Advanced Linear Algebra, Lecture 2.5: The transpose of a linear map

Advanced Linear Algebra, Lecture 2.5: The transpose of a linear map

If T is a linear map from X to U, then this induces a linear map T' from U' to X', that "precomposes" linear scalar functions with T. That is, it maps ℓ to ℓT. In our scalar product notation, this means that (T'ℓ,x)=(ℓ,Tx). We unpack this definition in the context of systems of equation, and how it relations to column and row vectors. We give a very straightforward proof that the annihilator of the range of T is the nullspace of its transpose. This is the abstract version of the fact that the column space is orthogonal to the left nullspace, and that the row space is orthogonal to the null space. An easy corollary is that the range T and its transpose T' have the same dimension, which is the abstract version of the fact that the column space and row space have the same dimension, the rank of T.

[squarerootofpi]

In slide 5 I suppose you could technically factor out the x on the left because all the li, tij, xj are elements in a field that commute. But I'd go for putting x on the right where they came from so it's easier to follow. And because matrices usually don't commute. This one happens to because it is simplified to a dot product.

My 2 cents. Thanks for the videos.

[fsaldan]

The penultimate expression on 15'25 has i and j swapped.

[dacianbonta]

So much effort; so much handwaving;
to tiptoe around and obfuscate
the fundamental tensorial nature of matrices

[dacianbonta]

without the scalar product smoke'n'mirrors students would have a chance at REAL understanding,
instead of replacing conceptual insight with infantile trickery
this obfuscation-by-simplification, otoh, very american.

[catdanceable]

thank you tor such a detailed eloquent lesson.