Practice question · Put in order
Order the steps of the subspace test applied to a subset W of a vector space V, cheapest check first.
- Take arbitrary u and v in W and check that u + v is still in W
- Take an arbitrary v in W and scalar c and check that cv is still in W
- Check that W is nonempty by finding the zero vector in it
- Conclude that the remaining axioms are inherited from V, so W is a subspace
Hints
- The whole point of the test is to avoid checking all the axioms.
- One check disqualifies a set instantly and costs nothing, do that one first.
Show the answer
- Check that W is nonempty by finding the zero vector in it
- Take arbitrary u and v in W and check that u + v is still in W
- Take an arbitrary v in W and scalar c and check that cv is still in W
- Conclude that the remaining axioms are inherited from V, so W is a subspace
Why
The zero-vector check is free and rejects most impostors on sight. The two closure checks then do all the real work, and every other axiom comes free from the surrounding space V, which is why the test is only two conditions long.
Practise Subspaces
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