Recent content by evlyn
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Graduate Question on the trace of two matrices
It is used to find the eigenvalues of the matrix- evlyn
- Post #16
- Forum: Linear and Abstract Algebra
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Graduate Question on the trace of two matrices
The determinant would be ad-bc the trace would be a+d and the characteristic polynomial I am not entirely sure about ... (t-a)(t-d)-bc ... t^2-at-td+ab-bc- evlyn
- Post #14
- Forum: Linear and Abstract Algebra
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Graduate Question on the trace of two matrices
the dimensions of B must be the same as A It is most likely assumed that the general population in my class knows what the minimal polynomial is- evlyn
- Post #12
- Forum: Linear and Abstract Algebra
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Graduate Question on the trace of two matrices
I don't understand how the minimal polynomial is helpful if I just have a general matrix A? I don't no anything about it except that AB = -BA and that A^2=1 and B^2=1- evlyn
- Post #10
- Forum: Linear and Abstract Algebra
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Graduate Question on the trace of two matrices
Does i here refer to I (the identity matrix)? Maybe I'm being really dumb but where does that come from? Is it finding the eigenvalues of (a+b)^2- evlyn
- Post #8
- Forum: Linear and Abstract Algebra
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Graduate Question on the trace of two matrices
I don't know what size the matrices are I just know that they are two matrices A and B. I guess I'm still not understanding the connection of the trace to anticommuting and to the squares- evlyn
- Post #6
- Forum: Linear and Abstract Algebra
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Graduate Question on the trace of two matrices
trace(A) = trace(B) = 0 and the problem said A^2 =1- evlyn
- Post #4
- Forum: Linear and Abstract Algebra
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Graduate Question on the trace of two matrices
How does knowing that two matrices anticommute AB=-BA and that A^2=1 and B^2=1 help me to know how to find the trace of the matrices. I am supposed to show that their traces equal each other which equals 0 but I am not sure exactly how the given information helps me determine the trace?- evlyn
- Thread
- Matrices Trace
- Replies: 16
- Forum: Linear and Abstract Algebra
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Show Hermitian Identity: (AB)^+ = A^+ B^+
That helped. So now using the definition (usually a good thing) I have: (AB)^+_ij = [AB_ji]^* = A_ji ^* B_ji^* = B_ij^+ A_ij^+ = B^+ A^+ I know that the complex conjugate is distributive can I just assume that for the proof?- evlyn
- Post #3
- Forum: Calculus and Beyond Homework Help
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Show Hermitian Identity: (AB)^+ = A^+ B^+
Homework Statement Show that (AB)^+ = A^+ B^+ using index notation Homework Equations + is the Hermitian transpose The Attempt at a Solution I know that AB = Ʃa_ik b_kj summed over k so (AB)^+ = (Ʃa_ik b_kj)^+ = Ʃ (a_ik b_kj)^+ = Ʃ (a_ik)^+(b_kj)^+ = A^+ B^+ I am not...- evlyn
- Thread
- Hermitian Identity
- Replies: 3
- Forum: Calculus and Beyond Homework Help
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Verifying that a tensor is isotropic
So I figured out that I need to rotate the tensor and from there show that it is the same in the new rotation. So if I have A_ijkl=(δ_ij)(δ_kl) In order to transform it I'm not really sure how to proceed.- evlyn
- Post #4
- Forum: Calculus and Beyond Homework Help
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Verifying that a tensor is isotropic
I'm not even sure how to go about doing that. I am taking math methods but have been out of school for a while and just trying to relearn things and I never took a proof class before. My book does not give any definition other than the one in English but there is no math that I can find that has...- evlyn
- Post #3
- Forum: Calculus and Beyond Homework Help
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Verifying that a tensor is isotropic
I'm supposed to verify that this fourth-rank tensor is isotropic assuming cartesian coordinates: [A]_{}[/ijkl]=[δ]_{}[/kl][δ_{}[/kl] from what I gathered being isotropic means that it stays the same no matter what the rotation is I have no clue how to even start this problem or what I am...- evlyn
- Thread
- Isotropic Tensor
- Replies: 3
- Forum: Calculus and Beyond Homework Help
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Math Methods Question about triple product
"when does this calculation come up in physics, and with what slight modification?" a' = [b x c]/[a*[(b x c)]], b' = [c x a]/[a*[(b x c)]], c' = [a x b]/a*[(b x c)]] a* (b x c) does not equal 0 (* is dot product and (x) is cross product) [b]2. Homework Equations Show that...- evlyn
- Thread
- Math methods Product
- Replies: 1
- Forum: Advanced Physics Homework Help