Question

Solve: x'=-y, y'=x
Please, help

Answer 1

My question is on eigenvector.
It becomes
\[\left[\begin{matrix}x'\\y'\end{matrix}\right]=\left[\begin{matrix}0&-1\\1&0\end{matrix}\right]\left[\begin{matrix}x\\y\end{matrix}\right]\]

Answer 2

Hence, characteristics equation is \(\lambda^2=-1\rightarrow \lambda =\pm i\)

Answer 3

you can also do this problem similar to what we did the other day

Answer 4

but if this is the way you want to go about it
I can look it up because I have forgotten this way

Answer 5

For eigenvalue \(-i\)
I have \\\[\left[\begin{matrix}i&-1\\1&i\end{matrix}\right]\left[\begin{matrix}x\\y\end{matrix}\right]=\left[\begin{matrix}0\\0\end{matrix}\right]\]

Answer 6

My question is from here

Answer 7

If I pick the first equation to solve for eigenvector, then my simplest eigenvector is from
\(ix =y\), choose \(\left[\begin{matrix}i\\-1\end{matrix}\right]\)

Answer 8

Question: do I have to manipulate the matrix or just randomly pick one of them to solve for eigenvector?

Answer 9

eigenvector for eigenvalue -i needs to be one that satisfies the above equation without you choosing the eigenvector (0,0)
so your choice of eigenvector seems fine to me

Answer 10

When I took linear algebra course, my Prof told me that I have to manipulate the matrix to get the same form for both rows, then solve for eigenvector. Like this
\[\left[\begin{matrix}i&-1\\1&i\end{matrix}\right]\]
multiply row 1 by (-i) I get \(\left[\begin{matrix}1&i\\1&i\end{matrix}\right]\), then solve

Answer 11

but to this course, my current Prof told me just pick either of them to solve.

Answer 12

They do confuse me by different methods.

Answer 13

I don't think I understand :p

Answer 14

@ganeshie8 @Kainui when you get a chance will you look at this one

Answer 15

It is like asking whether to pick the equation of lineas `y = (1/2)x` or `x-2y = 0`
the answer is, it doesn't matter.

Answer 16

If \(\dbinom{i}{-1}\) is an eigenvector, then any scalar multiple of this is also an eigenvector. You should probably ask your professor to "define" clearly what he means by the "simplest eigenvector".

Answer 17

This is such a popular system of equations, I think it's hard not to memorize it specifically because that matrix is the rotation by 90 degrees matrix. If you sketch the vector field you'll see that it creates a circular flow around the origin! So when I look at i vs -i it is basically telling me that there's an angular momentum vector pointing out of the 2D plane, and one is representing clockwise and the other the counter clockwise rotation.

Intuitively this should make sense, since there are no nonzero real eigenvectors of a rotation matrix in 2D. Just imagine rotating by 90 degrees, there are no vectors that get mapped to a scalar multiple of themselves!

That being said, I would be highly suspicious if your answer did not contain sine or cosine.

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A little about the method. Since your eigenvector equations are now determined with complex numbers, instead of needing two equations to solve, you only need one of them since you have a real and imaginary part which must be correspondingly equal. So this is why your differential equations professor is telling you not to worry about the other equation.

Answer 18

I think I got it. Thanks all. :)