Solving Linear Equations • Trying to solve the equation ○ Ax=y ○ where x∈V is sought, y∈W is given ○ V,W vector spaces ○ T:V→W linear transformation • Example 1 ○ {█(a_11 x_1+a_12 x_2+…+a_1n x_n=y_1@⋮@a_m1 x_1+a_m2 x_2+…+a_mn x_n=y_m )┤ ○ Let § x=(█(x_1@⋮@x_n ))∈Rn § y=(█(y_1@⋮@y_n ))∈Rn § A: Rn→Rn § A(█(x_1@⋮@x_n ))=(█(a_11 x_1+…+a_1n x_n@⋮@a_m1 x_1+…+a_mn x_n )) § A(█(x_1@⋮@x_n ))=(■8(a_11&⋯&a_1n@⋮&⋱&⋮@a_m1&⋯&a_mn ))(█(x_1@⋮@x_n )) § A with respect to standard bases of Rn, Rm ○ Then the linear equations could be represented as § Ax=y • Theorem 1 ○ Statement § If A:V→W is linear § and if u,v∈V are solutions to Ax=y § (i.e. if Au=y, and Av=y) § Then u−v∈N(A) ○ Proof § A(u−v)=Au−Av=y−y=0 ○ Text version § If 〖Ax〗_p=y then for all x∈V with Ax=y § There is an x_ℎ∈N(A) with x=x_p+x_ℎ • Theorem 2 ○ Statement § If u is a solution to Ax=y § and if w∈N(A) § then u+w is also a solution of Ax=y ○ Proof § A(u+w)=Au+Aw=y+0=y ○ Text version § For all x_p with 〖Ax〗_p=y and for all x_ℎ∈N(A) § A(x_p+x_h)=y • General solution ○ Homogeneous equation Ax=0 ○ Inhomogeneous equation § Ax=y, where y≠0 ○ The general solution to Ax=y is of the form § x_gen=x_p+x_ℎ, where § x_p is a particular solution § x_h is the general solution to the homogeneous equation ○ Set of all solutions § {x∈V│Ax=y}={x_p+x_h■8(Ax_p=y@x_hN(A) )} ○ Proof § We are given one solution x_p of Ax=y § If x_ℎ∈N(A) § then by definition 〖Ax〗_ℎ=0 § and hence A(x_p+x_h)=y § ⇒x_p+x_ℎ∈{x∈V│Ax=y} § Conversely if Ax=y then § A(x−x_p )=Ax−Ax_p=y−y=0 § So x_ℎ≝x−x_p∈N(A) • Example 2 ○ Solve the linear equation{█(x_1+2x_2−x_3=7@2x_1−x_2+x_3=4)┤ ○ Setup § V=R3⇒x=(█(x_1@x_2@x_3 )) § W=R2⇒y=(█(7@4)) § A: R3→R2 is matrix multiplication with [■8(1&2&−1@2&−1&1)] ○ Range(A) § ={Ax│x∈R3 } § ={all possible y∈R2 for which Ax=y has a solution} ○ By Rank–nullity theorem § dim〖N(A)+dim〖Range(A)〗=dim〖R3 〗=3〗 dim〖Range(A)〗 dimN(A) 0 3 1 2 2 1 ○ Solving the equation by Gaussian Elimination § [■8(1&2&−1@2&−1&1) │ ■8(7@4)]→[■8(1&0&1/5@0&1&−3/5) │ ■8(3@2)] § {█(x_1+1/5 x_3=3@x_2−3/5 x_3=2)┤ § Let x_3=5c § Then{█(x_2=2+3/5 x_3=2+3c@x_2=3−1/5 x_3=3−c)┤ § Therefore the general solution is § x=[█(3−c@2+3c@5c)]=⏟([█(3@2@0)] )┬(x_p )+c⏟([█(−1@3@5)] )┬(x_h) • Example 3 ○ Given § V=W={functions y:[a,b]→R § A:V→W where Af=f^′+xf ○ Question § Solve dy/dx+xy=x ○ The general solution is in form of § x+x_p+x_ℎ ○ It
