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In the page 2 of Linear algebra explained in four pages reference, it has a box describing the relationship between functions and linear transformation. It states that the set of zeroes of a function ($f(x)=0$) is analogous to the null space of a matrix ($A \vec x=\vec 0$), and the function inverse ($f^{-1}(x)$) is analogous to the matrix inverse ($A^{-1} \vec x$).

Is there a functional analogy of the matrix transpose ($A^\text{T}$), and the functional analogy of the null space of a transpose ($A^\text{T} \vec x = \vec 0$)?

Ming-Tang
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1 Answers1

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Yes these analogues exist, but you will probably not find them very helpful. If $g:X\to Y$ is a function, then the analogue of the transpose for $g$ is the right-composition $f\mapsto f\circ g$, mapping any $f:Y\to\Bbb R$ to the composite function $f\circ g:X\to\Bbb R$, defined as usual by $f\circ g:y\mapsto f(g(y))$.

The analogue of the nullspace of the transpose is the set of all $f$ such that $f\circ g=0$ identically, i.e., those $f:Y\to\Bbb R$ that take the value $0$ everywhere on the image of $g$.

One reason this is more useful for linear functions than for functions in general is that the image of a linear map is always a linear subspace of the codomain. So as soon as the linear map fails to be surjective (have its whole codomain as image) it fails so rather dramatically (the image has lower dimension that the codomain). Since kernels of linear maps are also subspaces, the set of linear functions vanishing on the image of $g$ now becomes an interesting thing to study.

As an aside, the text you are citing from seems to be orthogonal to the intent of your question. It's abstract says:

A lot of knowledge buzz awaits you if you choose to follow the path of understanding, instead of trying to memorize a bunch of formulas.

I don't know exactly how to read the intention of that, but the text focusses heavily on formulas to memorize, and explains virtually nothing (in spite of its title). For instance multiple formulas for dot and cross products are given, stated as if they are defined in all vector spaces, which they are not; and there is a lot more like that. I therefore assume "buzz" to be used in a pejorative and dismissive way here, much like the "bullshit" in the title of the book this is taken from. In any case, I find the text quite a bad start for anyone who would choose to follow the path of understanding in linear algebra.

Marc van Leeuwen
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    nice explanation. like your views on how to learn linear algebra. anyone or anything that claims that you can reduce linear algebra to four pages of symbols is not telling you much. they are new snake oil salesmen. – abel Nov 30 '14 at 15:16
  • In the first paragraph I think it should be "If $g:X\to Y$ is a function, then the analogue of the transpose for $g$ is the right-composition $f\mapsto f\circ g$, mapping any $f:Y\to\Bbb R$ to the composite function $f\circ g:X\to\Bbb R$, defined as usual by $f\circ g:x\mapsto f(g(x))$" – rych Jan 20 '15 at 01:33
  • @rych: Yes, of course. I've corrected it now; I cannot imagine how I managed to get $X$ and $Y$ mixed up where the transpose is the archetype of contravariance. – Marc van Leeuwen Jan 20 '15 at 06:04