(Sec:LinTrafo)= # Linear transformations (Subsec:LinTrafo:MatrixTrafo)= ## Introduction Until now we have used matrices in the context of linear systems. The equation $$ A\mathbf{x} = \mathbf{b}, $$ where $A$ is an $m \times n$-matrix, is just a concise way to write down a system of $m$ linear equations in $n$ unknowns. A different way to look at this matrix equation is to consider it as an input-output system: the left-hand side $A\mathbf{x}$ can be seen as a mapping that sends an "input" $\mathbf{x}$ to an "output" $\mathbf{y}= A\mathbf{x}$. For instance, in computer graphics, typically points describing a 3D object have to be converted to points in 2D, to be able to visualise them on a screen. Or, in a dynamical system, a matrix $A$ may describe how a system evolves from a "state" $\mathbf{x}_{k}$ at time $k$ to a state $\mathbf{x}_{k+1}$ at time $k+1$ via $$ \mathbf{x}_{k+1} = A\mathbf{x}_{k}. $$ A "state" may be anything ranging from a set of particles at certain positions, a set of pixels describing a minion, concentrations of chemical substances in a reactor tank, to population sizes of different species. Thinking mathematically we would describe such an input-output interpretation as a transformation (or: function, map, mapping, operator, etc. ). $$ T: \mathbb{R}^n \to \mathbb{R}^m. $$ We will see that these matrix transformations have two characteristic properties which makes them the protagonists of the more general linear algebra concept of a **linear transformation**. (Subsec:MatrixTrafo)= ## Matrix transformations Let $A$ be an $m\times n$-matrix. We can in a natural way associate a transformation $T_A:\mathbb{R}^n \to \mathbb{R}^m$ to the matrix $A$. ::::::{prf:definition} :label: Def:LinTrafo:MatrixTransformation The transformation $T_A$ corresponding to the $m\times n$-matrix $A$ is the mapping defined by $$ T_A(\mathbf{x}) = A\mathbf{x} \quad \text{or } \quad T_A:\mathbf{x} \mapsto A\mathbf{x}, $$ where $\mathbf{x} \in \mathbb{R}^n$. We call such a mapping a **matrix transformation**. Conversely we say that the matrix $A$ **represents** the transformation $T_A$. :::::: As a first example consider the following. ::::::{prf:example} :label: Ex:LinTrafo:FirstMatrixTrafo The transformation corresponding to the matrix $A = \begin{pmatrix} 1 & 2 & 0\\ 1 & 2 & 1 \end{pmatrix}$ is defined by $$ T_A(\mathbf{x}) = \begin{pmatrix} 1 & 2 & 0\\ 1 & 2 & 1 \end{pmatrix}\mathbf{x}. $$ We have, for instance $$ \begin{pmatrix} 1 & 2 & 0\\ 1 & 2 & 1 \end{pmatrix} \begin{pmatrix} 1\\1\\1 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix} \quad \text{and} \quad \begin{pmatrix} 1 & 2 & 0\\ 1 & 2 & 1 \end{pmatrix} \begin{pmatrix} 2\\-1\\0 \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix}. $$ According to the definition of the matrix-vector product we can also write :::{math} :label: Eq:LinTrafo:AxIsLinearCombination A\mathbf{x} = \begin{pmatrix} 1 & 2 & 0\\ 1 & 2 & 1 \end{pmatrix} \begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix} = x_1 \begin{pmatrix} 1\\ 1 \end{pmatrix}+ x_2 \begin{pmatrix} 2 \\ 2 \end{pmatrix}+ x_3 \begin{pmatrix} 0\\ 1 \end{pmatrix}. ::: :::::: We recall that for a transformation $T$ from a domain $D$ to a codomain $E$ the range $R= R_T$ is defined as the set of all images of elements of $D$ in $E$: $$ R_T = \{\text{ all images } T(x), \, \text{ for } x \text{ in }D\}. $$ ::::::{prf:remark} :label: Ex:LinTrafo:FirstMatrixTrafoContinued From Equation {eq}`Eq:LinTrafo:AxIsLinearCombination` it is clear that the range of the matrix transformation in {prf:ref}`Ex:LinTrafo:FirstMatrixTrafo` consists of all linear combinations of the three columns of $A$: $$ \operatorname{Range}(T_A) = \Span{ \begin{pmatrix} 1\\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 2 \end{pmatrix}, \begin{pmatrix} 0\\ 1 \end{pmatrix}}. $$ In a later section ({numref}`Sec:SubspacesRn`) we will call this the **column space** of the matrix $A$. :::::: The first example leads to a first property of matrix transformations: ::::::{prf:proposition} :label: Prop:LinTrafo:RangeTA Suppose $$ A = \begin{pmatrix} \mathbf{a}_1 & \mathbf{a}_2 & \cdots & \mathbf{a}_n \end{pmatrix} $$ is an $m\times n$-matrix. Then the range of the matrix transformation corresponding to $A$ is the span of the columns of $A$: $$ \operatorname{Range}(T_A) = \Span{\mathbf{a}_1, \mathbf{a}_2,\ldots,\mathbf{a}_n }. $$ :::::: ::::::{prf:example} :label: Ex:LinTrafo:SecondMatrixTrafo The matrix $$ A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix} $$ leads to the transformation $$ T: \mathbb{R}^2 \to \mathbb{R}^3, \quad T \left(\begin{pmatrix} x \\ y \end{pmatrix}\right)= \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \\0 \end{pmatrix}. $$ This transformation "embeds" the plane $\mathbb{R}^2$ into the space $\mathbb{R}^3$, as depicted in {numref}`Figure %s `. ```{applet} :url: linear_transformation/embed_r2_r3 :fig: Images/Fig-LinTrafo-EmbedR2R3.svg :name: Fig:LinTrafo:EmbedR2R3 :status: reviewed :class: dark-light $T$ embeds $\mathbb{R}^2$ into $\mathbb{R}^3$. ``` The range of this transformation is the span of the two vectors $$ \mathbf{e}_1 = \begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix} \quad \text{and} \quad \mathbf{e}_2 = \begin{pmatrix} 0\\ 1 \\ 0 \end{pmatrix}, $$ which is the $xy$-plane in $\mathbb{R}^3$. :::::: For $2\times2$ and $3\times3$-matrices the transformations often have a geometric interpretation, as the following example illustrates. ::::::{prf:example} :label: Eq:LinTrafo:SkewProjection The transformation corresponding to the matrix $$ A = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} $$ is the mapping $$ T: \mathbb{R}^2 \to \mathbb{R}^2, \quad T\left(\begin{pmatrix} x \\ y \end{pmatrix}\right)= \begin{pmatrix} x +y \\ 0 \end{pmatrix}. $$ First we observe that the range of this transformation consists of all multiples of the vector $ \begin{pmatrix} 1 \\ 0 \end{pmatrix} $, i.e. the $x$-axis in the plane. Second, let us find the set of points/vectors that is mapped to an arbitrary point $\begin{pmatrix} c \\ 0 \end{pmatrix}$ in the range. For this we solve $$ A\mathbf{x} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} c \\ 0 \end{pmatrix}, \quad \text{so} \quad \begin{pmatrix} x+y \\ 0 \end{pmatrix} = \begin{pmatrix} c \\ 0 \end{pmatrix}. $$ The points whose coordinates satisfy this equation all lie on the line described by the equation $$ x + y = c. $$ So what the mapping does is to send all points on a line $\mathcal{L}:x + y = c$ to the point $(c,0)$, which is the intersecting of this line with the $x$-axis. An alternative way to describe it: it is the skew projection, in the direction $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ onto the $x$-axis. See {numref}`Figure %s `. ::::{figure} Images/Fig-LinTrafo-SkewProjection.svg :name: Fig:LinTrafo:SkewProjection :class: dark-light The transformation of {prf:ref}`Eq:LinTrafo:SkewProjection`. :::: :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/2db421fe-0649-4c7e-9624-9fdf0123565d?id=122271 :label: grasple_exercise_3_1_1T :dropdown: :description: Finding out which vectors are in the range of a linear transformation. :::::: We close this subsection with an example of a matrix transformation representing a very elementary dynamical system. ::::::{prf:example} :label: Ex:LinTrafo:MigrationModel Consider a model with two cities between which over a fixed period of time migrations take place. Say in a period of ten years 90\% of the inhabitants in city $A$ stay in city $A$ and 10\% move to city $B$. From city $B$ 20\% of the citizens move to $A$, so 80\% stay in city $B$. The following table contains the relevant statistics: $$ \begin{array}{c|cc|} & A & B \\ \hline A & 0.9 & 0.2 \\ B & 0.1 & 0.8 \\ \hline \end{array} $$ For instance, if at time $0$ the population in city $A$ amounts to $50$ (thousand) and in city $B$ live $100$ (thousand) people, then at the end of one period the population in city $A$ amounts to $$ 0.9 \times 50 + 0.2 \times 100 = 65. $$ Likewise for city $B$. If we denote the population sizes after $k$ periods by a vector $$ \mathbf{x}_k = \begin{pmatrix} x_k \\ y_k \end{pmatrix} $$ it follows that $$ \begin{pmatrix} x_{k+1} \\ y_{k+1} \end{pmatrix} = \begin{pmatrix} 0.9x_k + 0.2y_k \\0.1x_k + 0.8y_k \end{pmatrix}, \quad \text{i.e., } \mathbf{x}_{k+1} = \begin{pmatrix} 0.9 & 0.2 \\ 0.1 & 0.8 \end{pmatrix} \begin{pmatrix} x_k \\ y_k \end{pmatrix} = M \mathbf{x}_{k}. $$ The $M$ stands for migration matrix. Obviously this model can be generalised to a "world" with any number of cities. :::::: (Subsec:LinTrafo:LinTrafo)= ## Linear transformations In the previous section we saw that the matrix transformation $\mathbf{y}=A\mathbf{x}$ can also be seen as a mapping $T(\mathbf{x}) = A\mathbf{x}$. This mapping has two characteristic properties on which we will focus in this section. ::::::{prf:definition} :label: Dfn:LinTrafo:LinTrafo A **linear transformation** is a function $T$ from $\mathbb{R}^n$ to $\mathbb{R}^m$ that has the following properties
  1. For all vectors $\mathbf{v}_1,\,\mathbf{v}_2$ in $\mathbb{R}^n$: $$ T(\mathbf{v}_1+\mathbf{v}_2) = T(\mathbf{v}_1) + T(\mathbf{v}_2). $$
