Question

If y1,y2,.....,yn are linear functionals on F^n such that each of the coordinate projections belongs to their span, does it always follow that the linear transformation T from F^n to F^n defined by T(x) = (y1(x),....., yn(x) is invertible?
Please explain me what does "the coordinate projections belongs to their span" mean

Answer 1

The solution is on the book, I don't get how they use the information
they says
Suppose x =(x1,...,xn) is a vector belongs to kernel T, so that (y1(x),....yn(x) =(0,....,0)
Since the coordinate projections \(p_j\) belong to the span of y1,....,yn
It follows that for each j there exist scalars alpha1,...,alphan such that
\(p_j=\sum_k \alpha_k y_k \) consequence \(p_j(x) =\sum_k \alpha_ky_k =0\)

Answer 2

@zzr0ck3r

Answer 3

for each j, which implies, of course, that x =0; in other words, the ker (T)=0---> T is invertible

Answer 4

@raffle_snaffle Thank you

Answer 5

yes, that is true.

Answer 6

@kirbykirby

Answer 7

Are they talking about orthogonal projections?

Answer 8

general projection, not orthogonal

Answer 9

Oh I would imagine that since the projections are "constructed" in the same vector space as the vectors resulting from that projection, then the projection would belong to the span of the vectors. Like if you had 2 vectors in \(\mathbb{R}^2\), then the projection will not lie in R3, R4, etc. It says in R2. So, say you have vectors \(\bf{a}\) and \(\bf{b}\) (not linearly dependent), then R2 is the span of vectors a and b. Thus, the projection resulting from a and b also lies in R2, and so it belongs to the span of a and b.

Answer 10

That makes sense to me. Thank you very much. :)

Answer 11

:)