Question

2. (8 points) Let Vị and V2 be the subspaces of R3 given respectively by the equations: V : x1 + 22 + x3 = 0 V2: 2x1 – 22 – x3 = 0 (a) (2 points) Find a basis for Vił. (b) (2 points) Find a basis for Vit.
Was wondering if someone could show me how to solve these. I am stuck... :/

FYI, this is linear algebra.

Answer 1

Note that (11000) & spa , toto to sop pn = ((10,0) - Co - (1) A (1010) - 1 (1101-1) - - - 2 (27) (10,0) +/4,0, ) +(5 -32 T Thus (tit, t) 1 is a basis for it 2n, -H2-Ryao 27 V : nithet az=0 V2 : 2x, -2₂-213=0 @nthetn3=0 nasoninne So, (nereins) = (nime, an, he) - (170,-1) + m2 (0,1,-1) minecir . ? (1,0,-1), (0.11-Dy is a basis for vi we find orthogonal basis for vi by Cram Schmidt method. take, &,= (1,0,-1) 12 = (0,1-1) B. de - 22 A = (0,1,-1) - 221481 RiiMiM, = (0,1,-1) - 1 (110,-1) = (0,1,-1) + (-2,0, ) = (- / 2,1,- ) thus ? (1,0,-1), (Ž 11, - 2) is an orthogonal basis for prov, 7 ng=anian2 (running) = (none, 22. - 2.) ani (110,2) the (0, 1,-1), mer take, 2,= (1,0,2),4₂ = (0-1). Miax = (1,0,2) 2A) = (0,1,-1) - (-2) (11012) = (0,1,-1) + (24,10, 1/2) = (2,2, - 2) so, & my is orthug and bring P = 22 - A. for Vzi

note that, (1,000) & V₂ as 2-0-070 = (10.) – 7 / (10.2) - (2/2) (1, 2) (10) - 6,0,?)-G5 = ) so, the basis for vzt is, (110,2). ( 211, - ²) K uch - T - 70-) - (10,2) - (92(190,--((by - (11012) + (Ž 10, ²) + (-211, - 2) - key (1,1,1) so, {(1,0,-1), (- 2,1,- ), (1,1)y is the required set of orthogonal baris normalizing { (110,-1), t (1-2, 1) , & (!!!!!)Y I is the orthenoome) basis of R3 Where { t (110,-1), to (11-2, 1)) is a basis of Vi and to (11-2, 1), 7(1/10) - is a basis it v₂. o note that, (1,0,2) & V as Hutato so, addition of (1,0,2) with the basis of vi will provide a basis of R {(1,0,-1), (0,1,-1), (1,0,21T is a basis of R² .. where ? (110,-1), (0.1,-1) is a basis for V and }(6.1,-1), ((10,27 is a basis for de he normalize (11012) as follows,
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