The existence part of this result means that we can find a linear map that takes on whatever values we wish on the vectors in a basis. The uniqueness part of this result means that a linear map is completely determined by its values on a basis.

Linear Map Lemma

Suppose is a basis of and . Then there exists a unique linear map such that

for each .

Proof.

The idea of this proof is that if we know what a linear map does the basis vectors of , then that linear map is completely determined for all of .

First, we show the existence of the desired linear map . We can explicitly define by

where are arbitrary elements of . The list is a basis of . Thus, the equation above does indeed define a function from to (because each element of can be uniquely written in the form ).

Every vector can be written uniquely as a linear combination of basis vectors, so is well-defined because the coefficients in are uniquely determined. For one of the basis vectors , taking and the other ‘s equal to in the equation above shows that .

Now, we seek to show additivity and homogeneity (properties of a linear map) If with and , then

Similarly, if and , then

Thus, is a linear map from to .

To prove uniqueness, now suppose that and that for each . Let . Then, the homogeneity of implies that for each . The additivity of now implies that

Thus, is uniquely determined on by the equation above. Because is a basis of , this implies that is uniquely determined on , as desired.