michele_Laino

the value \lambda =2 is double

- Michele_Laino

yes! we have to substitute \lambda=2 into that equation, and find the component of v, namely v_1 and v_2

- UsukiDoll

that's what drives me nuts
0 1 l 0
0 0 l 0
0v_1+v_2=0
0_v1+0v_2=0
we only have one equation
0v_1+v_2=0
and that's just
v_2 = 0

- Michele_Laino

we got this:
\[\left( {\begin{array}{*{20}{c}}
0&1 \\
0&0
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{{v_1}} \\
{{v_2}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{{v_2}} \\
0
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
0 \\
0
\end{array}} \right)\]

- UsukiDoll

yeah

- Michele_Laino

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

- UsukiDoll

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

- UsukiDoll

can v_1 be 1? but how?!

- Michele_Laino

namely our eigenvectors are these ones:
\[\left( {\begin{array}{*{20}{c}}
k \\
0
\end{array}} \right),\quad k \in \mathbb{R}\]

- UsukiDoll

where k belong in a set of reals.

- Michele_Laino

yes! that's right!

- UsukiDoll

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

- Michele_Laino

yes! more precisely, since our matrix can be viewed as an operator between real vector spaces, then k has to be real

- UsukiDoll

like a 1?

- Michele_Laino

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

- UsukiDoll

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

- Michele_Laino

k can be any real number

- UsukiDoll

so it can be rational or irrational.

- Michele_Laino

both rational and irrational

- UsukiDoll

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

- Michele_Laino

yes!

- UsukiDoll

ok but does it count as another eigenvector?

- Michele_Laino

the algebraic multiplicity is 2, whereas the geometric multiplicity is 1, namely the dimension of the subspace is 1, since it is a line, so our matrix can not be diagonalized

- Michele_Laino

more precisely we hav an infinity of eigenvectors, since we can write this:
\[\left( {\begin{array}{*{20}{c}}
2&1 \\
0&2
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
k \\
0
\end{array}} \right) = 2\left( {\begin{array}{*{20}{c}}
k \\
0
\end{array}} \right),\quad k \in \mathbb{R}\]

- Michele_Laino

geometrically speaking, we have a line of eigenvectors, and that line has to pass at the origin of the cartesian system

- phi

a few thoughts:
an eigenvector is only defined up to a constant. What we care about is its direction; its magnitude can be any value (except zero)
for example if we have this eigen equation
\[ A \vec{v} = \lambda \vec{v} \]
and we scale the eigenvector by a scalar c:
\[ A (c\vec{v} ) = \lambda (c\vec{v} ) \]
the equation is still satisfied.
a triangular matrix has its eigenvalues on the diagonal. We immediately see that we have a repeated eigenvalue of 2
According to Prof Strang
see http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-21-eigenvalues-and-eigenvectors/
at the end of the lecture he points out that a version of your matrix (his has 3 where yours has 2) has *only one* eigenvector. (the [0 0 ] vector does not carry any info for us)
thus if you are solving the differential equation
\[ {\dot{x}}= Ax\] with eigenvectors, you will run into trouble.
notice you can solve the 2nd equation
\[ \frac{d \ x_2}{dt}= 2x_2 \]
using separation of variables to get
\[ x_2 = c_2 e^{2t} \]
substitute that into the first equation and solve "normally?" to get
\[ x_1 = c_1 e^{2t} + c_2 t e^{2t} \]

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