# Search results

Thank you!
2. ### Polar decompositions

About the equation T=Ssqrt(T*T), what purpose does the sqrt(T*T) serve?
3. ### Isometries math problem

Oh right. :rofl: Ok here's plan B. (sqrt(2/3), sqrt(1/3) ). Now let T's 2x2 matrix be of rank 1 with sqrt(3/2) at the top left hand corner and 0 everywhere else. Now T(sqrt(2/3), sqrt(1/3))=(1, 0). Now applying T again, we get (sqrt(2/3), 0). ll(sqrt(2/3), 0)ll=sqrt(2/3).
4. ### Isometries math problem

I digress, but to clear up any misconceptions I have about isometries, lets say we have a vector (2, 2). apply T (2x2 matrix of rank 1 with sqrt(8)/2 at the far upper left corner and 0 everywhere else) to (2, 2) to get (sqrt(8), 0). (2, 2) is not an eigenvector of T and T(2, 2)=/=1*(2, 2). But...
5. ### Isometries math problem

Homework Statement Prove or give a counterexample: if S ∈ L(V) and there exists an orthonormal basis (e1, . . . , en) of V such that llSejll = 1 for each ej , then S is an isometry. Homework Equations The Attempt at a Solution Can't think of a counterexample. I am assuming that...
6. ### Les Paul

If it wasn't for him, there'd be no metal. Among various other things. rip
7. ### Another positive operator proof

Ok suppose <Ts, s>=0. Now we have a1lc1l^2+...+anlcnl^2. Since we are dealing with a nonzero s, assume there must be at least one lcil^2=/=0. Knowing that T is a positive operator, all eigenvalues (ai) are >=0. Now, considering that ai is >=0, we know that if ai is not 0, then ai is >0 which...
8. ### Another positive operator proof

Alright let me fix it. Ok suppose <Ts, s>=0. Now we have a1lc1l^2+...+anlcnl^2. Since we are dealing with a nonzero s, assume there must be at least one ci=/=0. Knowing that T is a positive operator, all eigenvalues (ai) are >=0. Now, considering that ai is >=0, we know that if ai is not 0...
9. ### Another positive operator proof

Oh whoops. Here let me make this more clear. Ok suppose <Ts, s>=0. Now we have a1lc1l^2+...+anlcnl^2. Let 1<=n<=dimV. Since we are dealing with nonzero eigenvectors, assume lcil^2 is >0. Knowing that T is a positive operator, all eigenvalues (ai) are >=0. Now, considering that ai is >=0, we...
10. ### Another positive operator proof

oh right. read you wrong when you said compute the inner product, and nothing else. sorry about that. Ok <Ts, s>=a1lc1l^2+...+anlcnl^2. Let 1<=n<=dim V and let lcil>0. If <Ts, s>=0, then a1lc1l^2+...+anlcnl^2=0 if all eigenvalues ai of the eigenvectors adding to s are 0. Therefore there...
11. ### Another positive operator proof

Ok s=c1e1+...+cnen. Now Ts=a1c1e1+...+ancnen. Now <Ts, s>=<a1c1e1+...+ancnen,s> =<a1c1e1, s>+...+<ancnen, s>. each ai=0 now <0*c1e1+...+0*cnen, s> =<0*c1e1, s>+...+<0*cnen>=0*c1^2+...+0*cn^2=0.
12. ### Another positive operator proof

Ok let <Ts, s>=0. Let s be a linear combination of k linearly independent vectors in our orthonormal basis. If <Ts, s>=0, and each eigenvalue of the eigenvectors of B is an entry on T's diagonal (such that Tckek=ak,kckek) >=0, then <Ts, s>=0 if each eigenvector adding to s corresponds to an...
13. ### Another positive operator proof

Ok let <Ts, s>=0. Let s be a linear combination of k linearly independent vectors in our orthonormal basis. If <Ts, s>=0, and each eigenvalue of the eigenvectors of B is an entry on T's diagonal (such that Tek=ck,kek) >=0, then <Ts, s>=0 if each eigenvector adding to s corresponds to an...
14. ### Another positive operator proof

Ok let <Ts, s>=0. Let s be a linear combination of k linearly independent vectors in our orthonormal basis. If <Ts, s>=0, and each eigenvalue of the eigenvectors of B is an entry on T's diagonal (such that Tek=ck,kek) >=0, then <Ts, s>=0 if Ts=0*s where 0 is an eigenvalue. Therefore there are...
15. ### Another positive operator proof

