Conversation started Apr 30, 2023 at 21:15.
D.C. the III
D.C. the III
Apr 30, 2023 21:15
$$\left(\begin{matrix} m_{11} & m_{12} \\ m_{21} & m_{22} \\ m_{31} & m_{32} \end{matrix} \right) \left(\begin{matrix} a_1 \\ a_2 \end{matrix} \right) \left( \begin{matrix} h_1 \\ h_2 \end{matrix} \right)$$
had to take a quick bathroom break then type this. So this would be $T(a)h$, now $Ma$ will be a vector
leslie townes
leslie townes
dc: the "(a)" in Df(a) is a label telling you what point you're considering Df at. of course Df(a) is also a linear map and hence takes a vector as an argument. but this aspect of the "(a)" notation isn't too compatible with actually writing out what a is when you're applying Df(a) to a given h
D.C. the III
D.C. the III
but I am not in the place where I could consider the two vectors to have a product of $(Ma)^t h$ right?
leslie townes
leslie townes
all of this being made more confusing by what Df actually is in the linear case
sigh
D.C. the III
D.C. the III
You said the key thing in "applying Df(a) to a given h"....
leslie townes
leslie townes
in particular, the components of a in "Df(a)" are not something with a role in the kind of matrix-vector product you might write down for "Df(a) h"
D.C. the III
D.C. the III
Apr 30, 2023 21:21
I kind of get what your saying broadly, but I am confused by it....
leslie townes
leslie townes
i don't think it's a good idea to signal the potential dependence of Df(a) on the point a by writing out the components of a in the middle of expression involving Df(a) or Df(a) applied to something else
Ted Shifrin
Ted Shifrin
How are you multiplying a 3-vector by a 2-vector?
leslie townes
leslie townes
so up above, i would scrub out the column vector "a" between the 3x2 matrix (presumably representing Df at the point you're fixing, which you might be calling a) and the vector h
just understand that for general f, the 3x2 matrix Df(a) has entries that depend on a
Ted Shifrin
Ted Shifrin
Leslie’s point is: Think back to how you worked with $a$ when you typically found the matrix $Df(a)$.
D.C. the III
D.C. the III
I was doing that right now, thinking about it. I should've put some ellipses to show me contemplating
so you guy's know I'm not wasting your time
leslie townes
leslie townes
Apr 30, 2023 21:26
for example if f(x) = x^2 in calc 1, and i evaluate f' at for example at a = 3 i get that f'(3) = 2*3 = 6. but i wouldn't write this as 6(3), even though 6 definitely is what f' is when evaluated at 3, and even though in general, what f' is would depend on what point i'm evaluating it at
i don't need to expressly note the dependence on my choice of a by writing the value of a next to what the derivative is
Ted Shifrin
Ted Shifrin
@leslie I no longer have any hair left.
D.C. the III
D.C. the III
Sorry to stress you so much sir.
Ok I do have a better idea of what you guys were talking about. Just going to need a minute to really reconcile the notion
leslie townes
leslie townes
it's a problem of juxtaposition being used to denote both "evaluation at" in general, and also, with linear maps, that's matrix multiplication with the thing you're evaluating at
in something like Df(a)(h), the juxtaposition of Df with (a) means "Df evaluated at a" [which is not, in general, a product of some matrix, on the one hand, and the column vector a, on the other - in particular, Df(a) has type "matrix" and not type "vector"], while the juxtaposition of Df(a) with (h) means both evaluation of the linear map Df(a) at h, and a matrix-vector multiplication of the matrix Df(a) with the column vector h
D.C. the III
D.C. the III
the way you summed it up is what I aspire to be able to do
leslie townes
leslie townes
it looks like maybe sometimes your book up above is trying to keep this a little separate by not putting "evaluation at" in parentheses when doing matrix multiplication. they write Df(a)h and not Df(a)(h)
but i wouldn't count on other books to do the same
Ted Shifrin
Ted Shifrin
Apr 30, 2023 21:36
I typically wrote $Ah$ for matrix product, as opposed to $T(h)$ for the value of a linear function.
D.C. the III
D.C. the III
well at least I have a better comprehension of this now so when I do encounter it again in higher courses
@TedShifrin Right. It was me using $T$ and not thinking about it as a matrix in the derivative form that messed things up......but it was a good learning lesson
Ted Shifrin
Ted Shifrin
Well, we do say that a linear map is its own derivative at any point .
Parse the sentence carefully.
D.C. the III
D.C. the III
It's how I ended up here. Needed to parse more finely than I did....now I can get back to your lectures.
 
Conversation ended Apr 30, 2023 at 21:39.