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Linear Algebra II - Change of Basis

  1. Aug 24, 2015 #1
    1. The problem statement, all variables and given/known data

    From Linear Algebra with applications 7th Edition by Keith Nicholson.
    Chapter 9.2 Example 2.

    Let T: R3 → R3 be defined by T(a,b,c) = (2a-b,b+c,c-3a).
    If B0 denotes the standard basis of R3 and B = {(1,1,0),(1,0,1),(0,1,0)}, find an invertible matrix P such that P-1MB0(T)P=MB(T).

    2. Relevant equations



    3. The attempt at a solution


    I know to find P, I have to first find MB0(T) and MB(T).

    MB0(T) is easy because I have to just find each a, b, and c as linear combinations of B0 and the coefficients are MB0(T).

    However, I'm not sure how to find MB(T). In the textbook, they write:

    MB(T) = [CB(1,1,-3) CB(2,1,-2) CB(-1,1,0)]

    I cannot figure out where they came up with these numbers (1,1,-3), (2,1,-2), and (-1,1,0).

    Please help. Thank you!

     
  2. jcsd
  3. Aug 24, 2015 #2

    RUber

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    Homework Helper

    ##M_B(T) ## is the 3x 3 matrix where the columns are the result of T on each of the basis vectors.
    T(1,1,0) = (1,1,-3).
    So your matrix will look like:
    ##\begin{bmatrix} 1& 2& -1 \\ 1& 1& 1 \\ -3 & -2 & 0 \end{bmatrix}##
     
  4. Aug 24, 2015 #3
    Oh wow thanks! That was simple.

    But in the textbook ##M_B(T)## = ##\begin{bmatrix} 4&4&-1\\-3&-2&0\\-3&-3&2\end{bmatrix}##

    From trying to figure out where (1,1,-3) came from, I figured out that (4,-3,-3) came from the coefficients of the linear combination of each of the basis vectors. But is that wrong?
     
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