Math for Non-Geeks/Vector space of a linear map

oben Here, we consider the vector space of linear mappings between two given vector spaces. The vector space of linear maps mini|400px|Forming the vector space Let be a field and two -vector spaces. We want to investigate the set of all -linear mappings from to . We give this set (like "homeomorphism", which is another word for "linear map"). So: We will now prove that this set is again a -vector space, so it has some very convenient properties. More precisely, is a subspace of the set of functions from to , called . Let be a field and and two -vector spaces. Then the set of linear maps from to is again a -vector space. It is sufficient to show that is a -subspace of . For this we have to show: If then also . If and then also . We show that is a subspace of . The map is -linear, so . If then also . We have to show that the map is -linear. Additivity of Let . Then Homogeneity of Let and . Then If and then . Let be a linear map and . We must show that the map is -linear. Additivity of Let . Then Homogeneity of Let and . Then Therefore, is a subspace of . The dimension of the vector space of linear mappingsDimension In the following, we calculate the dimension of the vector space of linear maps between two finite-dimensional vector spaces. Dimension of the vector space of linear maps Let be a field and two vector spaces with . Then . To find the dimension of , we have to construct a basis of this space. If we want to define a map from to , we have to define its image in for different basis vectors of . This can again be represented as a linear combination of the different basis vectors of . At first, we chose two bases: Let be a basis of , Let be a basis of . For each we want to define a linear map . By the principle of linear continuation, we can define these uniquely by their values on the basis : Let . This set has elements. To prove the statement of the theorem, we must therefore justify that is a basis of . To do sop, we have to show that is linearly independent and a generating system. is linearly independent Let , such that . We must now show that for all . Let . Then: As the form a basis of , they are linearly independent. Therefore, already follows for all and our fixed . Since this was chosen arbitrarily, now follows for all . The are therefore linearly independent. is a generating system Let . For every we have . Since is a basis of , there is a unique decomposition such that . We will now show that Due to the linearity, this can be verified on the basis vectors . Let be arbitrary. Then, Therefore, is a generating system. This proves the statement of the theorem. A similar statement also applies to infinite-dimensional vector spaces: If or , and , then we also have . However, if and is arbitrary (or vice versa), then we always get . unten