ed84c Posted November 19, 2007 Posted November 19, 2007 M = -M^t where M is 3*3 prove that det(M - kI) = -det(M+kI) Ok for starters its a square matrix => M^t = M => M = -M for our matrix So to get to our proof my solution would be; det(M) = det(-M) det(M)(I-kM^-1) = det(M)(I-kM^-1) det(M - k I) = det(-M + k I) Unfortunately as you can see this doesn't give the answer... any ideas?
ed84c Posted November 20, 2007 Author Posted November 20, 2007 I cant see how to use that? ed84c said: M = -M^t where M is 3*3 prove thatdet(M - kI) = -det(M+kI) Ok for starters its a square matrix => M^t = M => M = -M for our matrix So to get to our proof my solution would be; det(M) = det(-M) det(M)(I-kM^-1) = det(-M)(I-kM^-1) det(M - k I) = det(-M + k I) Unfortunately as you can see this doesn't give the answer... any ideas? btw it should look like that
timo Posted November 20, 2007 Posted November 20, 2007 Your first step is wrong, already. M^t = -M, not M.
ed84c Posted November 20, 2007 Author Posted November 20, 2007 So your suggesting I put (M - k I) to the power of T? And then take into account that; -M = M^t and I = I^t?
timo Posted November 20, 2007 Posted November 20, 2007 Yes, but it's not taking something to the power of t but taking the transposed. If M^t = -M and I^t = I, what is (M-kI)^t ?
ed84c Posted November 20, 2007 Author Posted November 20, 2007 M^t + (kI)^t = -M + I^t k^T = -M + Ik^t Its wierd that the text book doesnt seem to have explained any of the ^t notation or rules thereof. Maybe it was assumed knowledge from FP1 but I can't remember it.
timo Posted November 20, 2007 Posted November 20, 2007 I wouldn't assume anyone knowing what a matrix and a determinant is not knowing what the transpose of a matrix is, either. And I've never seen a different notation for the transpose, either. So yes, I strongly assume it was assumed knowledge. Anyways, - your transformation is correct. The k^t is redundant. The transpose of a scalar is, depending on point of view, either not defined (in which case you wouldn't get the k^t in the first place but just k) or equals k. Either way, you end up with -M + kI. - I later edited my post to ask what (M-kI)^t is, rather than (M+kI)^t. Of course, figuring that out goes exactly the same way as (M+kI)^t (alternatively, you could just replace k with -k). The reason why I later edited it to (M-kI)^t of course it that (M-kI) is the expression that appears on the left-hand side of your original question. - Since you didn't know the notation of the transpose: Do you know what det(...) is and do you also know some calculation rules for it (you will need two of them)?
ed84c Posted November 20, 2007 Author Posted November 20, 2007 I know what det(...) is and I know various rules relating to eigenvalue, eigenvectors and finding powers. Or are you talking about something else?
timo Posted November 20, 2007 Posted November 20, 2007 You can probably get the rules I meant from those. Just see how far you get with your proof.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now