Mathematics, 21.08.2019 19:30 clarissa2006

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.

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Answer from: limelight11

a) Assume that $x,y\in\ker L$ , and $\alpha$ is a scalar (a real or complex number).

First. Let us prove that $\ker L$ is not empty. This is easy because L(0_V)=0_W , by linearity. Here, 0_V stands for the zero vector of V, and 0_W stands for the zero vector of W.

Second. Let us prove that $\alpha x\in\ker L$ . By linearity

$L(\alpha x) = \alpha L(x)=\alpha 0_W=0_W$ .

Then, $\alpha x\in\ker L$ .

Third. Let us prove that $y+ x\in\ker L$ . Again, by linearity

L(x+y)=L(x)+L(y) = 0_W + 0_W=0_W .

And the statement readily follows.

b) Assume that and are in range of . Then, there exist $x,y\in V$ such that L(x)=u and L(y)=v .

First. Let us prove that range of is not empty. This is easy because L(0_V)=0_W , by linearity.

Second. Let us prove that $\alpha u$ is on the range of .

$\alpha u = \alpha L(x) = L(\alpha x) = L(z)$ .

Then, there exist an element $z\in V$ such that $L(z)=\alpha u$ . Thus $\alpha u$ is in the range of .

Third. Let us prove that u+v is in the range of .

u+v = L(x)+L(y) = L(x+y)=L(z) .

Then, there exist an element $z\in V$ such that L(z)= u +v . Thus u +v is in the range of .

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.

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Answer from: Quest

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L: v --> w is a linear transformation. prove each of the following (a) ker l is a subspace of v....

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