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Mimi_x3
 one year ago
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Mimi_x3
 one year ago
something is wrong

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Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0I'm trying to find the eigenvector

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0so 0v1 + 1v2 = 0 v2 = 0 v1 = 0

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0https://www.wolframalpha.com/input/?i=eigenvectors+ {{2%2C1}%2C{0%2C2}} BUT LOOK HERE

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1did you compute the eigenvalues first?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439100292165:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1then using \[\det(A \lambda I) = 0\] dw:1439100348042:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439100402741:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1\[(\lambda)(\lambda) v_2(0) = 0 \] \[\lambda^20=0\] \[\lambda^2 = 0 \] \[\lambda = \sqrt{0} \] \[\lambda = 0 \] both of our eigenvalues so be 0 \[\lambda_1 = 0, \lambda_2=0 \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1we will have 2 eigenvectors... and one of them all 0's.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1we 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
 one year ago
Best ResponseYou've already chosen the best response.1we have a row of zeros, so we're on the right track. We have a single equation which is v_2 = 0

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1eigenvectors are in the form of \[[v_1,v_2...v_n]\] except it's written vertically. dw:1439101413091:dw

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0my eigen values is 2 from dw:1439101590357:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1huh? what was the original question?

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0I found the eigen value => 2

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0but the eigen vector does not seem to be right

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1ok ... so we have to find eigenvalues and eigenvectors of 2 1 0 2 is this the right 2 x 2 matrix?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1ok gonna take a shortcut dw:1439101734333:dw ah! we have a repeated eigenvalue \[\lambda = 2 \]

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0yeah so one eigen value

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.0there's only one eigen vector..

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.0so i get this at the end dw:1439101932129:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1there 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
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year ago
Best ResponseYou've already chosen the best response.0ok dw:1439102632011:dw

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.00v1 + v2 = 0 0v1 + 0v2 = 0

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2what is the starting matrix?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2or 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
 one year ago
Best ResponseYou've already chosen the best response.1?! 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
 one year ago
Best ResponseYou've already chosen the best response.2ok! 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
 one year ago
Best ResponseYou've already chosen the best response.1isn't v is the right eigenvector and w the left eigenvector?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so, 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
 one year ago
Best ResponseYou've already chosen the best response.2we have to request this condition: \[\det \left( {\begin{array}{*{20}{c}} {2  \lambda }&1 \\ 0&{2  \lambda } \end{array}} \right) = 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2namely: \[{\left( {2  \lambda } \right)^2} = 0\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1yeah where there is a repeated eigenvalue and that's 2.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year ago
Best ResponseYou've already chosen the best response.2the value \lambda =2 is double

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! we have to substitute \lambda=2 into that equation, and find the component of v, namely v_1 and v_2

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1that'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
 one year ago
Best ResponseYou've already chosen the best response.2we got this: \[\left( {\begin{array}{*{20}{c}} 0&1 \\ 0&0 \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{v_1}} \\ {{v_2}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{v_2}} \\ 0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2right! v_2=0, and v_1 can be any real number, so we have a line of eigenvector

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1I did matrix multiplication way earlier got dw:1439107968867:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1can v_1 be 1? but how?!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2namely our eigenvectors are these ones: \[\left( {\begin{array}{*{20}{c}} k \\ 0 \end{array}} \right),\quad k \in \mathbb{R}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1where k belong in a set of reals.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! that's right!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1so k belongs in a set of reals k must be a real number

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! more precisely, since our matrix can be viewed as an operator between real vector spaces, then k has to be real

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2k is varying in the real line, namely k is not fixed

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1:/ stuck! k can be anything 1 2 3 4 5 6 7 8 9

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2k can be any real number

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1so it can be rational or irrational.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2both rational and irrational

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1so what are we doing ? just leaving the other vector as dw:1439108587318:dw ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1ok but does it count as another eigenvector?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the 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
 one year ago
Best ResponseYou've already chosen the best response.2more 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
 one year ago
Best ResponseYou've already chosen the best response.2geometrically speaking, we have a line of eigenvectors, and that line has to pass at the origin of the cartesian system

phi
 one year ago
Best ResponseYou've already chosen the best response.0a 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/1806linearalgebraspring2010/videolectures/lecture21eigenvaluesandeigenvectors/ 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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