  2. For all vectors $\mathbf{v}$ in $\mathbb{R}^n$ and all scalars $c$ in $\mathbb{R}$: $$ T(c\mathbf{v}) = c\,T(\mathbf{v}). $$
:::::: ::::::{exercise} :label: Exc:LinTrafo:ImageofZeroVector Show that a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$ always sends the zero vector $\mathbf{0}_n$ in $\R^n$ to the zero vector $\mathbf{0}_m$ in $\R^m$. Thus, if $ T:\mathbb{R}^n \to\mathbb{R}^m$ is a linear transformation, then $T(\mathbf{0}_n) = \mathbf{0}_m$. :::::: ::::::{admonition} Solution to {numref}`Exc:LinTrafo:ImageofZeroVector` :class: solution, dropdown If $ T:\mathbb{R}^n \to\mathbb{R}^m$ is linear, and $\vect{v}$ is any vector in $\R^n$, then $\mathbf{0}_n = 0\vect{v}$. From the second property in {prf:ref}`Dfn:LinTrafo:LinTrafo` it follows that $$ T(\mathbf{0}_n) = T(0\vect{v}) = 0\,T(\vect{v}) = \mathbf{0}_m. $$ :::::: ::::::{prf:example} :label: Ex:LinTrafo:FirstLinearMap Consider the map $T:\mathbb{R}^2\rightarrow\mathbb{R}^3$ that sends each vector $\begin{pmatrix} x \\ y \end{pmatrix}$ in $\mathbb{R}^2$ to the vector $\begin{pmatrix} x \\ y \\ 0 \end{pmatrix}$ in $\mathbb{R}^3$. Let us check that this a linear map. For that, we need to check the two properties in the definition. For property (i) we take two arbitrary vectors $$ \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} \quad \text{ and }\quad \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} \quad \text{in} \quad \mathbb{R}^2, $$ and see: $$ T\left(\begin{pmatrix} x_1 \\ y_1 \end{pmatrix} + \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} \right)= T \left(\begin{pmatrix} x_1+x_2 \\ y_1+y_2 \end{pmatrix}\right)= \begin{pmatrix} x_1 + x_2 \\ y_1 + y_2 \\ 0 \end{pmatrix} = \begin{pmatrix} x_1 \\ y_1 \\ 0 \end{pmatrix} + \begin{pmatrix} x_2 \\ y_2 \\ 0 \end{pmatrix}. $$ This last vector indeed equals $$ T\left(\begin{pmatrix} x_1 \\ y_1 \end{pmatrix}\right)+ T\left(\begin{pmatrix} x_2 \\ y_2 \end{pmatrix}\right). $$ Similarly, for the second property, given any scalar $c$, $$ T\left(c \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}\right)= T \left(\begin{pmatrix} c x_1 \\ cy_1 \end{pmatrix}\right)= \begin{pmatrix} c x_1 \\ c y_1 \\ 0 \end{pmatrix} = c \begin{pmatrix} x_1 \\ y_1 \\ 0 \end{pmatrix}= cT \left(\begin{pmatrix} x_1 \\ y_1 \end{pmatrix}\right). $$ So indeed $T$ has the two properties of a linear transformation. :::::: ::::::{prf:example} :label: Ex:LinTrafo:NotLinearMap Consider the mapping $T:\mathbb{R}^2\rightarrow\mathbb{R}^2$ that sends each vector $ \begin{pmatrix} x \\ y \end{pmatrix}$ in $\mathbb{R}^2$ to the vector $\begin{pmatrix} x+y \\ xy \end{pmatrix}$: $$ T: \begin{pmatrix} x \\ y \end{pmatrix} \mapsto \begin{pmatrix} x+y \\ xy \end{pmatrix} $$ This mapping is **not** a linear transformation. $$ T \left(\begin{pmatrix} 1 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ 2 \end{pmatrix}\right)= T \left(\begin{pmatrix} 2 \\ 3 \end{pmatrix}\right) = \begin{pmatrix} 5 \\ 6 \end{pmatrix}, $$ whereas $$ T \left(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\right)+ T \left(\begin{pmatrix} 1 \\ 2 \end{pmatrix}\right)= \begin{pmatrix} 2 \\ 1 \end{pmatrix} + \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix} \,\neq\, \begin{pmatrix} 5 \\ 6 \end{pmatrix} . $$ The second requirement of a linear transformation is violated as well: $$ T\left(3 \begin{pmatrix} 1 \\ 1 \end{pmatrix}\right)= T \left(\begin{pmatrix} 3 \\ 3 \end{pmatrix}\right)= \begin{pmatrix} 6 \\ 9 \end{pmatrix} \,\,\neq\,\, 3\,T \left(\begin{pmatrix} 1 \\ 1 \end{pmatrix} \right)= 3 \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \end{pmatrix}. $$ :::::: ::::::{exercise} :label: Exc:LinTrafo:T(x)=x+p Let $\mathbf{p}$ be a non-zero vector in $\mathbb{R}^2$. Is the translation $$ T\!:\mathbb{R}^2 \to \mathbb{R}^2, \quad \mathbf{x} \mapsto \mathbf{x} + \mathbf{p} $$ a linear transformation? :::::: ::::::{admonition} Solution to {numref}`Exc:LinTrafo:T(x)=x+p` :class: solution, dropdown The transformation defined by $T(\vect{x}) = \vect{x} + \vect{p}$, with $\vect{p}\neq \vect{0}$ does not have any of the two properties of a linear transformation. For instance, since $\vect{p}+\vect{p} \neq \vect{p}$, $$ T(\vect{x}+\vect{y}) = \vect{x}+\vect{y} + \vect{p} \neq T(\vect{x})+T(\vect{y}) = \vect{x}+ \vect{p} +\vect{y} + \vect{p}. $$ :::::: Note that {prf:ref}`Ex:LinTrafo:FirstLinearMap` was in fact the first example of a matrix transformation in the Subsection {ref}`Subsec:LinTrafo:MatrixTrafo`: $$ \begin{pmatrix} x \\ y \end{pmatrix} \mapsto \begin{pmatrix} x \\ y \\ 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0&1 \\ 0&0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} $$ As we will see: **any** linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$ is a matrix transformation. The converse is true as well. This is the content of the next proposition. ::::::{prf:proposition} :label: Prop:LinTrafo:MatrixTrafoIsLinear Each matrix transformation is a linear transformation. :::::: ::::::{admonition} Proof of {prf:ref}`Prop:LinTrafo:MatrixTrafoIsLinear` :class: tudproof This is a direct consequence of the two properties of the matrix-vector product ({prf:ref}`Prop:MatVecProduct:Linearity`) that say $$ A\,(\mathbf{x}+\mathbf{y} ) = A\mathbf{x} + A\mathbf{y} \quad \text{and} \quad A\,(c\mathbf{x}) = c\,A\mathbf{x}. $$ :::::: ::::::{prf:proposition} :label: Prop:LinTrafo:CompositionLintrafos Suppose $T: \mathbb{R}^n\to\mathbb{R}^m$ and $S:\mathbb{R}^m\to\mathbb{R}^p$ are linear transformations. Then the transformation $S\circ T:\mathbb{R}^n\to\mathbb{R}^p $ defined by $$ S\circ T(\mathbf{x}) = S(T(\mathbf{x})) $$ is a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^p$. :::::: ::::::{prf:remark} :label: Rmk:LinTrafo:Composition The transformation $S\circ T$ is called the **composition** of the two transformations $S$ and $T$. It is best read as _"$S$ after $T$"_. :::::: ::::::{admonition} Proof of {prf:ref}`Prop:LinTrafo:CompositionLintrafos` :class: tudproof Suppose that $$ T(\mathbf{x}+\mathbf{y}) = T(\mathbf{x})+T(\mathbf{y})\quad \text{and} \quad T(c\mathbf{x}) = cT(\mathbf{x}), \quad \text{for}\,\, \mathbf{x}, \mathbf{y} \quad \text{in } \mathbb{R}^n, \,\, c \text{ in } \mathbb{R} $$ and likewise for $S$. Then $$ \begin{array}{rl} S\circ T(\mathbf{x}+\mathbf{y}) = S(T(\mathbf{x}+\mathbf{y})) = S( T(\mathbf{x})+T(\mathbf{y})) \!\!\!\!& = S(T(\mathbf{x})) + S(T(\mathbf{y})) \\ & = S\circ T(\mathbf{x}) + S\circ T(\mathbf{y}) \end{array} $$ and $$ S\circ T(c\mathbf{x}) = S(T(c\mathbf{x})) = S(cT(\mathbf{x})) = c\,S(T(\mathbf{x})) = c\,S\circ T(\mathbf{x}). $$ Hence $S\circ T$ satisfies the two requirements of a linear transformation. :::::: In words: the composition/concatenation of two linear transformations is itself a linear transformation. ::::::{exercise} :label: Exc:LinTrafo:CombiningLinTrafos There are other ways to combine linear transformations. The sum $S = T_1 + T_2$ of two linear transformation $T_1,T_2: \mathbb{R}^n \to \mathbb{R}^m$ is defined as follows $$ S: \mathbb{R}^n \to \mathbb{R}^m, \quad S(\mathbf{x}) = T_1(\mathbf{x}) + T_2(\mathbf{x}). $$ And the (scalar) multiple $T_3 = cT_1$ is the transformation $$ T_3: \mathbb{R}^n \to \mathbb{R}^m, \quad T_3(\mathbf{x}) = cT_1(\mathbf{x}). $$ Show that $S$ and $T_3$ are again linear transformations. :::::: ::::::{admonition} Solution to {numref}`Exc:LinTrafo:CombiningLinTrafos` :class: solution, dropdown The properties of the linear transformation $T_1$ and $T_2$ carry over to $S$ and $T_3$ in the following way. We check the properties one by one. For the sum $S$ we have
  1. For all vectors $\mathbf{v}_1,\,\mathbf{v}_2$ in $\R^n$ $$ \begin{array}{rcl} S(\mathbf{v}_1+\mathbf{v}_2) &=& T_1(\mathbf{v}_1+\mathbf{v}_2) + T_2(\mathbf{v}_1+\mathbf{v}_2)\\ &=& T_1(\mathbf{v}_1) + T_1(\mathbf{v}_2) + T_2(\mathbf{v}_1) + T_2(\mathbf{v}_2)\\ &=& T_1(\mathbf{v}_1) + T_2(\mathbf{v}_1) + T_1(\mathbf{v}_2) + T_2(\mathbf{v}_2)\\ &=& S(\mathbf{v}_1)+S(\mathbf{v}_2). \end{array} $$
  2. Likewise, for all vectors $\mathbf{v}$ in $\mathbb{R}^n$ and all scalars $c$ in $\mathbb{R}$: $$ \begin{array}{rcl} S(c\mathbf{v}) &=& T_1(c\mathbf{v})+T_2(c\mathbf{v}) \\ &=& cT_1(\mathbf{v})+cT_2(\mathbf{v}) \\ &=& c \big(T_1(\mathbf{v})+cT_2(\mathbf{v})\big)\\ &=& cS(\mathbf{v}). \end{array} $$
The linearity of $T_3$ is verified in a similar manner. :::::: ::::::{exercise} :label: Exc:LinTrafo:LineSegments This exercise sheds some light on the geometry behind linear transformations. We restrict ourselves to linear transformations in the plane, but the ideas can be generalised. Suppose $T$ is a linear transformation from $\R^2$ to $\R^2$. Show that the image of a line segment $\mathcal{S}$ between two points $P$ and $Q$ is either a line segment or a single point. :::::: ::::::{admonition} Solution to {numref}`Exc:LinTrafo:LineSegments` :class: solution, dropdown Suppose $\vect{p} = \overrightarrow{OP}$ and $\vect{q} = \overrightarrow{OQ}$ are the position vectors, and let $\vect{r} = \overrightarrow{PQ} = \vect{q} - \vect{p}$. Then the line segment between $\vect{p}$ and $\vect{q}$ consist of all vectors $\vect{v} = \vect{p} + t\vect{r}$, where $t$ runs from $0$ to $1$. By the linearity of $T$ it follows that the image of $\mathcal{S}$ consist of all vectors $T(\vect{p} + t\vect{r}) = T(\vect{p}) + tT(\vect{r})$, with $0 \leq t \leq 1$. This describes the line segment between the points $T(\vect{p})$ and $T(\vect{p}) + T(\vect{r})$. Note that $T(\vect{p}) + T(\vect{r}) = T(\vect{p} + \vect{r}) = T(\vect{q})$. If by any chance $T(\vect{r}) = \mathbf{0}$, the segment 'shrinks' to a single point. :::::: And now, let us return to matrix transformations. (Subsec:LinTrafo:LinTrafoeqMatrixTrafo)= ## Standard matrix for a linear transformation We have seen that every matrix transformation is a linear transformation. In this subsection we will show that conversely every linear transformation $T:\mathbb{R}^n \to \mathbb{R}^m$ can be represented by a matrix transformation. The key to construct a matrix that represents a given linear transformation lies in the following proposition. ::::::{prf:proposition} :label: Prop:LinTrafo:ExtendedLinearity Suppose $T:\mathbb{R}^n\rightarrow\mathbb{R}^m$ is a linear transformation. Then the following property holds: for each set of vectors $\mathbf{x}_1, \ldots, \mathbf{x}_k$ in $\mathbb{R}^n$ and each set of numbers $c_1,\ldots,c_k$ in $\mathbb{R}$: :::::{math} :label: Eq:LinTrafo:LinComb T(c_1\mathbf{x}_1+c_2 \mathbf{x}_2+ \cdots +c_k \mathbf{x}_k) = c_1T(\mathbf{x}_1)+c_2T(\mathbf{x}_2)+ \cdots +c_kT( \mathbf{x}_k). ::::: :::::: In words: for any linear transformation _the image of a linear combination of vectors is equal to the linear combination of their images_. ::::::{admonition} Proof of {prf:ref}`Prop:LinTrafo:ExtendedLinearity` :class: tudproof Suppose $T:\mathbb{R}^n\rightarrow\mathbb{R}^m$ is a linear transformation. So we have $$ \text{(i) } T(\mathbf{x}+\mathbf{y}) = T(\mathbf{x})+T(\mathbf{y}) \quad\text{and} \quad \text{(ii) } T(c\mathbf{x}) = c T(\mathbf{x}). $$ First apply rule (i) to split the term on the left in Equation {eq}`Eq:LinTrafo:LinComb` into $k$ terms: $$ \begin{array}{ccl} T(c_1\mathbf{x}_1+c_2 \mathbf{x}_2+ \cdots +c_k \mathbf{x}_k) &=& T(c_1\mathbf{x}_1)+T(c_2 \mathbf{x}_2+ \cdots +c_k \mathbf{x}_k) \\ &\vdots& \\ &=& T(c_1\mathbf{x}_1)+T(c_2 \mathbf{x}_2)+ \cdots + T(c_k \mathbf{x}_k) \end{array} $$ and then apply rule (ii) to each term. :::::: ::::::{prf:example} :label: Ex:LinTrafo:ExtendedLinearity Suppose $T: \mathbb{R}^3 \to \mathbb{R}^2$ is a linear transformation, and we know that for $$ \vect{a}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \quad \vect{a}_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \quad \vect{a}_3 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} $$ the images under $T$ are given by $$ T(\vect{a}_1) = \vect{b}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}, \quad T(\vect{a}_2) = \vect{b}_2 = \begin{pmatrix} 3 \\ -1 \end{pmatrix}, \quad \text{and} \quad T(\vect{a}_3) = \vect{b}_3 = \begin{pmatrix} 2 \\ -2 \end{pmatrix}. $$ Then for the vector $$ \vect{v} = \begin{pmatrix} 4 \\ 1 \\ -1 \end{pmatrix} = 3 \vect{a}_1 + 2 \vect{a}_2 - 1 \vect{a}_3 $$ it follows that $$ T(\vect{v}) = 3 \vect{b}_1 + 2 \vect{b}_2 + (-1) \vect{b}_3 = 3 \begin{pmatrix} 1 \\ 2 \end{pmatrix} + 2 \begin{pmatrix} 3 \\ -1 \end{pmatrix} + (-1) \begin{pmatrix} 2 \\ -2 \end{pmatrix}= \begin{pmatrix} 7 \\ 6 \end{pmatrix}. $$ :::::: The central idea illustrated in {prf:ref}`Ex:LinTrafo:ExtendedLinearity`, which is in fact a direct consequence of {prf:ref}`Prop:LinTrafo:ExtendedLinearity`, is the following: a linear transformation $T$ from $\mathbb{R}^n$ to $\mathbb{R}^m$ is completely specified by the images $ T(\mathbf{a}_1), T(\mathbf{a}_2), \ldots , T(\mathbf{a}_n)$ of a set of vectors $\{\mathbf{a}_1, \mathbf{a}_2, \ldots, \mathbf{a}_n\}$ that spans $\mathbb{R}^n$. The simplest set of vectors that spans the whole space $\mathbb{R}^n$ is the standard basis for $\mathbb{R}^n$ which was introduced in the Section {ref}`Sec:LinearCombinations`. Recall that this is the set of vectors :::{math} :label: Eq:LinTrafo:StandardBasis \left(\vect{e}_1,\mathbf{e}_2, \ldots, \mathbf{e}_n\right)= \left(\begin{pmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \quad \cdots \quad , \begin{pmatrix} 0 \\ 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix}\right). ::: The next example gives an illustration of the above, and it also leads the way to the construction of a matrix for an arbitrary linear transformation. ::::::{prf:example} :label: Ex:LinTrafo:StandardMatrixIntro Suppose $T$ is a linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$ for which $$ T(\mathbf{e}_1) = \mathbf{a}_1 = \begin{pmatrix} 1 \\2 \end{pmatrix}, \quad T(\mathbf{e}_2) = \mathbf{a}_2 = \begin{pmatrix} 4 \\3 \end{pmatrix}. $$ Then for an arbitrary vector $$ \mathbf{x} = \begin{pmatrix} x_1\\x_2 \end{pmatrix} = x_1 \begin{pmatrix} 1\\0 \end{pmatrix} + x_2 \begin{pmatrix} 0\\1 \end{pmatrix} = x_1\mathbf{e}_1 + x_2\mathbf{e}_2, $$ it follows that $$ \begin{array}{rcl} T(\mathbf{x}) &=& x_1T(\mathbf{e}_1) + x_2T(\mathbf{e}_2) \\ &=& x_1 \begin{pmatrix} 1 \\2 \end{pmatrix} + x_2 \begin{pmatrix} 4 \\3 \end{pmatrix} \,\,=\,\,\, \begin{pmatrix} 1 &4 \\2 &3 \end{pmatrix}\mathbf{x}. \end{array} $$ So we see that $$ T(\mathbf{x}) = A \mathbf{x}, \quad\text{where} \quad A = \begin{pmatrix} T(\mathbf{e}_1) & T(\mathbf{e}_2) \end{pmatrix}. $$ :::::: ::::::{exercise} :label: Exc:LinTrafo:MatrixForFirstExample Show that the procedure of {prf:ref}`Ex:LinTrafo:StandardMatrixIntro` applied to the linear transformation of {prf:ref}`Ex:LinTrafo:FirstLinearMap` indeed yields the matrix $$ A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}. $$ :::::: ::::{admonition} Solution to {numref}`Exc:LinTrafo:MatrixForFirstExample` :class: solution, dropdown Consider the linear transformation $T:\mathbb{R}^2\rightarrow\mathbb{R}^3$ that sends each vector $ \begin{pmatrix} x \\ y \end{pmatrix}$ in $\mathbb{R}^2$ to the vector $\begin{pmatrix} x \\ y \\ 0 \end{pmatrix}$. It holds that $$ T(\vect{e}_1) = T\left(\begin{pmatrix} 1\\ 0 \end{pmatrix}\right) = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \quad T(\vect{e}_2) = T\left(\begin{pmatrix} 0\\ 1 \end{pmatrix}\right) = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}. $$ We find that for an arbitrary vector $\begin{pmatrix} x\\ y \end{pmatrix} = x\begin{pmatrix} 1\\ 0 \end{pmatrix}+y\begin{pmatrix} 0\\ 1 \end{pmatrix}$ it holds that $$ T\left(\begin{pmatrix} x\\ y \end{pmatrix}\right) = xT(\vect{e}_1) + yT(\vect{e}_2) = x\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}+ y\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1\\ 0 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}. $$ :::: The reasoning of {prf:ref}`Ex:LinTrafo:StandardMatrixIntro` can be generalised. This is the content of the next theorem. ::::::{prf:theorem} :label: Thm:LinTrafo:LinTrafo=MatrixTrafo Each linear transformation $T$ from $\mathbb{R}^n$ to $\mathbb{R}^m$ is a matrix transformation. More specific, if $T: \mathbb{R}^n \to \mathbb{R}^m$ is linear, then for each $\mathbf{x}$ in $\mathbb{R}^n$ :::::{math} :label: Eq:Lintrafo:StandardMatrix T(\mathbf{x}) = A\mathbf{x}, \quad \text{where} \quad A = \begin{pmatrix} T(\mathbf{e}_1) & T(\mathbf{e}_2) & \cdots & T(\mathbf{e}_n) \end{pmatrix}. ::::: :::::: ::::::{admonition} Proof of {prf:ref}`Thm:LinTrafo:LinTrafo=MatrixTrafo` :class: tudproof We can more or less copy the derivation in {prf:ref}`Ex:LinTrafo:StandardMatrixIntro`. First of all, any vector $\mathbf{x}$ is a linear combination of the standard basis: $$ \mathbf{x} = \begin{pmatrix} x_1\\x_2\\ \vdots \\ x_n \end{pmatrix} = x_1 \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} + x_2 \begin{pmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{pmatrix} + \cdots + x_n \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix}, $$ i.e., $$ \mathbf{x} = x_1 \mathbf{e}_1 + x_2 \mathbf{e}_2 + \cdots + x_n \mathbf{e}_n. $$ From {prf:ref}`Prop:LinTrafo:ExtendedLinearity` it follows that $$ T( \mathbf{x}) = x_1 T(\mathbf{e}_1) + x_2 T(\mathbf{e}_2) + \cdots + x_n T(\mathbf{e}_n). $$ The last expression is a linear combination of $n$ vectors in $\mathbb{R}^m$, and thus can be written as a matrix-vector product: $$ x_1 T(\mathbf{e}_1) + x_2 T(\mathbf{e}_2) + \cdots + x_n T(\mathbf{e}_n) = \begin{pmatrix} T(\mathbf{e}_1) & T(\mathbf{e}_2) & \cdots & T(\mathbf{e}_n) \end{pmatrix} \mathbf{x}. $$ :::::: ::::::{prf:definition} :label: Dfn:LinTrafo:StandardMatrix For a linear transformation $T:\mathbb{R}^n \to \mathbb{R}^m$, the matrix :::{math} :label: Eq:LinTrafo:StandardMatrix2 \begin{pmatrix} T(\mathbf{e}_1) & T(\mathbf{e}_2) & \cdots & T(\mathbf{e}_n) \end{pmatrix} ::: is called the **standard matrix** of $T$. :::::: In the Section {ref}`Sec:GeomLinTrans` you will learn how to build standard matrices for rotations, reflections and other geometrical mappings. For now let us look at a more "algebraic" example. ::::::{prf:example} :label: Ex:LinTrafo:MatrixToLinearMap Consider the transformation $$ T: \begin{pmatrix} x \\ y \\ z \end{pmatrix} \mapsto \begin{pmatrix} x-y \\ 2y+3z \\ x+y-z \end{pmatrix}. $$ It can be checked that the transformation has the two properties of a linear transformation according to the definition. Note that $$ T(\mathbf{e}_1) = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \quad