How to Calculate Transition Matrices for Symmetric 2x2 Matrices?

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Homework Help Overview

The discussion revolves around calculating transition matrices for symmetric 2x2 matrices within the context of linear algebra. The original poster presents a problem involving two bases of a vector space of symmetric matrices and seeks to find the transition matrix Ps,b and its inverse Pb,s.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to approach the problem, noting a lack of experience with 2x2 matrices compared to previous vector-related questions. They attempt to relate the problem to their understanding of vectors.
  • Another participant provides a potential answer for Ps,b and suggests using an identity matrix to find Pb,s.
  • There is a mention of a follow-up question regarding another part of the assignment, indicating ongoing exploration of the topic.

Discussion Status

Contextual Notes

Participants are navigating the complexities of transitioning between bases of symmetric matrices, with some expressing confusion about the format and calculations involved. There is an indication of a larger assignment context, which may impose additional constraints or expectations on the participants.

schmiggy
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Homework Statement


Let V be the vector space of all symmetric 2x2 matrices, and consider the bases.
S = {
[1 0] [0 1] [0 0]
[0 0],[1 0],[0 1]}

B = {
[1 1] [-1 1] [1 0]
[1 2],[ 1 1],[0 1]}
of V.
Find the transition matrix Ps,b. Use your answer to calculate Pb,s.

Homework Equations


a = [itex]\alpha[/itex][itex]_{1}[/itex]b[itex]_{1}[/itex] + [itex]\alpha[/itex][itex]_{2}[/itex]b[itex]_{2}[/itex] + [itex]\alpha[/itex][itex]_{3}[/itex]b[itex]_{3}[/itex] +... + [itex]\alpha[/itex][itex]_{k}[/itex]b[itex]_{k}[/itex]

The Attempt at a Solution


I honestly don't know where to start. All previous questions like this we've dealt with vectors and not 2x2 matrices..

ie B = (1,3),(2,1)

(1,3) = 1(1,0) + 3(0,1) and (2,1) = 2(1,0) + 1(0,1)

So Ps,b = [1 2] and Pb,s is just the inverse of Ps,b = -1/5[1 -2]
[3 1] [-3 1]

But I don't know how to even start when I'm given 2x2 matrices..
 
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haha I am doing the same linear assignment. Thought id be awesome and let you know that Ps,b is {(1, -1, 1), (1, 1, 0), (2, 1, 1)}. make an identity matrix to solve the inverse to get Pb,s.
 
Haha yeah, I ended up figuring it out - I never usually get a response from this website, I don't know why, yours is the first one I've gotten so thanks.

Let me know if you need a hand with either of the other questions.
 
No worries mate, and actually I am pretty good for the other two, was just a bit unsure of Q1 b). did you manage to sus that?
 
For 1b) I got
[ 1 2 -1] [-1] = [0]
[-1 -1 1] [ 3] [3]
[-1 -3 2] [ 5] [2]

That's written out pretty garbagety, but hopefully you can decipher it.. it's Pb,s multiplied by (-1,3,5)
 

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