I'm at a complete loss for dual space right now, and linear functionals by association. My notes from lecture and my book are completely unhelpful, and I'm finding myself making up solutions to homework from patterns I'm gleaning from answers I find online or the examples in the book.

The current problem I need to do reads:

Define f (some special non-italicised notation for a linear functional, whatever that is -- I have no idea how to denote that in text and will just toss it into the \$'d limits) $\in(\Bbb R^2)^*$ by f$(x,y) = 2x + y$ and $T:\Bbb R^2 \rightarrow \Bbb R^2$ by $T(x,y) = (3x + 2y, x)$.
(a) Compute $T^t(f)$.
(b) Compute $[T^t]_{\beta^*}(f)$, where $\beta$ is the standard ordered basis for $\Bbb R^2$ and $\beta^* = \{f_1, f_2\}$ is the dual basis, by finding scalars $a, b, c$, and $d$ such that $T^t(f_1) = af_1 + cf_2$ and $T^t(f_2) = bf_1 + df_2$.
(c) Compute $[T]_\beta$ and $([T]_\beta)^t$, and compare your results with (b).

I have no clue where to even start. I thought maybe I could consider f as some sort of thing I could plug into the transformation (like $f = \{2, 1\}$ then $T(2,1) = (8,2)$) but that doesn't get me anywhere, let alone a matrix to transpose. I don't think I understand anything regarding this topic; I can't find any examples in the book that make any sense, my notes are equally cryptic (and most of my lecture time I'm frantically scribbling what the professor writes on the board with no concept of what's going on), and I'm not seeing anything online that's of help either. Usually there are a decent amount of pdf's to search... but not this time.

  • $\begingroup$ What does the superscript $t$ stands for? $\endgroup$ – Jacky Chong Oct 22 '16 at 7:44
  • $\begingroup$ I'm assuming the transpose, e.g. $[T]^{-1}$ $\endgroup$ – Asinine Oct 22 '16 at 7:45
  • $\begingroup$ I don't know if this helps, but one way to think of dual spaces (for finite-dimensional spaces) is the following: any orthonormal coordinate system for $V$ can be defined by the action of an orthonormal dual basis of $V^*$. I.e. the coordinate representation of $v \in V$ is $(f_1(v), f_2(v), \dots, f_n(v) )$, where $(f_1, \dots, f_n)$ is the orthonormal dual basis of $V^*$. Having two bases allows us to compare the before and after of a linear transformation: Given $T: V \to V$, if we want to compare $v$ and $Tv$, the fact that $T$ doesn't affect $V^*$ means that our coordinate system $\endgroup$ – Chill2Macht Oct 22 '16 at 8:27
  • $\begingroup$ doesn't change after applying $T$ to the vectors of $V$. Thus we can compare $v$ and $Tv$ from the perspective of a "neutral observer", $V^*$. In contrast, if we tried to define a coordinate system for $V$ in terms of a decomposition with respect to vectors in $V$, then upon applying $T$, not only would the vectors $v \mapsto Tv$ change, but our coordinate system would too, making us unable to compare $v$ and $Tv$. Keep in mind that any dot product can be thought of as the action of a dual vector on a vector, i.e. $v \cdot w := v^*(w)$ -- this is the difference between row and column vectors. $\endgroup$ – Chill2Macht Oct 22 '16 at 8:32
  • $\begingroup$ In other words it helps to think of dual spaces as a generalization of orthonormal bases: en.wikipedia.org/wiki/Orthonormal_basis#Basic_formula $\endgroup$ – Chill2Macht Oct 22 '16 at 8:35

Let me first give some general context before considering the question at hand. If $V$ is a real finite dimensional vector space, the dual space of $V$ is the space of all linear maps $f \colon V \rightarrow \mathbb{R}$. Such maps $f$ are called linear functional - you feed them vectors in $V$ and they spit out scalars. Given a basis $\beta = (v_1, \dots, v_n)$ for $V$, one can construct a basis $\beta^{*} = (f_1, \dots, f_n)$ for the dual space $V^{*}$ that satisfies $f_i(v_j) = \delta_{ij}$ (where $\delta_{ij} = 1$ if $i = j$ and $0$ otherwise). The dual basis $\beta^{*}$ is determined uniquely by the original basis $\beta$. In particular, this shows that if $V$ is $n$-dimensional then so is the dual space $V^{*}$.

If $T \colon V \rightarrow V$ is a linear map, it induces a linear map $T^{*} \colon V^{*} \rightarrow V^{*}$ between the dual spaces by the formula $T(f)(v) = f(Tv)$ (that is, the linear functional $T(f)$ eats a vector $v \in V$, applies $T$ to it and then applies $f$ to the result).

In part $(a)$, you are asked to compute the linear functional $T^{*}(f)$ (you denote it by $T^{t}(f)$ but I think it is best to reserve this notation only for matrices in order to avoid some confusion). Let us try and do that. The expression $T^{*}(f)$ should be a linear functional on $\mathbb{R}^2$ so let us try and feed it with a vector $(x,y)$:

$$ (T^{*}(f))(x,y) = f(T(x,y)) = f(3x + 2y, x) = 2(3x + 2y) + x = 7x + 4y.$$

Thus, if we set $g(x,y) = 7x + 4y$, we see that $T^{*}(f) = g$.

