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We could express any vector as
So any vector would have coefficients
that satisfy
from which we can show that any such vector is a linear combination of some other vectors:
More explicitly, we've shown that is a linear combination of the vectors
and
; in other words,
, and in fact these two vectors form a basis for
. But this set does not span all of
because there's no combination of these two vectors that can be used to obtain a constant. We want the transformation to be usable for any vector in
, so we need to add an additional vector extend the basis. We can do this simply by appending
into the spanning set. (Do check that the vectors remain linearly independent.)
Now we want the transformation to map those polynomials for which
to the zero vector. We know which vectors belong to the basis of
, so we need
where the choice of the assignment for is arbitrary, so long as it's non-zero.