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Finding Transition Matrix from Bases

  1. Apr 22, 2014 #1
    1. The problem statement, all variables and given/known data
    Let So = {v1,v2,v3,v4} be a basis of the vector space V.
    S= {u1,u2,u3,u4} is a set of vectors defined as follows:
    u1 = 80v1 + 106v2 + 120v3 +164v4
    u2 = 80v1 + 146v2 + 136v3 + 91v4
    u3 = 90v1 + 143v2 + 122v3 + 70v4
    u4 = 80v1 + 56v2 + 80v3 + 48v4

    Find the Transition Matrix A from the basis So to S. That is find PS->So. This is the transition matrix from S coordinates to So coordinates.



    2. Relevant equations
    I know that to get from one transition matrix to another transition matrix you can take inverse.
    Also, for PS->So I know that means you want to turn So into identity matrix.

    I have solved problems like this which small numbers, not equations so I am confused.



    3. The attempt at a solution

    I think the easiest way to solve this (this might be wrong) is to find PSo->S and then take inverse. because the equations are in terms of So. So does that mean I take inverse of this matrix:
    80 106 120 164
    80 146 136 91
    90 143 122 70
    80 56 80 48

    OR

    80 80 90 80
    106 146 143 56
    120 136 122 80
    164 91 70 48

    -Or is this not even the way to approach this problem? Any advice would be greatly appreciated. Thanks in advance.
     
  2. jcsd
  3. Apr 23, 2014 #2

    Mark44

    Staff: Mentor

    Yes.
    Yes, take the inverse of this matrix. Let's call this matrix A.
    No. What you have here is AT, the transpose of matrix A. You get the transpose of a matrix by writing its rows as columns. For some matrices, the transpose is equal to the inverse, but I doubt that's the case here.
     
  4. Apr 23, 2014 #3
    Okay thats what I thought, just wanted to make sure since when solving something Ax=0 you write the vectors as columns, but I guess here that wouldn't be true since the vector is a multiple of the other vector components.

    Well, I took the inverse of matrix A and the answer was not correct.
    I got the following as inverse:


    \begin{array}{cc}11/2025 & -839/20250 & 11167/324000&2801/288000\\1/270& -16/675&199/5400&-103/4800\\-5/324 & 53/810 & -829/12960 & 253/11520\\1/81 & -1/81 &1/162 & -1/144\end{array}
     
    Last edited: Apr 23, 2014
  5. Apr 23, 2014 #4

    Mark44

    Staff: Mentor

    There's always the possibility that you made a mistake or two along the way. That would prevent your answer from being correct.

    Using an online calculator, this is what I got:
    $$ \frac 1 {2592000}\begin{bmatrix}14080 & -107392 & 89336 & 25209 \\
    9600 & -61440 & 95520 & -55620 \\
    -40000 & 169600 & -165800 & 56925 \\
    32000 & -32000 & 16000 & -18000 \end{bmatrix}$$

    The size of the matrix and the numbers in it make me think that you're not expected to calculate the inverse by hand. Unless you're extremely careful, it's very easy to make a mistake that screws up all the work that follows.
     
  6. Apr 24, 2014 #5
    Yes definitely possible I will cross check my answer with yours and see. I also used an online inverse calculator and got my answer that way since doing it by hand could get messy. I'll let you know.
     
  7. Apr 28, 2014 #6
    Was the inverse of the matrix A correct? I have a similar problem on my online homework and I have spent countless hours trying to get it right.
     
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