Section 1.10

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3. Let be a linear map and a subspace. Show that: is a subspace of .

Proof:
Suppose . Then . But is a subspace and so . But is linear so that so that . Thus, is closed under vector addition. Now suppose and . Then and since is a subspace, . But again is linear so . This means . Hence is closed under scalar multiplication. Therefore is a subspace of .


10. Show that if and are subspaces, then is also a subspace.

Proof:
Suppose . Then and . But is a subspace and so . Also is a subspace so . This means . On the other hand . Thus, is closed under vector addition. Now suppose and . Then and . But and are subspaces so and . That means . This means . Hence is closed under scalar multiplication. Therefore is a subspace of .


12. Let be a linear map and consider the graph (a) Show that is a subspace.

Proof:
Suppose . Then . Here I used the fact that is linear which means . Thus, is closed under vector addition. Now suppose and . Then . Again I used the linearity property to conclude . Hence is closed under scalar multiplication. Therefore is a subspace of .


(b) Show that the map that sends to is an isomorphism.

Proof:
Call this map . That is . First I will show this map is linear:

and . Thus is linear. Now to show is bijective. If , then so is trivially onto. In fact, we essentially chose to the codomain of our function to just be the image/range of the map to ensure it was onto. Now to show is one-to-one. Suppose . Then . But two ordered pairs are equal if and only if both components are equal. That is, . Thus is one-to-one. Therefore is an isomorphism.