Organizatio nal• Interested in joining class committe e?meet∼3 times to discuss ideas you may have for improving classNext: bases, dimension and suchhttp://abstrusegoose.com/235Armin [email protected] Ax = 0 and Ax = bColumn spacesDefinition 1. The column space Col(A) of a matrix A is the span of t he columns of A.IfA = [a1an], then Col(A) = span{a1,, an}.• In other words, b is in Col(A) if and only if Ax = b has a solution.Why? BecauseAx =x1a1++xnanis the linear combination of columns of A with coefficientsgiven byx.• If A is m × n, th e n Col(A) is a subspac e of Rm.Why? Because any span is a space.Example 2. Find a matrix A such that W = Col(A) whereW =2x − y3y7x + y: x, y in R.Solution. Note that2x − y3y7x + y= x207+ y−131.Hence,W = span207,−131= Col2 −10 37 1.Armin [email protected] spacesDefinition 3. T he null space of a matrix A isNul(A) = {x : Ax = 0}.In other words, if A is m × n, then its null space consists of those vectors x ∈ Rnwhich solve thehomogeneous equationAx = 0.Theorem 4. If A is m × n, then Nul(A) is a subspace of Rn.Proof. We chec k that Nul(A) satisfies the conditions of a subspace:• Nul(A) contains 0 because A0 = 0.• If Ax = 0 and Ay = 0, then A(x + y) = Ax + Ay = 0.Hence, Nul(A) is closed under addition.• If Ax = 0, then A(cx) = cAx = 0.Hence, Nul(A) is closed under scalar multiplication. Armin [email protected] Ax = 0 yields an explicit des c r iption of Nu l(A).By that we mean a descripti on as the span of some vectors.Example 5. Find an explicit descrip tion of Nul(A) whereA =3 6 6 3 96 12 13 0 3.Solution.3 6 6 3 96 12 13 0 3>R2→ R2−2R13 6 6 3 90 0 1 −6 −15>R1→13R11 2 2 1 30 0 1 −6 −15>R1→ R1−2R21 2 0 13 330 0 1 −6 −15From the RREF we read off a parametric description of the solutions x to Ax = 0. Notethatx2, x4, x5ar e free.x =x1x2x3x4x5=−2x2− 1 3x4− 3 3x5x26x4+ 1 5x5x4x5= x2−21000+ x4−130610+ x5−3301501In other words,Nul(A) = span−21000,−130610,−3301501.Note. The number of vectors in the s panning set for Nul(A) as derived above (whichis as small as possib le) equals the number o f free variables inAx = 0.Armin [email protected] look at solutions t o Ax = bTheorem 6. Let xpbe a solution of the equati on Ax = b.Then ever y solution toAx = b is of the form x = xp+ xn, where xnis a solution tothe homogeneous equation Ax = 0.• In other words, {x : Ax = b} = xp+ Nul(A).• We often call xpa particul ar solution.The theorem then says that every solution toAx = b is the sum of a fixed chosen particularsolution and some solution toAx = 0.Proof. Let x be an ot her solutio n to Ax = b.We need to show thatxn= x − xpis in Nul(A).A(x − xp) = Ax − Axp= b − b = 0 Example 7. Let A =1 3 3 22 6 9 7−1 −3 3 4and b =155.Using the RREF, fin d a parametric description of the solutions toAx = b:1 3 3 2 12 6 9 7 5−1 −3 3 4 5>R2→ R2−2R1R3→R3+R11 3 3 2 10 0 3 3 30 0 6 6 6>R3→ R3−2R21 3 3 2 10 0 3 3 30 0 0 0 0>R2→13R21 3 3 2 10 0 1 1 10 0 0 0 0>R1→ R1−3R21 3 0 −1 −20 0 1 1 10 0 0 0 0Every solution to Ax = b is therefore of the form:x =x1x2x3x4=−2 − 3x2+ x4x21 − x4x4Armin [email protected]=−2010xp+ x2−3100+ x410−11elements of N ul(A)We can see nicely how every solution is the sum of a particular solution xpand solution sto Ax = 0.Note. A conven ient way to just find a particular solution is to set all free variables tozero (here, x2= 0 and x4= 0).Of course, any other choice for the free variables will result in a particular solution.For instance, x2= 1 and x4= 1 we would get xp=−4101.Practice problems• True or fals e ?◦ The solutions to the equation Ax = b form a vector s pace.No, with the only exception ofb = 0.◦ The solutions to the equation Ax = 0 form a vector s pace.Yes. This is the null spaceNul(A).Example 8. Is the given set W a vector space?If possible, expressW as the column or null space of s ome ma trix A.(a) W =(xyz: 5x = y + 2z)(b) W =(xyz: 5x − 1 = y + 2z)(c) W =(xyx + y: x, y in R)Example 9. Find an explicit descrip tion of Nul(A) whereA =1 3 5 00 1 4 −2.Armin
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