Exercises 3.11 Matching Exercises
1. Matching Problem, Dates.
2. Matching Problem, Derivatives.
3. Matching Problem, Linear Algebra.
Match each subspace with a basis for that subspace. (You may assume that each set is really a basis for at least one of the subspaces.)
Each putative basis is a subset of exactly one of the three subspaces. So for each subspace, two of the three sets can be ruled out by simply testing that the vectors of the basis are members of the subspace, via the membership criteria.
- \(\left\{\langle x,y,z\rangle\mid - y + z = 0\right\}\)
- \(\left\{\langle -4, 3, 3\rangle, \langle 3, -2, -2 \rangle\right\}\)
- \(\left\{\langle x,y,z\rangle\mid -3x - 5y + z = 0\right\}\)
- \(\left\{\langle -4, 3, 3\rangle, \langle 5, -4, -5 \rangle\right\}\)
- \(\left\{\langle x,y,z\rangle\mid -2x - 5y + 2z = 0\right\}\)
- \(\left\{\langle 3, -2, -2 \rangle, \langle 5, -4, -5 \rangle\right\}\)
Hint.
For openers, a basis for a subspace must be a subset of the subspace.
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