Null Spaces And Ranges

Definitions of null space, injectivity, range, and surjectivity. Fundamental theorem of linear maps. Consequences for systems of linear equations.

[ArteomKorotchenya]

Thank you for video, it reminds, stresses concepts and useful when doing exercises.

[apaksoy]

In the proof of a linear map T from V to W not being injective when dim V > dim W, it is stated (similar to the way it is stated in pp. 64 of the book) that:

dim V > dim W
≥ dim range T.

It is explained in the video that what is meant exactly here is dim V > dim W ≥ dim range T. 

But without the explanation in the video, these inequalities are easily interpreted as

dim V > dim W and dim V ≥ dim range T

instead and this leads to some head scratching!

I wish what is meant were expressed in the related inequalities more clearly!

[monislamba]

Hello Prof. Axler
Example 2 of Page No 61 of the book and 5:00 of this video mentions that (2, 0, 1) and (0, 5, 1) form a basis for the Range of T (which is basically R^3)

My question is how can a vector list of length 2 span or form a basis of R^3

[muratcan__]

So I think for proof at 8:50 to make sense for me, the dim of {0} should be zero.

[kavitawagh]

Dear prof., I have one doubt. If I have a set of linearly independent linear maps then will the ranges of all these maps be disjoint subspaces?

[twisthz]

Dear Prof. Axler, on p.61 of the textbook you mention that the differentiation map D:P(R) -> P(R) is surjective. If I am not mistaken, P(R) is the vector space of polynomials with degree m with coefficients in R. However, given that applying D to a polynomial reduces its degree by 1, it seems to me the range of D is of dimension "m" rather than "m+1", the latter being the dimension of P(R). If that is the case, then the range and vector space do not coincide, and D is not surjective. Could you please point out what I am missing. Thank you.