Preparation problems for the discussion sections on February 20th and 21st1. Determine which of the following sets are subspaces and give reasons:(a) W1= {abc: a − 2b = c, 4a + 2c = 1},(b) W2= {a − bca + ca − 2b − c: a, b, c ∈ R},(c) W3= {ab: a · b ≥ 0}.(d) W4= {ab: a2+ b2≤ 1}.Solution: a) W1is not a subspace, since the zero vector is not in W1. The zero vector is notin W1, because4 · 0 + 2 · 0 6= 1.b) Since{a − bca + ca − 2b − c: a, b, c ∈ R} = span1011,−100−2,011−1and every span is a subspace, this set is a subspace as well.c) W3is not a vector subspace. Consider the two vectors10and0−1. Both are inW3, because 1 · 0 = 0 · (−1) ≥ 0. But10+0−1=1−1and 1 · (−1) = −1 < 0. Hence1−1is not in W3. Hence W3is not closed under addition.d) This set is not a vector subspace. Consider the vector10. Since 12≤ 1, we have that10is in W4. However, 210=20is not in W4, since 22= 4 > 1. Hence W4is notclosed under scalar multiplication.12. Is H =a + 1a: a in Ra subspace of R2? Why or why not?Is K =a + 1b: a and b in Ra subspace of R2? Why or why not?Solution: The set H is not a subspace, because it does not contain the zero vector00.(Why? Because if there is a in R such thata + 1a=00,then a + 1 = 0 and a = 0. Such an a can not exist). While H is not a subspace, K is asubspace. It is enough to realize thata + 1b: a and b in R=cb: c and b in R= span{10,01} = R2.Note that the first equality holds, because you can take c to be a − 1.3. Is the set H of all matrices of the form2a b3a + b 3ba subspace of M2×2? Explain.Solution: Let A be a matrix in H. There are a, b in R such thatA =2a b3a + b 3b= a2 03 0+ b0 11 3.Hence A is a linear combination (in M2×2) of2 03 0and0 11 3. HenceH = span{2 03 0,0 11 3},and so H is a subspace of M2×2.4. A matrix B is called symmetric if BT= B. Let V be the set of all symmetric 2×2-matrices.Is V a subspace of M2×2?Solution: Yes, V is a subspace of M2×2. We have to check that it contains0 00 0and isclosed under addition and scalar multiplication. First note that0 00 0is in V , because it isobviously symmetric.Now take two matrices A, B in V . So we have AT= A and BT= B. Then we have(A + B)T= AT+ BT= A + B.Hence A + B is in V . So V is closed under addition.Now take a matrix A in V and a scalar r. Since A is in V , we have AT= A. Then we have(rA)T= rAT= rA.Hence rA is in V . So V is closed under scalar
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