Question

How do I do these linear algebra questions?

The question is:

Consider the Vector Space V and its subset W given below. Determine whether W forms a subspace of V. If your answer is negative then you must provide which subspace requirement is violated.

(b). V is P5, the vector space of all polynomials in x of degree s5 and W is the set of all polynomials divisible by x – 3. (c). V is P5, the vector space of all polynomials in x of degree s 5 and W is the set of all polynomials having leading coefficient equal to 1. (d). V is the set of all nxn matrices over the field of real numbers and W is the set of all singular matrices. (e). Mnn is the set of all nxn matrices over the field of real numbers and W = {A E Mnn] A2 = A}.

Answer 1

v=ps Then W = { fons Ev: fred is divisble by (x-3) Let fra), gca) EW. Then foxy = h, (x). [(x- ] where hi (a) is C a polynomial of degree less than 5. g(x) = (2-3) h2(a) Where hala) is a boty romial of competits. degree less than 5.

Now Now fees +9(x) = (2-3) [hi(*) the cas] Chicas thecw] is a folynomial of degree less than 5. Hence we see that (x-3) divided fror+gra). This f, gfW ftg tu. Let eff. then efan) = chica). 69-) where child) is a polynomial of degree less than 5. from © wwe see than cfles is divisble by (0-3). 1. Hence of tw. when eEF 1 from ③ and ④ we see that w is a subspace of v = Ps ev=Pg w={set of all polynomial having leading coefficient ?? This take for exth tw and g(x)= x+2fW. we see that fen) tg (2) = 20 +3 €w. as Hore 2 is the leading coefficient. Hence aestito w is not a Subspace of Vaps

> V is the set of all nxn Matrices over the field TR we is the set of all singular Matrices. ve w-{AE R2X2 I det 1=0}. .. take A - [: 7 thon det (A) –o so, meW. take Az = To :). thn. dit (n.) -0. So, y ew. . Now, (41+A3) - [] . ..but det (A1 + A2) = 70. So. (A1+A2) & W. Here we see A, EW, A2 EN but (A1+A₂) & W. Iso, Wis not a subspace of V. e) Man be the set of all Matrices over the field IR. and w={A E Mon 1 42= A } take An- [1 o 1 then A = A. 50. AtW. take B = [1 07.tun B?=B, so Bew. Now, A+B = [37] . Now, (A +)? - (A+B)(4+2) = [:. • ] + (A+B). ..50. (A+B)&W. Here we see that A, BEW but (A+B)¢w.so. Wis not a subspace of W. Man 1