Powers of a Matrix
ฝัง
- เผยแพร่เมื่อ 27 ต.ค. 2024
- MIT 18.06SC Linear Algebra, Fall 2011
View the complete course: ocw.mit.edu/18...
Instructor: Ben Harris
A teaching assistant worksthrough a problem on powers of a matrix.
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu
You can get to the result the moment you have the diagonal matrix since that's just -E (a=b=-1). Since C^k = X * S^k * X^-1, and S = -E it will become A = X * (-E)^k * X^-1 = X * E * X^-1 = X * X^-1 = E. Chosing a different a,b would make the whole calculation more rewarding
my eigenvalue matrix was different so my first thought was that i had a mistake but it works as long as its columns is are linear combination of the eigenvalues in the video. stating the obvious kinda, but in case someone has a brainfart like i did. as usual thanks to MIT for content
I really enjoyed solving this one .. thank you
Thank you Ben and MIT!
Find formula for c^k nhi PTA lekin C^100 = identity Matrix elements are 1001 hoga esko C^2 nikalo, C^3 nikalo phele put the value of a=b=-1 uske baad nikalo
Very good video.
I selected different eigenvectors for both of the eigenvalues (for eig.value = b : [-1;-1] and for eig. value = a : [-1/2;-1], but at the end of the calculations I got a different C^k matrix expression which does not yield the original C matrix when I plug in k=1 in my expression. How should we determine the eigenvector matrix so that we can get the true expression at the end by considering the fact that eigenvectors are not unique?
I have another question also when we plug a=b=-1, we have repeated eigenvalues, so the expression we get at the end should not be true all the time( I mean generally, for other matrices for example) even it is true for this particular matrix because we calculated this expression according to the diagonalizable feature of the matrix.
@@berkayyaldz9580 i wonder this as well, did you get an explanation?
@@richea8947 for the first one, I had written eigenvector matrix as b = [-beta; -beta] and a = [-alpha/2; -alpha] and made all calculations accordingly and I got the correct result at the end. For a=b=-1, I think we can plug it in because we have distinct eigenvectors. (we need to first calculate eigenvector matrix and see that it has distinct eigenvector matrix)
Thank you! It's very helpful! I wish I could do the calculation as fast as you :)
how did he compute S inverse so fast. can someone explain.
take the negative of non diagonal elements and exchange the positions of diagonal elements, this is the trick for finding adjoint of 2by2 matrix
@@atharvaggarwal1764 and divide by det.
Damn I made a silly mistake, took the 1/det(S) (-2 for me since different eigenvectors) to the left of the eq and when multiplying S and S^-1 forgot that I had taken that out and got -2*identity oof
C for a=b=-1 is just minus the Identity. So C^100 is the Identity !
How much do I get for getting the 2nd part right ?? 😉
The problem only asks for the C(-1,-1)^100, and doesn't say you have to diagonalize C. I'd give you full credit!
@@wontpower no. the problem asks for a formula for c^k for random a's and b's. the second part is worthless imo.
It's all fun and games until you make me do the algebra in the first step.😜
Just substitute a=b=-1 into C first, then this will be 100 times easier. Calculating a determinant with letters on the matrix is really a nightmare.
I don't trust this guy skipping steps after last time.
"you know how to multiply, I'm not going to do it"
gets it wrong
mistakes do happen, otherwise this guy is also good. By the way, you can report that video's error to MIT OCW, I did that.
@Joseph Levine I also got an email from OCW team saying that they have sent my message to the concerned team of the course. Right now I have completed 26 lectures and hope to finish this series by the end of this month. Brace up and restart :D Godspeed mate!
Your comment made me fall out of my chair
@@bridge5189 so uh did you finish it in that month?
@@theblinkingbrownie4654 Yes, I did