Ok I will use the same basis and T. Now if <Tv, v>=0 for some v, then we have <Tv, v>=<Te1, v>+...+<Tek, v>=<0e1, v>+...+<0ek, v>=0 where 1<=k. If this is true, then at least one eigenvalue would be 0, thus T is not invertible.
16. ### Another positive operator proof

The idea that if <Tv, v>=0 for some nonzero v then T is not invertibel was what I had in mind. Suppose <Tv, v> is >0 for all v in V. Now let the matrix of T be a diagonal matrix containing eigenvalues for each eigenvector within our orthonormal basis B (in V)={e1......en}. Now let v=ek...
17. ### Another positive operator proof

Ok just so we're clear....lets say that our diagonal matrix has one zero eigenvalue on its diagonal. This IS positive. But the problem is that we cannot guarantee that the v T is mapping is not going to map to 0. So lets say that v=ek. If v=ek and 0 is the eigenvalue at entry k,k on our...
18. ### Another positive operator proof

ok I took one post you have given me for granted. 0 is not positive. I've been moping around about a zero eigenvalue when in reality, T (as a positive operator) is not even supposed to have 0 in the first place. It is supposed to have eigenvalues with postive square roots. Square roots will...
19. ### Another positive operator proof

<Tv, v> is >0, v is not a zero vector, and T is a positive operator, then T is invertible.
20. ### Another positive operator proof

Ok let <Tv, v> be >0. Now let v=e1+...+en where ek is an eigenvector of T from an orthonormal basis on V. Since T is positive and self adjoint, let T be a diagonal matrix with one positive eigenvalue for each ek. Should I assume this? wait nevermind, the eigenvalues should be nonnegative.
21. ### Another positive operator proof

Lets say v=(1 1 1). Now Tv=(1 1 0 ) with (0 0 1) being in nullT. Now this is an orthogonal projection since the inner product between (1 1 0) and (0 0 1) is 0. But the inner product between (1 1 0 ) and (1 1 1) or <Tv, v> is greater than 0. Orthogonal projections are not invertible. But the...
22. ### Another positive operator proof

The problem is that when given <Tv, v> is >0, T doesn't have to be invertible. T can be an orthogonal projection with all eigenvalues >0 (except for say, one eigenvalue being 0) and still be >0. I mean we know that <Tv, v> is >0 but we don't know whether or not it is invertible. But it is...
23. ### Another positive operator proof

Its like this: Say the domain has a basis {(1 0 0), (0 1 0), (0 0 1)}. Now say T maps from here to a range with basis {(1 0 0) (0 1 0)}. So the matrix is 3x3 with a 0 row at the bottom. This implies that there is a 0 eigenvalue. Now lets say that the vector we map is v=a(1 1 1) (a=/=0). T(1 0 0...
24. ### Another positive operator proof

no. If Tv=a1v1+...+0*vn, you're not going to get v back . If Tv=a1v1+...+anvn (no eigenvalue is 0) then you can get v back by dividing each eigenvalue by itself. Sorry for being lazy, its a hot day. I gotta pull my weight here.
25. ### Another positive operator proof

what should I do now?
26. ### Another positive operator proof

All of the eigenvalues should be >=0 so it follows that each eigenvalue has a positive square root. I take it that this is important because if an eigenvalue is negative, then the square root would be imaginary. So we'd have 0-i with its conjugate 0+i. This wouldn't allow self adjointness.
27. ### Another positive operator proof

Alright for the first part the only thing that I will change is the definition of <Tv, v>. Since we know T is self adjoint, and postive, T has a positive square root. So <S^2v, v>=<v, S^2v>. I have no idea how to do the other direction.
28. ### Another positive operator proof

I don't know how to handle the other direction. All we know is that <Tv, v> is >0. T doesn't have to be invertible. It can be a projection and still be >0.
29. ### Medical Marijuana Overdose

I thought I was the only one, lol. Yes I remember I was at a party, I smoked some mj then consumed ONE beer. Yes, thats right, ONE beer. I wasn't even drunk. Next thing I know, I started feeling sick and was on the couch. Then I figured I needed to walk around a few times to feel better...
30. ### Medical Marijuana Overdose

I second that. Not everybody CAN handle drugs. Though I tend to panic from time to time after taking some pulls, I am always with a friend (at least) when I smoke and he doesn't get high very easily.