OpenStudy (loser66):

If \(A= \left[\begin{matrix}a&0\\b&c\end{matrix}\right]\) 1)compute \(e^{At}\) 2) Find the eigenvalues and eigenvectors of \(e^{-A}\) Please, help

2 years ago
OpenStudy (loser66):

@oldrin.bataku I got \[e^{At}= \left[\begin{matrix}e^{at}(a-c)&0\\b(e^{at}-e^{ct})&e^{ct}(a-c)\end{matrix}\right]\]

2 years ago
OpenStudy (loser66):

hence if t = 1, we have \(e^{A}= \left[\begin{matrix}e^{a}(a-c)&0\\b(e^{a}-e^{c})&e^{c}(a-c)\end{matrix}\right]\)

2 years ago
OpenStudy (loser66):

for part 2), we know that \(e^{-A}=(e^{A})^{-1}\), so that we can calculate its eigenvalues and eigenvectors. However, it takes a long time to work on it. I would like to know there is any link between the two's eig.values and eig.vectors? Since we both work on the same A. Moreover, we have theorem for eigenvalue, it says if \(\lambda\) is one of eigenvalue of A, then \(\lambda^n\) is one of eigenvalue of \(A^n\) But I don't know whether we can apply for \(e^{A}\) and \(e^{-A}\) or not, please, explain me

2 years ago
OpenStudy (loser66):

One more thing I got confuse: Surely \(-A \neq A^{-1}\), If I calculate \(- A\) by putting - sign in the front of A, then, \(-A = \left[\begin{matrix}-a&0\\-b&-c\end{matrix}\right]\), while \[A^{-1}= \left[\begin{matrix}1/c &0\\b/ac & 1/a\end{matrix}\right]\] How can \(e^{-A}=(e^A)^{-1}\) ???

2 years ago
OpenStudy (anonymous):

by definition \(-A,A\) commute so \(e^Ae^{-A}=e^{A+-A}=e^0=I\)

2 years ago
OpenStudy (anonymous):

since \(e^{-A}=(e^A)^{-1}\) it follows the eigenvectors are identical but the eigenvalues are related by \(\lambda_{e^{-A},i}=1/\lambda_{e^A,i}\)

2 years ago
OpenStudy (empty):

Hmmm is there a reason why this method I'm using here doesn't work? First I separate A into a sum of the diagonal matrix D and the corner matrix C. \[A=D+C\] \[e^{At}=e^{Dt+Ct}=e^{Dt}e^{Ct}\] Then I compute them individually: \[e^{Dt} = \left[\begin{matrix}e^{at} & 0 \\ 0 & e^{ct} \end{matrix}\right]\] \[e^{Ct} = I+tC=\left[\begin{matrix}1 & 0 \\ bt & 1 \end{matrix}\right]\] Of course multiplying these together doesn't give the matrix I wanted since addition is commutative in the exponents but multiplying these two matrices together is easy to check that it's not commutative. I guess my most important question is, if \(C\) had been a nonsingular matrix, would this be valid, or is there more to it than that?

2 years ago
OpenStudy (anonymous):

well, you need \(C,D\) to commute to compute it in that way, and both being invertible is definitely insufficient -- consider a change of basis followed by a scaling along the coordinate axes. these very clearly do not commute as linear transformations and yet both are invertible

2 years ago
OpenStudy (anonymous):

the standard way to compute \(\exp(A)\) for diagonalizable \(A\) is to diagonalize \(A=P^{-1}DP\) where \(P\) rotates into an eigenbasis and \(D\) describes the scaling, since \(A^n=(P^{-1}DP)^n=P^{-1}D^nP\) so: $$\exp(A)=P^{-1}\left(\sum_{n=0}^\infty\frac1{n!}D^n\right)P$$ and \(D^n\) is trivial for diagonal matrices

2 years ago
OpenStudy (anonymous):

@empty if you're curious as to when two matrices A,B commute: https://en.wikipedia.org/wiki/Commuting_matrices#Characterization_in_terms_of_eigenvectors

2 years ago
OpenStudy (empty):

Yeah I'm only able to know when matrices commute when I understand their geometric interpretation such as rotation matrices will commute with each other and with scalar matrices, things like that. Thanks @oldrin.bataku

2 years ago
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