Solving Transition Matrix: Find Coefficients of B to B

In summary, the conversation is about finding the Transition Matrix P from the basis B={t+1, 2t, t-1} to B'={4t^{2}-6t, 2t^{2}-2, 4t} for the space R[t]. The example from the book shows that 4t^{2}-6t can be represented as (-3, 2, -3) in basis B, and similarly for 2t^{2}-2 and 4t. However, there seems to be some confusion about the notation and the basis vectors, so further clarification is needed.
  • #1
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Im trying to figure out how to do this question. This is an example in the book i have. I am not sure how they got the answer.

Here is the example from the book:

Find the Transition Matrix P from the basis B={t+1, 2t, t-1} to B'={4t[tex]^{2}[/tex]-6t, 2t[tex]^{2}[/tex]-2, 4t} for the space R[t].
A little computataion shows that 4t[tex]^{2}[/tex]-6t:(-3, 2, -3), 2t[tex]^{2}[/tex]-2: (-1, 1, 1) and 4t:(2, 0, 2). Therefore

[tex]P=\left(\begin{array}{ccc}-3 & -1 & 2 \\ 2 & 1 & 0\\ -3 & 1 & 2\end{array}\right)[/tex]

I'd like to know how they found 4t[tex]^{2}[/tex]-6t to be(-3, 2, -3), 2t[tex]^{2}[/tex]-2 to be (-1, 1, 1) and 4t to be(2, 0, 2). Any help is apprectated.
 
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  • #2
I'm afraid you are going to have to clarify some things first. What is "R[t]"? My first thought would be "functions in thr variable t on the real numbers" but all you give for your basis are powers of t. Polynomials over the real numbers with variable t? But "B" only has linear polynomials so that can't be right. And, for that matter, B' has quadratic polynomials so B and B' can't possibly span the same space! Are you sure you didn't miss at least one square in B?

The only way I could interpret "4t2- 6t: (-3, 2, -3)" is that 4t2- 6t, one of the basis vectors in B', can be written -3(t+1)+ 2(2t)- 3(t-1), in terms of the basis B. But that is equal to -3t- 3+ 4t- 3t+ 3. Trying to find f1, f2, f3 so that -3f1+ 2f2- 3f3= 42- 6t, -f1+ f2+ f3= 2t2- 2, and 2f1+ 2f3= 4t, I arrive at f1= t+ 1, just as you have, f3= t- 1, just as you have, but f2= 2t2, not 2t- you dropped the "square".

Now, what they are saying is: 4t2- 6t can be represented as (-3, 2, -3) in basis B because -3(t+1)+ 2(2t2)- 3(t-1)= 4t2- 6t, etc.
 

1. How do I find the coefficients of B to B in a transition matrix?

The coefficients of B to B in a transition matrix can be found by solving a system of equations using the given transition matrix and the initial and final states. The coefficients represent the probability of transitioning from one state to another.

2. What is a transition matrix?

A transition matrix is a mathematical tool used to model the probability of transitioning from one state to another in a system. It is commonly used in fields such as economics, biology, and engineering.

3. What is the purpose of solving for the coefficients of B to B?

The coefficients of B to B provide valuable information about the behavior and stability of a system. They can help predict the likelihood of a system transitioning from one state to another and can be used to make informed decisions in various fields.

4. Are there any specific methods for solving transition matrices?

Yes, there are several methods for solving transition matrices, including Gaussian elimination, inverse matrix, and eigenvector methods. The choice of method may depend on the complexity of the matrix and the desired accuracy of the solution.

5. How can I use the coefficients of B to B in real-life applications?

The coefficients of B to B can be applied in various real-life scenarios, such as predicting stock market trends, analyzing population growth, and studying chemical reactions. They provide a quantitative understanding of the behavior of a system and can help make informed decisions and predictions.

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