Mimi_x3
  • Mimi_x3
something is wrong
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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Mimi_x3
  • Mimi_x3
|dw:1439098716867:dw|
Mimi_x3
  • Mimi_x3
I'm trying to find the eigenvector
Mimi_x3
  • Mimi_x3
so 0v1 + 1v2 = 0 v2 = 0 v1 = 0

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Mimi_x3
  • Mimi_x3
https://www.wolframalpha.com/input/?i=eigenvectors+{{2%2C1}%2C{0%2C2}} BUT LOOK HERE
UsukiDoll
  • UsukiDoll
did you compute the eigenvalues first?
UsukiDoll
  • UsukiDoll
|dw:1439100292165:dw|
UsukiDoll
  • UsukiDoll
then using \[\det(A- \lambda I) = 0\] |dw:1439100348042:dw|
UsukiDoll
  • UsukiDoll
|dw:1439100402741:dw|
UsukiDoll
  • 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
  • UsukiDoll
we will have 2 eigenvectors... and one of them all 0's.
UsukiDoll
  • 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
  • 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
  • UsukiDoll
eigenvectors are in the form of \[[v_1,v_2...v_n]\] except it's written vertically. |dw:1439101413091:dw|
Mimi_x3
  • Mimi_x3
wait
Mimi_x3
  • Mimi_x3
my eigen values is 2 from |dw:1439101590357:dw|
UsukiDoll
  • UsukiDoll
huh? what was the original question?
Mimi_x3
  • Mimi_x3
https://gyazo.com/3f46ed9dd85e81fc5b637b27dcfbe1b6
Mimi_x3
  • Mimi_x3
I found the eigen value => 2
Mimi_x3
  • Mimi_x3
but the eigen vector does not seem to be right
UsukiDoll
  • 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
  • Mimi_x3
yeah
UsukiDoll
  • UsukiDoll
ok gonna take a shortcut |dw:1439101734333:dw| ah! we have a repeated eigenvalue \[\lambda = 2 \]
Mimi_x3
  • Mimi_x3
yeah so one eigen value
UsukiDoll
  • 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
  • Mimi_x3
there's only one eigen vector..
Mimi_x3
  • Mimi_x3
so i get this at the end |dw:1439101932129:dw|
UsukiDoll
  • 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
  • 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
  • Mimi_x3
ok |dw:1439102632011:dw|
Mimi_x3
  • Mimi_x3
0v1 + v2 = 0 0v1 + 0v2 = 0
UsukiDoll
  • UsukiDoll
@Michele_Laino
Michele_Laino
  • Michele_Laino
what is the starting matrix?
UsukiDoll
  • UsukiDoll
https://gyazo.com/3f46ed9dd85e81fc5b637b27dcfbe1b6
Michele_Laino
  • 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
  • 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
  • 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
  • UsukiDoll
isn't v is the right eigenvector and w the left eigenvector?
Michele_Laino
  • 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
  • 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
  • Michele_Laino
namely: \[{\left( {2 - \lambda } \right)^2} = 0\]
UsukiDoll
  • UsukiDoll
yeah where there is a repeated eigenvalue and that's 2.
UsukiDoll
  • 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
  • Michele_Laino
the value \lambda =2 is double
Michele_Laino
  • 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
  • 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
  • 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
  • UsukiDoll
yeah
Michele_Laino
  • Michele_Laino
right! v_2=0, and v_1 can be any real number, so we have a line of eigenvector
UsukiDoll
  • UsukiDoll
I did matrix multiplication way earlier got |dw:1439107968867:dw|
UsukiDoll
  • UsukiDoll
can v_1 be 1? but how?!
Michele_Laino
  • Michele_Laino
namely our eigenvectors are these ones: \[\left( {\begin{array}{*{20}{c}} k \\ 0 \end{array}} \right),\quad k \in \mathbb{R}\]
UsukiDoll
  • UsukiDoll
where k belong in a set of reals.
Michele_Laino
  • Michele_Laino
yes! that's right!
UsukiDoll
  • UsukiDoll
so k belongs in a set of reals k must be a real number
Michele_Laino
  • 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
  • UsukiDoll
like a 1?
Michele_Laino
  • Michele_Laino
k is varying in the real line, namely k is not fixed
UsukiDoll
  • UsukiDoll
:/ stuck! k can be anything 1 2 3 4 5 6 7 8 9
Michele_Laino
  • Michele_Laino
k can be any real number
UsukiDoll
  • UsukiDoll
so it can be rational or irrational.
Michele_Laino
  • Michele_Laino
both rational and irrational
UsukiDoll
  • UsukiDoll
so what are we doing ? just leaving the other vector as |dw:1439108587318:dw| ?
Michele_Laino
  • Michele_Laino
yes!
UsukiDoll
  • UsukiDoll
ok but does it count as another eigenvector?
Michele_Laino
  • 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
  • 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
  • Michele_Laino
geometrically speaking, we have a line of eigenvectors, and that line has to pass at the origin of the cartesian system
phi
  • 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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