MATH SOLVE

2 months ago

Q:
# L :V --> W is a linear transformation. Prove each of the following (a) ker L is a subspace of V. (b) range L is a subspace of W.

Accepted Solution

A:

Answer:a) Assume that [tex]x,y\in\ker L[/tex], and [tex]\alpha[/tex] is a scalar (a real or complex number).First. Let us prove that [tex]\ker L[/tex] is not empty. This is easy because [tex]L(0_V)=0_W[/tex], by linearity. Here, [tex]0_V[/tex] stands for the zero vector of V, and [tex]0_W[/tex] stands for the zero vector of W.Second. Let us prove that [tex]\alpha x\in\ker L[/tex]. By linearity[tex]L(\alpha x) = \alpha L(x)=\alpha 0_W=0_W[/tex].Then, [tex]\alpha x\in\ker L[/tex].Third. Let us prove that [tex]y+ x\in\ker L[/tex]. Again, by linearity[tex]L(x+y)=L(x)+L(y) = 0_W + 0_W=0_W[/tex].And the statement readily follows.b) Assume that [tex]u[/tex] and [tex]v[/tex] are in range of [tex]L[/tex]. Then, there exist [tex]x,y\in V[/tex] such that [tex]L(x)=u[/tex] and [tex]L(y)=v[/tex].First. Let us prove that range of [tex]L[/tex] is not empty. This is easy because [tex]L(0_V)=0_W[/tex], by linearity.Second. Let us prove that [tex]\alpha u[/tex] is on the range of [tex]L[/tex]. [tex] \alpha u = \alpha L(x) = L(\alpha x) = L(z) [/tex].Then, there exist an element [tex]z\in V[/tex] such that [tex]L(z)=\alpha u[/tex]. Thus [tex]\alpha u[/tex] is in the range of [tex]L[/tex].Third. Let us prove that [tex] u+v[/tex] is in the range of [tex]L[/tex].[tex] u+v = L(x)+L(y) = L(x+y)=L(z) [/tex].Then, there exist an element [tex]z\in V[/tex] such that [tex]L(z)= u +v[/tex]. Thus [tex]u +v[/tex] is in the range of [tex]L[/tex].Notice that in this second part of the problem we used the linearity in the reverse order, compared with the first part of the exercise.Step-by-step explanation: