Question

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.

Answer 1

Answer:

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

First. Let us prove that range of $L$ 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 $L$.

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

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

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

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

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

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: