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I'm confused on the notation for $\mathbb{R}^{m}$ and $\mathbb{R}^{n}$ and its relationship to linear transformations. We say that $\mathbb{R}^{m}$ is just the space of column vectors since there are n columns, right?

If so, then what do we mean when we say T: $\mathbb{R}^{m} \mapsto \mathbb{R}^{n}$ is equivalent to $T_A(X) = AX$? I watched a video on linear algebra (essence of Linear Algebra) that (only if I'm remembering correctly) that all matrix multiplication was is the change of the standard basis vectors such as $\hat{\imath} \space \hat{\jmath}$ landed on different spots in the same dimension.

What is exactly going on here? Can you have linear transformations that don't depend on the number of rows or columns to be the same? Geometrically, what is going on? It seems like I have no idea what is going on in a matrix at all.

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  • $\begingroup$ Is your question "what is a linear transformation", or "what is matrix multiplication"? $\endgroup$ – GFauxPas Oct 16 '16 at 3:07
  • $\begingroup$ @Jerry When you type "definition linear transformation" in google, you get very precise definitions... $\endgroup$ – imranfat Oct 16 '16 at 4:31
  • $\begingroup$ Sorry, I should have been more clear: what I mean to ask is, what exactly does matrix multiplication imply about a linear transformation geometrically? $\endgroup$ – Jerry Qu Oct 16 '16 at 6:55
  • $\begingroup$ @imranfat When you read the rest of the text, you understand that that is not exactly the question the OP had in mind. $\endgroup$ – Billy Rubina Oct 16 '16 at 7:48
  • $\begingroup$ @OppaHilbertStyle. OP clearified, you are pretty good in ready people's mind :) $\endgroup$ – imranfat Oct 16 '16 at 17:34
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Well. To avoid confusion, let us say a linear transformation is mapping $T:V\rightarrow W$ between two abstract vector spaces over $\mathbb{C}$, $V$ and $W$, with $\dim V =n$ and $\dim W= m$.

We say $T$ is a linear transformation if it satisfies the following properties

  • $T(x+y) = T(x)+T(y) \ \ \forall \ \ x, y \in V$
  • $T(\lambda x) = \lambda T(x) \ \ \forall \ \ \lambda \in \mathbb{C}$.

The definition is very general and has nothing to do with matrices (not yet).

Now, suppose we choose a basis for $V$, say $B_1=\{v_1, \ldots, v_n\}$, and a basis for $W$, say $B_2=\{w_1, w_2, \ldots, w_m\}$. Hence each vectors $v \in V$ will have a unique representation with respect to the basis $B_1$, i.e. we can write \begin{align} v = \alpha_1v_1+\alpha_2v_2+\ldots + \alpha_nv_n. \end{align} If we collect the coefficient into an array say $(\alpha_1, \ldots, \alpha_n)$, then this array uniquely determines $v$ since the representation is unique. In this process of identifying the abstract vector $v$ with its array of coefficients, we have essentially identified $V$ with $\mathbb{R}^n$. Likewise, we can identify $W$ with $\mathbb{R}^m$.

Well. Once we have an identification, we would also like to see how $T$ works under this new identification, i.e. how would the coefficients transform with respect to the linear map $T$? Observe by linearity of $T$, we have \begin{align} T(v) = \alpha_1T(v_1) + \ldots +\alpha_nT(v_n). \ \ (\ast) \end{align} Since $T(v_i) \in W$, then they could be represented in terms of the basis $B_2$, i.e. \begin{align} T(v_j) = \beta_{1j}w_1 + \beta_{2j}w_2 + \ldots + \beta_{mj}w_m = \sum^m_{i=1} \beta_{ij}w_i. \ \ \ (\ast\ast) \end{align} Substitue $(\ast\ast)$ back into $(\ast)$ yields \begin{align} T(v) = \sum^n_{j=1}\alpha_j\sum^m_{i=1}\beta_{ij} w_i = \sum^m_{i=1}\left( \sum^n_{j=1} \beta_{ij}\alpha_j\right)w_i \end{align} which means the coefficients transform as follow \begin{align} (\alpha_1, \ldots, \alpha_n) \in \mathbb{R}^n \longmapsto \left(\sum^n_{j=1} \beta_{1j}\alpha_j, \sum^n_{j=1} \beta_{2j}\alpha_j, \ldots, \sum^n_{j=1} \beta_{mj}\alpha_j \right) \in \mathbb{R}^m \end{align} One should note that the above transformation of the coefficients is actually given by \begin{align} \begin{pmatrix} \beta_{11} & \beta_{12} & \ldots & \beta_{1n}\\ \beta_{21} & \ddots & \ldots & \vdots\\ \vdots & \ldots & \ddots & \vdots\\ \beta_{m1} & \ldots & \ldots & \beta_{mn} \end{pmatrix} \begin{pmatrix} \alpha_1\\ \alpha_2\\ \vdots\\ \alpha_n \end{pmatrix}. \end{align} Hence we could identify $T$ with the transformation of the coefficients via multiplication by a matrix $(b_{ij})$. In general, textbooks like to denote $(b_{ij})$ by $[T]_{B_1, B_2}$.

Example Let us consider the following example to conclude our discussion. Consider $V = P_2$, the space of polynomials with degree at most $2$, and $W = P_1$, the space of polynomials with degree at most $1$. Let us define the linear $T$ given by \begin{align} T(a_0+a_1 x+a_2x^2):= \frac{d}{dx}(a_0+a_1 x+a_2x^2)= a_1+2a_2 x. \end{align} It's clear that $T$ is a linear map from $P_2$ to $P_1$.

Let us choose a basis $B_1=\{1, x, x^2\}$ for $V$ and a basis $B_2=\{1, x\}$ for $W$, then we see \begin{align} \left[ \frac{d}{dx}\right]_{B_1, B_2} = \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 2\\ \end{pmatrix}. \end{align}

Edit: A fast way to determine the matrix representation of a linear transformation is to use the following identity \begin{align} [T]_{B_1, B_2} = ( [T(v_1)]_{B_2}, [T(v_2)]_{B_2},\ldots, [T(v_n)]_{B_2}). \end{align}

For the above example, we see that \begin{align} T(1) = 0= 0\cdot 1 + 0\cdot x, \ \ T(x) = 1 = 1\cdot 1 + 0\cdot x \ \ \text{ and } \ \ T(x^2) = 2x = 0\cdot 1 + 2\cdot x \end{align} where $\{1, x, x^2\}$ is the basis for $V$ which mean \begin{align} [T(1)]_{B_2} = \begin{pmatrix} 0 \\ 0\\ \end{pmatrix}, \ \ [T(x)]_{B_2} = \begin{pmatrix} 1 \\ 0\\ \end{pmatrix} \ \ \text{ and } \ \ [T(x^2)]_{B_2} = \begin{pmatrix} 0 \\ 2\\ \end{pmatrix}. \end{align} So together we have \begin{align} [T]_{B_1, B_2} =( [T(v_1)]_{B_2}, [T(v_2)]_{B_2},\ldots, [T(v_n)]_{B_2})= \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 2\\ \end{pmatrix}. \end{align}

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  • $\begingroup$ This is a great explanation! I like how you showed that the coefficients transform exactly from the basis of V and the basis of W. It is EXACTLY the matrix multiplication with the column vector R^n, which makes perfect sense that you are transforming from R^n to R^m! I'm not exactly clear about the example. How did you get that bij matrix to be exactly the one at the very end? Could you possibly show the multiplication for some coefficients once, just to make it clear? Thanks. $\endgroup$ – Jerry Qu Oct 16 '16 at 7:45
  • $\begingroup$ I will leave it as an exercise. =) $\endgroup$ – Jacky Chong Oct 16 '16 at 7:45
  • $\begingroup$ Oh I see. You simply multiply a0 by zero, a1 by one, and a2 by 2. .00 $\endgroup$ – Jerry Qu Oct 16 '16 at 7:48
  • $\begingroup$ Yup. Try to see if you could reconstruct what I did for this example. $\endgroup$ – Jacky Chong Oct 16 '16 at 7:49
  • $\begingroup$ Here's a question: how did you determine which spot to put 1 and 2 in? I'm still totally confused... $\endgroup$ – Jerry Qu Oct 16 '16 at 8:03
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I know it as "linear map". It is a kind of function between vector spaces.

If $v,w$ are vectors of the vector space $(X,K)$ and $\lambda,\mu$ are elements of the field $K$ then we say that the function $f:(X,K)\to (Y,K)$ between vectors space over the same field is a linear map if

$$f(\lambda v+\mu w)=\lambda f(v)+\mu f(w)$$

In your case $\Bbb R^m$ and $\Bbb R^n$ are the vector spaces, both over the field $\Bbb R$. There are a lot of linear maps, by example the limit function defined as

$$\lim:c(\Bbb R)\to \Bbb R,\quad (x_n)\mapsto x$$

where $c(\Bbb R)$ is the vector space of all convergent sequences composed by elements of $\Bbb R$. For convergent sequences $(x_n)$ and $(y_n)$ and $\alpha,\beta\in\Bbb R$ we have that

$$\lim(\alpha x_n+\beta y_n)=\alpha\lim (x_n)+\beta\lim (y_n)$$

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  • $\begingroup$ I'm not quite sure what a field is, so this explanation is a little beyond me. $\endgroup$ – Jerry Qu Oct 16 '16 at 7:00
  • $\begingroup$ @JerryQu common fields are $\Bbb R$, $\Bbb C$ or $\Bbb Q$. $\endgroup$ – Masacroso Oct 16 '16 at 8:24
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Instead of thinking of vectors as columns of numbers, take the example of points on a plane. These are vectors and to add them use parallelogram law of addition. If you interpret a a vector here as a force with length (distance from the origin) representing the strength of the force and the direction of line from origin to the point as direction of force, you can see that this parallelogram law of addition of vectors comes from physics.

Now in this setup a linear transformation $T$ is a function from the plane to itself that has the property that (i) for two vectors on the plane if you apply the function and then add the resulting two vectors, or add the two vectors first and then apply the function $T$ to the sum both yield the same result.

And further, $T(v)$ scaled by a constant $k$ is the same as $T$ applied to $v$ scaled by $k$.

A very desirable geometrical property linear transformations have is : Image of a parallelogram under $T$ is again a parallelogram.

Consider the transformation of the plane that rotates all the points of the plane by a fixed angle $\theta$. It carries a parallelogram to again a parallelogram (actually a congruent one).

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