T(\mathbf{e}_2) = \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}, \quad \text{and} \quad T(\mathbf{e}_3) = \begin{pmatrix} 0 \\ 3 \\ -1 \end{pmatrix}. $$ So we find that the standard matrix $[T]$ of $T$ is given by $$ [T] = \begin{pmatrix} 1 & -1 & 0 \\ 0 &2&3 \\ 1 & 1 & -1 \end{pmatrix}. $$ :::::: ::::::{exercise} :label: Exc:LinTrafo:FillBlanks In the previous example we could have found the matrix just by inspection. For the slightly different transformation $T:\R^3 \to \R^3$ given by $$ T: \begin{pmatrix} x \\ y \\ z \end{pmatrix} \mapsto \begin{pmatrix} 3x-z \\ y+4z \\ x-y+2z \end{pmatrix}, $$ can you fill in the blanks in the following equation? $$ \begin{pmatrix} 3x-z \\ y+4z \\ x-y+2z \end{pmatrix} = \begin{pmatrix} \ldots & \ldots & \ldots \\ \ldots & \ldots & \ldots \\ \ldots & \ldots & \ldots \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}. $$ If you can, you will have shown that $T$ is a matrix transformation, and as a direct consequence $T$ is a linear transformation. :::::: :::{admonition} Solution to {numref}`Exc:LinTrafo:FillBlanks` :class: solution, dropdown $$ \begin{pmatrix} 3x-z \\ y+4z \\ x-y+2z \end{pmatrix} = \begin{pmatrix} 3 & 0 & -1\\ 0 & 1 & 4 \\ 1 & -1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}. $$ ::: To conclude we consider an example that refers back to {prf:ref}`Prop:LinTrafo:CompositionLintrafos`, and which will to a large extent pave the road for the product of two matrices. ::::::{prf:example} :label: Ex:LinTrafo:ProductOfMatrices Suppose $T:\mathbb{R}^2 \to \mathbb{R}^3$ and $S:\mathbb{R}^3 \to \mathbb{R}^3$ are the matrix transformations given by $$ T(\mathbf{x}) = A\mathbf{x} = \begin{pmatrix} 1&2 \\ 3&4 \\ 1&0 \end{pmatrix} \mathbf{x} \quad \text{and} \quad S(\mathbf{y}) = B\mathbf{y} = \begin{pmatrix} 1&0 &1 \\ 1 & -1 &2 \\ -1&1&-3 \end{pmatrix} \mathbf{y}. $$ From {prf:ref}`Prop:LinTrafo:CompositionLintrafos` we know that the composition $S\circ T: \mathbb{R}^2 \to \mathbb{R}^3$ is also a linear transformation. What is the (standard) matrix of $S\circ T$? For this we need the images of the unit vectors $\mathbf{e}_1$ and $\mathbf{e}_2$ in $\mathbb{R}^2$. For each vector we first apply $T$ and then $S$. For $\mathbf{e}_1$ this gives $$ T(\mathbf{e}_1) = \begin{pmatrix} 1&2 \\ 3&4 \\ 1&0 \end{pmatrix} \begin{pmatrix} 1\\0 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}, $$ and then $$ S (T(\mathbf{e}_1)) = \begin{pmatrix} 1&0 &1 \\ 1 & -1 &2 \\ -1&1&-3 \end{pmatrix} \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}= \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}. $$ Likewise for $\mathbf{e}_2$: $$ T(\mathbf{e}_2) = \begin{pmatrix} 1&2 \\ 3&4 \\ 1&0 \end{pmatrix} \begin{pmatrix} 0\\1 \end{pmatrix} = \begin{pmatrix} 2\\4\\0 \end{pmatrix} \,\,\Longrightarrow\,\, S (T(\mathbf{e}_2)) = \begin{pmatrix} 1&0 &1 \\ 1 & -1 &2 \\ -1&1&-3 \end{pmatrix} \begin{pmatrix} 2 \\ 4 \\ 0 \end{pmatrix}= \begin{pmatrix} 2 \\ -2 \\ 2 \end{pmatrix}. $$ So the matrix of $S\circ T$ becomes $$ [S\circ T] \,= \, \begin{pmatrix} S\circ T(\mathbf{e}_1)&S\circ T(\mathbf{e}_2) \end{pmatrix} \,\,=\,\, \begin{pmatrix} 2 &2 \\ 0&-2 \\ -1&2 \end{pmatrix}. $$ In the Section {ref}`Sec:MatrixOps` we will define the product of two matrices precisely in such a way that $$ \begin{pmatrix} 1&0 &1 \\ 1 & -1 &2 \\ -1&1&-3 \end{pmatrix} \begin{pmatrix} 1&2 \\ 3&4 \\ 1&0 \end{pmatrix} = \begin{pmatrix} 2 &2 \\ 0&-2 \\ -1&2 \end{pmatrix}. $$ :::::: ## Grasple exercises %::::::{grasple} %:iframeclass: dark-light %:url: https://embed.grasple.com/exercises/97a589a8-54f9-4688-bd4d-a17a9585813b?id=69465 %:label: grasple_exercise_3_1_1 %:dropdown: %:description: This is {prf:ref}`Ex:LinTrafo:SecondMatrixTrafo`. %:::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/3f14573a-1d4c-4a4b-ae48-ccb168005702?id=70373 :label: grasple_exercise_3_1_2 :dropdown: :description: To specify the domain and the codomain of a linear transformation. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/b80d9889-bd46-45c6-a9cb-d056aa315232?id=70374 :label: grasple_exercise_3_1_3 :dropdown: :description: To find the size of the matrix of a linear transformation. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/be6a768d-c60d-4ed6-81a7-5dea71b4a1a5?id=70375 :label: grasple_exercise_3_1_4 :dropdown: :description: To find the image of two vectors under $T(\vect{x}) = A\vect{x}$. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/c8bb24f6-d357-4571-adb3-39ea0fa9e4ee?id=70395 :label: grasple_exercise_3_1_5 :dropdown: :description: For a linear map $T$, find $T(c\vect{u})$ and $T(\vect{u}+\vect{v})$ if $T(\vect{u})$ and $T(\vect{v})$ are given. