Question

find exponential matrix \(e^{At}\) if \(A =\left(\begin{matrix}0&1\\0&0\end{matrix}\right)\)
Please, help

Answer 1

According to this: https://en.wikipedia.org/wiki/Matrix_exponential,
\[e^{At}=\sum_{k=0}^{\infty} \frac{ 1 }{ k! } (At)^k=\sum_{k=0}^{\infty} \frac{ t^k }{ k! } A^k\]
(I'm assuming t is a scalar value)

Answer 2

Hint: What is \(A^k\)?

Answer 3

^
compute for \(k=0,1,2\), then you will "see" the general form of \(A^k\).

Answer 4

I have a hard time to get this method when I check back to the old one.
Like this : if\( A=\left(\begin{matrix}0&1\\1&0\end{matrix}\right)\)
then \(e^{At}= I +At+\dfrac{A^2t^2}{2!}+\dfrac{A^3t^3}{3!}+\cdots\)

\(=\left(\begin{matrix}1&0\\0&1\end{matrix}\right)+\left(\begin{matrix}0&t\\t&0\end{matrix}\right)+\left(\begin{matrix}t^2/2!&0\\0&t^2/2!\end{matrix}\right)+\left(\begin{matrix}0&t^3/3!\\t^3/3!&0\end{matrix}\right)+\cdots\)

Answer 5

\(e^{At}=\left(\begin{matrix}1+t^2/2!+t^4/4!+\cdots&t+t^3/3!+\cdots\\t+t^3/3!+\cdots&1+t^2/2!+\cdots\end{matrix}\right)\)

Answer 6

\(e^{At}=\left(\begin{matrix}cosh t&sinht\\sinht&cosh t\end{matrix}\right)\)
That allows me to get the general solution for \(x'=Ax\) with IC x(0)=1, y(0)=-1 is the form
\(x(t) =e^{At}\vec c\) where \(\vec c=(1 -1)^T\)
Hence \(x(t) = \left(\begin{matrix}cosh t&sinht\\sinht&cosh t\end{matrix}\right)*\left(\begin{matrix}1\\-1\end{matrix}\right)=\left(\begin{matrix}cosh t-sinht\\sinht-cosh t\end{matrix}\right)\)

Replace \(cosh t=\dfrac{e^t+e^{-t}}{2}\) and \(sinh t=\dfrac{e^t-e^{-t}}{2}\)
I got \(x(t) =\left(\begin{matrix}e^{-t}\\-e^{-t}\end{matrix}\right)\) \)

Answer 7

Now, If I go to traditional way, solve \(x'=Ax\) with IC x(0)=1, y(0)=-1
characteristics equation of A is \(\lambda^2-1=0\), gives me \(\lambda =\pm 1\)

Answer 8

I have 2 eigenvalues ,right? so that on solution, I must have \(e^t \) and \(e^{-t}\)
But for exponential matrix method above, I have just \(e^{-t}\)
It's frustrating.

Answer 9

@phi

Answer 10

Wait a sec, what's your matrix?

\(A =\left(\begin{matrix}0&1\\0&0\end{matrix}\right)\)
and you were just using this one as an example right:
\(A =\left(\begin{matrix}0&1\\1&0\end{matrix}\right)\)

Answer 11

second one

Answer 12

the first one is just showing the easiest one so that i can get some help. :)

Answer 13

haha ok cool, yeah I guess I'm not sure where all you're at but I think you are sorta stuck cause you have:
\(\cosh t\) and \(\sinh t\) but you think your answer should be \(e^t\) and \(e^{-t}\) is that right? Luckily both of these span the same space of solutions since they're related by a linear combination

\[\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} \cosh t \\ \sinh t \end{pmatrix}= \begin{pmatrix} e^t \\ e^{-t} \end{pmatrix}\]
And hey, you can invert that matrix, left multiply both sides by it to get cosh and sinh in terms of e^t and e^-t. I don't know if that's what you're wondering about or not, but you can throw constants on those functions in the vectors as well to see how to transfer the constants from one representation ot the other.

Answer 14

where is your matrix from?

Answer 15

you mean my solution is correct? how? it should have both e^t and e^(-t) in. but I got just e^-t

Answer 16

I dunno I just know it's true from looking at their taylor series and because I use the (inverse matrix relation) a lot
\[\cosh t = \frac{e^t+e^{-t}}{2}\]\[\sinh t = \frac{e^t-e^{-t}}{2}\]
so if you add them together, you get
\[\cosh t+ \sinh t = e^t\]\[\cosh t - \sinh t = e^{-t}\]

which are the matrix equations I'm refererring to :D

Answer 17

Yes, but my solution is