Math 231 — Hw 16 solved

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In class, we continued to discuss how we to use matrix representations of linear maps between
vector spaces. Suppose we define the horrible map you came up with in class, T : R
4 → R
3
• T(1, 2, 3, 4) = (5, 6, 7)
• T(11, 10, 9, 8) = (2, 3, 1)
• T(1, 5, 7, 2) = (7, 8, 6)
• T(0, 0, 0, 1) = (9, 1, 1)
Our strategy for handling this is to define a map U : R
4 → R
4
that translates from a nice basis
{(1, 0, 0, 0),(0, 1, 0, 0), …} to the basis upon which T is defined: {(1, 2, 3, 4),(11, 10, 9, 8), …}.
We did this in class and defined the matrix representation for U to be
M(U) =


25
24
−17
6
15
8
0
1
24
1
6
−1
8
0
−1
2
1
−1
2
0
−7
2
8
−11
2
1


If we define the set of mapped elements in R
3
to be {(5, 6, 7),(2, 3, 1),(7, 8, 6),(9, 1, 1)}, then
the matrix representation of T is
M(T) =


1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1


But this is not enough to get the full answer. Why? Suppose there is a vector v for which
we apply M(U) and then M(T) and we get the result (1, 0, 1, 1). This vector represents:
1 · (5, 6, 7) + 0 · (2, 3, 1) + 1 · (7, 8, 6) + 1 · (9, 1, 1). Ideally we’d like the result to actually be
our answer directly and not have to do this extra step. So we can define a map S that goes
from {(5, 6, 7),(2, 3, 1),(7, 8, 6),(9, 1, 1)} to {(1, 0, 0),(0, 1, 0),(0, 0, 1)}.
Define M(S).