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Hw2 similar answer

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School
McGill University**We aren't endorsed by this school
Course
MATH 223
Subject
Mathematics
Date
Dec 23, 2024
Pages
4
Uploaded by SuperMorning8633
Problem 1. Suppose that three vectors u, v, and w are linearly independent in some vector space. (a) [5 points| Are the three vectorsu + v , u+ w, and v+ w also lincarly independent? Give a complete proof of your assertion. (b) [5 points| Is span{u,v,w} equal to span{u + v.u + w,v+ w}? Explain your reasoning. Solution (a) Let ai(u+v) + az(u + w) + az(v + w) = 0. We can write the above equation as (a1 + a2)u + (ay + az)v + (a2 + az)w = 0. Since u, v, and w are linearly independent, we have (aj +a2) = (a; + a3) = (a2 + a3) = 0. The only solution is a; = ap = ag = 0. Therefore, u + v , u + w, and v + w are linearly independent (b) Since u + v , u + w, and v + w are linear combinations of u, v, and w, we have span{u + v,u + w,v + w} C span{u, v, w}. 1. We have shown that both (u,v,w) and (u+ v, u + w,v + w) are linearly independent, then dim(spanf{u + v,u +w,v+w}) = 3 = dim(span{u, v, w}). 2. Or we can show u, v, and w are linear combinations of u + v , u + w, and v + w u=(u+v)+ (ut+w) - (v+w) v=(w+v)+@+w) - (ut+w) w=(u+w)+ (v+w) — (u+v) Then, span{u,v,w} C span{u + v, u + w,v + Therefore, span{u+ v,u + w,v + w} = span{u, v, w}. Problem 2. Let W, be the subsct of R" consisting of all vectors that read the same forwands as 1) € We but (1,2,1,1,2,0) ¢ W. backwards. For example, (1,2 (a) [3 points| Prove that W, is a subspace of R™. (b) |7 points| Find a basis for W, and find the dimension of W.
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Solution (a) We will show that for even n, the proof is similar for odd n.. Taking any — rz..n) € W, !/1».'/1) € Wh. Then a+b <.r, +y1, T2 + Y2 T2 + Y2, 71 + _:/,) € Wh. Therefore, W, is closure under vector addition. Let and ¢ be a scalar, then c-a= (er,cr Hence, W, is closure under scalar multiplication. Since the zero vector reads the same forwards and backwards given that 0 € w,. Therefore, ,, is a subspace of R", (b) e When n is even: we have any vector a = (.n.!‘z ------ T, Tg,..., T2, .n) € Wy can be written as a=x1(1,0,0,...0,0,1) + 22(0.1.0...0,1,0) + - + 22(0,0,0...1,1...0,0,0). Also, for vectors (1,0,0,...0,0,1),(0,1,0...0,1,0)...(0,0,0...1,1...0,0,0), non-zero entries appears in different positions, so we can prove they are linearly independent. Therefore basis = span{(1,0,0,...0,0,1),(0.1,0...0,1,0)...(0.0,0...1,1...0,0,0)} and dim(W,) = %. o When n is odd: the length of the vector is n = 2k+ 1 and we will have k vectors of the form (0,...,1,0...0,1,...,0) and the vector (0,0,0 ...1...0,0,0). Therefore, use the same argument as when 1 is even, we have basis = span{(1.0.0....0.0.1).(0.1.0...0.1.0)...(0.0....1.0.1.0....0.). (0,0,0...1...0.0.0)} and dim(W,,) L Problem 3. Recall that a square matriz A is called *symmetric* if AT = A, where AT is the transpose. Let S be the set of all symmetric 3 x 3 matrices with entries from R. Note that S is a subspace of Myy3(R) (you do not need to prove this). (a) [4 points] Find a basis of, and give the dimension of, S. (b) [6 points] Let W be any 4-dimensional subspace of Msys(R). Prove that W contains a non-zero symmetric matriz.
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Solution For any polynomial p(x) € P, (R), we can write it as pa) = P TP | p@) —p(x) v ple)tp(=2) AT p(z)=p(—2) o : Since 2P i an even function (in £) and 2225 g an odd function (in 0). We have proved Taking any p(z) € EN O, then p(x) = p(—x) = —p(x), which gives p(x) = 0, so EN O = {0}. Therefore, P,(R) = E - O. a
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