In part $(b)$, you are asked to compute the matrix representation of the dual operator $T^{*}$ with respect to a given basis $(f_1,f_2)$. The basis $(f_1,f_2)$ is said to be the dual basis to the standard basis $(e_1,e_2)$. Let us try and write $f_1,f_2$ explicitly. A general linear functional $f$ on $\mathbb{R}^2$ has the form $f(x,y) = ax + by$ for some $a,b \in \mathbb{R}$. Writing $f_1(x,y) = ax + by$, we see that it must satisfy

$$ f_1(e_1) = f_1(1,0) = a = 1, f_1(e_2) = f_1(0,1) = b = 0 $$

and so $f_1(x,y) = x$. Similarly, $f_2(x,y) = y$ and so the dual basis acts on a vector $(x,y)$ simply by returning the coordinates of the vector. Now, in order to compute the matrix representation of $T^{*}$ with respect to the basis $(f_1,f_2)$, we must compute $T^{*}(f_1),T^{*}(f_2)$ and express the result in terms of $f_1, f_2$:

$$ T^{*}(f_1) = a f_1 + c f_2, T^{*}(f_2) = bf_1 + c f_2. $$

Having done that, we will know that

$$ [T^{*}]_{\beta^{*}} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$

(this has nothing to do with dual spaces, it simply follows from the definition of what it means to represent an operator as a matrix with respect to a given basis). In our case,

$$ (T^{*}(f_1))(x,y) = f_1(T(x,y)) = f_1(3x + 2y, x) = 3x + 2y = (3f_1 + 2f_2)(x,y), \\ (T^{*}(f_2))(x,y) = f_2(T(x,y)) = f_2(3x + 2y, x) = x = f_1(x,y) $$

and so $T^{*}(f_1) = 3f_1 + 2f_2, T^{*}(f_2) = f_1$ and

$$ [T^{*}]_{\beta^{*}} = \begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix}. $$

Finally, for part $(c)$, we need to compute $[T]_{\beta}$ and so we need to compute $T(e_1),T(e_2)$ and express the result in terms of $e_1,e_2$:

$$ T(e_1) = T(1,0) = (3, 1) = 3e_1 + e_2, \\ T(e_2) = T(0,1) = (2, 0) = 2e_1 + 0 \cdot e_2 $$

and we get

$$ [T]_{\beta} = \begin{pmatrix} 3 & 2 \\ 1 & 0 \end{pmatrix}, \left( [T]_{\beta} \right)^{t} = \begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix}. $$

You might notice that we got $[T^{*}]_{\beta^{*}} = \left( [T]_{\beta} \right)^t$ and in fact you can prove that this will always be the case.

  • $\begingroup$ I'm trying to digest this, and will comment as I grasp (or attempt to) each section. The part following your general context (which lost me completely) seems to state that $T^*(f)$ is better written (or more clearly, at least, by my understanding) as $f(T(x,y))$? This is what frustrates me; why don't they just ask that? $\endgroup$ – Asinine Oct 22 '16 at 9:06
  • $\begingroup$ @Asinine: They just assume you are familiar already with the definition of $T^{*}(f)$. The definition of the linear functional $T^{*}(f)$ is such that $(T^{*}(f))(x,y) = f(T(x,y))$ so is just a matter of "unraveling the definition". Think of someone that doesn't know what is vector addition and tries to solve the exercise $(1,2) + (3,4) = ?$. If you know the definition, you immediately write $(1,2) + (3,4) = (1 + 3, 2 + 4) = (4, 6)$ but if you don't you might wonder "why they just don't ask to compute $(1 + 3, 2 + 4)$?" $\endgroup$ – levap Oct 22 '16 at 9:22
  • $\begingroup$ For part b, why not just state that I'm to set $f$ to the two options for the standard basis for $\Bbb R^2$? That is the one pattern I noticed (and hopefully did correctly in a previous problem): you just set the "answer" to each vector for the given basis and solve for $f$, right? $\endgroup$ – Asinine Oct 22 '16 at 9:24
  • $\begingroup$ That's my issue though; I have no clue as to the definition of $T^*(f)$! I mean, they "give" it in the book (I think?) but I honestly have no idea what they're saying. Most of what I've learned is from examples, and they're being skimpy with those. I can type out all they define things as, and I'm sure most folks answering these questions will understand, but it means nothing to me. $\endgroup$ – Asinine Oct 22 '16 at 9:26
  • $\begingroup$ @Asinine: You don't have to feel bad about it - the notion of a dual space and the dual map is quite abstract and it takes some time getting used to. If you want some intuition about linear functionals, you can think of linear functionals as a generalization of "linear equations". On $\mathbb{R}^2$, a single linear homogeneous equation has the form "$aX + bY = 0$". Throw away the "=0" part and you get an an expressions $aX + bY$ where you think of $X,Y$ not as "concrete numbers" numbers but formal variables in which you can plug numbers and get a result. This is a linear functional. $\endgroup$ – levap Oct 22 '16 at 9:34

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