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/93048f7c-b755-4445-a532-949f34136096?id=70398 :label: grasple_exercise_3_1_6 :dropdown: :description: For a linear map $T:\R^2 \to \R^2$, find $T((x_1,x_2))$ if $T(\vect{e}_1)$ and $T(\vect{e}_2)$ are given. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/2af6559f-8871-494d-abce-d4263d530c69?id=70381 :label: grasple_exercise_3_1_7 :dropdown: :description: Find all vectors $\vect{w}$ for which $T(\vect{w}) = \vect{u}$. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/ce6e4a52-c985-43ee-92cb-2762a467ac5a?id=70383 :label: grasple_exercise_3_1_8 :dropdown: :description: Find vectors $\vect{w}$ for which $T(\vect{w}) = \vect{u}$. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/37b6bd46-8cfc-4c98-a5e8-53aa41c87dcf?id=70384 :label: grasple_exercise_3_1_10 :dropdown: :description: Find vectors $\vect{w}$ for which $T(\vect{w}) = \vect{u}$. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/c5b2a642-fd50-43f6-9346-c37a0ffe1a40?id=70386 :label: grasple_exercise_3_1_10b :dropdown: :description: Find vectors $\vect{w}$ for which $T(\vect{w}) = \vect{u}$. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/c3d009c0-62d6-4ae3-8ca1-04a5d2730455?id=70406 :label: grasple_exercise_3_1_11 :dropdown: :description: To show that a given transformation is non-linear. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/b9a4b128-f2c2-4612-a7f5-271c4e69aa70?id=70418 :label: grasple_exercise_3_1_12 :dropdown: :description: Finding an image and a pre-image of $T:\R^2 \to \R^2$ using a picture. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/4058e54a-74f2-414e-9693-420abbc62677?id=70391 :label: grasple_exercise_3_1_13 :dropdown: :description: 'To give a geometric description of $T: \vect{x} \mapsto A\vect{x}$.' :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/990bf561-629e-430f-b8d0-e757c63fe15c?id=70392 :label: grasple_exercise_3_1_14 :dropdown: :description: To give a geometric description of $T: \vect{x} \mapsto A\vect{x}$. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/4e5d3f55-9257-4023-9739-5df0a1a9f277?id=70410 :label: grasple_exercise_3_1_15 :dropdown: :description: To find the matrix of the transformation that sends $(x,y)$ to $x\vect{a}_1 + y\vect{a}_2$. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/9efa96e2-483d-4b2c-a58a-ba197bc09a81?id=70411 :label: grasple_exercise_3_1_16 :dropdown: :description: To find the matrix of the transformation that sends $(x,y)$ to $x\vect{a}_1 + y\vect{a}_2$. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/729cba57-72d1-4d54-8cf9-c9946952bf9d?id=70412 :label: grasple_exercise_3_1_17 :dropdown: :description: To rewrite $T:\R^3 \to \R^2$ to standard form. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/b4bb3730-f14c-4a60-a8b8-6b895cf93ac5?id=70413 :label: grasple_exercise_3_1_18 :dropdown: :description: To find the standard matrix for $T:\R^4 \to \R$. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/34bb6386-7e7c-411b-83a1-09bbaf1106c5?id=70415 :label: grasple_exercise_3_1_19 :dropdown: :description: To find the standard matrix for $T:\R^2 \to \R^2$ if $T(\vect{v}_1)$ and $T(\vect{v}_2)$ are given. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/ce8ba17c-0a17-4d5e-b4b7-5c277c7e8df8?id=70416 :label: grasple_exercise_3_1_20 :dropdown: :description: To find the standard matrix for $T:\R^2 \to \R^3$ if $T(\vect{v}_1)$ and $T(\vect{v}_2)$ are given. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/2de4f8d1-ab3d-4d3a-94e4-5e414e2da3d9?id=70372 :label: grasple_exercise_3_1_21 :dropdown: :description: If $T(\vect{0}) = \vect{0}$, is $T$ (always) linear? :::::: %::::::{grasple} %:iframeclass: dark-light %%:url: https://embed.grasple.com/exercises/3f992e7a-19e3-4b83-8d90-db86e323ea94?id=69296 %:label: grasple_exercise_3_1_21 %:dropdown: %:description: To show that $T(\vect{0}) = \vect{0}$ for a linear transformation. %:::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/94d618e0-de21-491c-ad44-8e29974e0303?id=71098 :label: grasple_exercise_3_1_22 :dropdown: :description: True or false? If $\{\vect{v}_1,\vect{v}_2,\vect{v}_3\}$ is linearly dependent, then $\{T(\vect{v}_1),T(\vect{v}_2),T(\vect{v}_3)\}$ is also linearly dependent. :::::: ::::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/f983b627-10c2-4dd6-a273-2a33e99d0ded?id=71101 :label: grasple_exercise_3_1_23 :dropdown: :description: True or false? If $\{\vect{v}_1,\vect{v}_2,\vect{v}_3\}$ is linearly independent, then $\{T(\vect{v}_1),T(\vect{v}_2),T(\vect{v}_3)\}$ is also linearly independent. ::::::