For each of the following sets S. divide whether it is a subspace of R3. Circle

Question

For each of the following sets S. divide whether it is a subspace of R3. Circle 'Yes' or 'No'. middot If you circle 'Yes', do one (and only one) of the following (i) write down a system of homogeneous linear equations (in standard form) such that S is the solution set. or (ii) write down a small set of vectors of which the subspace is the span. middot If you circle 'No', write down an explicit example (using actual numbers!) of a violation of one of the closure axioms. S = {(a + ba -b2a +b)a,b any real numbers}. Yes No

Explanation / Answer

What you want to do is confirm that the set satisfies the closure axioms, that is the set is closed under scalar addition and multiplication and lies in R3. When attempting to solve a problem such as this, it may help to try and think intuitively what the solution may be before starting. Whether the set seems to be a subspace will dictate what you are trying to prove. For part a), our intuition would tell us that since we are merely applying arithmetic to real numbers, the Set is really just generating R1. To see this more clearly, let b = 0. We can do this since b can be any real number. Now the element of S is governed strictly by 'a'. Since 'a' can be any real number, we get all of R. R is obviously a subspace of R3, and now we must show either i) or ii). ii) is easier, since a basis for R1 is obviously .

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