Linear Transformations

The matrix (123456) takes elements of R3 and outputs elements of R2(123456)(abc)=(a+2b+3c4a+5b+6c)In other words, it defines a function‾ from R3 to R2\text{The matrix } \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} \text{ takes elements of } \mathbb{R}^3 \text{ and outputs elements of } \mathbb{R}^2 \\ \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} \begin{pmatrix}a \\ b \\ c\end{pmatrix} = \begin{pmatrix} a + 2b + 3c \\ 4a + 5b + 6c \end{pmatrix} \\ \text{In other words, it defines a \underline{function} from } \mathbb{R}^3 \text{ to } \mathbb{R}^2 \\

Linear functions

A function takes elements from one set (known as the domain) and outputs elements of another set (known as the range or codomain).

Multiplying by a m * n matrix (where m is the number of columns and n is the number of rows) defines a function that maps elements from Rn to Rm.

T:Rn→Rm is linear ifT(v→+w→)=T(v→)+T(w→) and T(λv→)=λT(v→)T: \mathbb{R}^n \rightarrow \mathbb{R}^m \text{ is linear if} \\ T(\overrightarrow{v} + \overrightarrow{w}) = T(\overrightarrow{v}) + T(\overrightarrow{w}) \text{ and } T(\lambda \overrightarrow{v}) = \lambda T(\overrightarrow{v})

Classic examples of linear functions

T(xy)=x+yT(v→)=Av→T\begin{pmatrix}x \\ y \end{pmatrix} = x + y \\ T(\overrightarrow{v}) = A\overrightarrow{v}

Properties of linear functions

Assume that T:Rn→Rm is linear. Then:T(0→)=0→T(av→+bw→)=a∗T(v→)+b∗T(w→)T(c1v→1+...+cnv→n)=c1∗T(v→1)+...+cn∗T(v→n)\text{Assume that } T: \mathbb{R}^n \rightarrow \mathbb{R}^m \text{ is linear. Then:} \\ T(\overrightarrow{0}) = \overrightarrow{0} \\ T(a\overrightarrow{v} + b\overrightarrow{w}) = a * T(\overrightarrow{v}) + b * T(\overrightarrow{w}) \\ T(c_1\overrightarrow{v}_1 + ... + c_n\overrightarrow{v}_n) = c_1 * T(\overrightarrow{v}_1) + ... + c_n * T(\overrightarrow{v}_n)

Composition of Linear Functions

Suppose we have linear functions f and g such that f:Rn→Rm and g:Rm→RkWe can compose these functions to transfer a vector from Rn to Rk by using the functions like f(g(x→)). This new function is denoted as g∘f.\text{Suppose we have linear functions f and g such that } f: \mathbb{R}^n \rightarrow \mathbb{R}^m \text{ and } g: \mathbb{R}^m \rightarrow \mathbb{R}^k \\ \text{We can compose these functions to transfer a vector from } \mathbb{R}^n \text{ to } \mathbb{R}^k \text{ by using the functions like } f(g(\overrightarrow{x})) \text{. This new function is denoted as } g \circ f \text{.} Theorem: If T:Rn→Rm and S:Rm→Rk are both linear, then so is S∘T:Rn→Rk(S∘T)(λv→)=S(λT(v→))=λS(T(v→))=λ(S∘T)(v→)\text{Theorem: If } T: \mathbb{R}^n \rightarrow \mathbb{R}^m \text{ and } S: \mathbb{R}^m \rightarrow \mathbb{R}^k \text{ are both linear, then so is } S \circ T: \mathbb{R}^n \rightarrow \mathbb{R}^k \\ (S \circ T)(\lambda \overrightarrow{v}) = S(\lambda T(\overrightarrow{v})) = \lambda S(T(\overrightarrow{v})) = \lambda (S \circ T)(\overrightarrow{v})

Matrix multiplication is a linear function that can be composed, as shown below

(01−10) and (10111−1)Multiplying a vector by the first matrix can be thought of as a function that spans R2→R2 with the second matrix being R3→R2(01−10)(10111−1)(abc)=(01−10)(a+ca+b−c)=(a+b−c−a−c)\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \text{ and } \begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & -1 \end{pmatrix} \\ \text{Multiplying a vector by the first matrix can be thought of as a function that spans } \mathbb{R}^2 \rightarrow \mathbb{R}^2 \text{ with the second matrix being } \mathbb{R}^3 \rightarrow \mathbb{R}^2 \\ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & -1 \end{pmatrix} \begin{pmatrix}a \\ b \\ c \end{pmatrix} = \\ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} a + c \\ a + b - c \end{pmatrix} = \\ \begin{pmatrix} a + b - c \\ -a - c \end{pmatrix}

You can add linear functions, as well:

If T:Rn→Rm and S:Rn→Rm are both linear, then S+T:Rn→Rm is linear and (S+T)(v→)=S(v→)+T(v→)\text{If } T: \mathbb{R}^n \rightarrow \mathbb{R}^m \text{ and } S: \mathbb{R}^n \rightarrow \mathbb{R}^m \text{ are both linear, then } \\ S + T: \mathbb{R}^n \rightarrow \mathbb{R}^m \text{ is linear and } (S + T)(\overrightarrow{v}) = S(\overrightarrow{v}) + T(\overrightarrow{v})

This can be used in matrix addition:

((01−10)+(20012))(ab)=(b−a)+(a2b2)=(b+a2b2−a)( \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} + \begin{pmatrix} \sqrt{2} & 0 \\ 0 & \frac{1}{\sqrt{2}} \end{pmatrix} ) \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} b \\ -a \end{pmatrix} + \begin{pmatrix} a\sqrt{2} \\ \frac{b}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} b + a\sqrt{2} \\ \frac{b}{\sqrt{2}} - a \end{pmatrix}

Unit Vector With Linear Functions

$$ \text{The unit vector is defined as } \hat{e}_i \in \mathbb{R}^n
\hat{e}_i = \begin{pmatrix} 0 \ \vdots \ 0 \ 1 \ 0 \ \vdots \ 0 \end{pmatrix} \text{ where the 1 is in the ith position} \

\begin{pmatrix} a_1 \ \vdots \ a \ \vdots \ a_n\end{pmatrix} = \begin{pmatrix} a_1 \ \vdots \ 0 \ \vdots \ 0\end{pmatrix} + \begin{pmatrix} 0 \ \vdots \ a \ \vdots \ 0\end{pmatrix} + \begin{pmatrix}0 \ \vdots \ 0 \ \vdots \ a_n \end{pmatrix} = a_1\hat{e}_1 + a_i\hat{e}_i + a_n\hat{e}_n $$

T:Rn→Rmv→=a1e^1+...+ane^nT(v→)=T(a1e^1+...+ane^n)=a1T(e^1)+...+anT(e^n)=(T(e^1)...T(e^n))(a1⋮an)T: \mathbb{R}^n \rightarrow \mathbb{R}^m \\ \overrightarrow{v} = a_1\hat{e}_1 + ... + a_n\hat{e}_n \\ T(\overrightarrow{v}) = T(a_1\hat{e}_1 + ... + a_n\hat{e}_n) = a_1T(\hat{e}_1) + ... + a_nT(\hat{e}_n) = \begin{pmatrix} T(\hat{e}_1) & ... & T(\hat{e}_n)\end{pmatrix} \begin{pmatrix}a_1 \\ \vdots \\ a_n \end{pmatrix}

Example of how to get matrix multiplication from a function

$$ T \begin{pmatrix}a \ b \ c\end{pmatrix} = \begin{pmatrix} a + 2b - c
3a - 4b + 2c
5a + 7b + c \end{pmatrix}
\hat{e}_1 = \begin{pmatrix}1\0\0\end{pmatrix}, \hat{e}_2 = \begin{pmatrix}0\1\0\end{pmatrix}, \hat{e}_3 = \begin{pmatrix}0\0\1\end{pmatrix}
T\begin{pmatrix}1\0\0\end{pmatrix} = \begin{pmatrix}1\3\5\end{pmatrix}, T\begin{pmatrix}0\1\0\end{pmatrix} = \begin{pmatrix}2\-4\7\end{pmatrix}, T\begin{pmatrix}0\0\1\end{pmatrix} = \begin{pmatrix}-1\2\1\end{pmatrix}
A_T = \begin{pmatrix} 1 & 2 & -1
3 & -4 & 2
5 & 7 & 1 \end{pmatrix} \begin{pmatrix}a\ b\ c\end{pmatrix}

$$

Matrix Multiplication

Let A represent a matrix.A∗(v1→...vn→)=(Av1→...Avn→)\text{Let A represent a matrix.}\\ A * \begin{pmatrix} \overrightarrow{v_1} & ... & \overrightarrow{v_n} \end{pmatrix} = \begin{pmatrix} A \overrightarrow{v_1} & ... & A \overrightarrow{v_n} \end{pmatrix}

Examples

(1724)(3352)=(1∗3+7∗51∗3+7∗22∗3+4∗52∗3+4∗2)=(38172614)\begin{pmatrix} 1 & 7\\ 2 & 4 \end{pmatrix} \begin{pmatrix} 3 & 3\\ 5 & 2 \end{pmatrix} = \begin{pmatrix} 1*3 + 7*5 & 1*3 + 7*2\\ 2*3 + 4*5 & 2*3 + 4*2 \end{pmatrix} = \begin{pmatrix} 38 & 17\\ 26 & 14 \end{pmatrix}

(13−401−1001)(1−162−31)=(A(16−3)A(−121))=((1∗1+3∗6+(−4)(−3)0∗1+1∗6+(−1)∗(−3)0∗1+0∗6+1∗(−3))(1∗(−1)+3∗2+(−4)∗10∗(−1)+1∗2+(−1)∗10∗(−1)+0∗2+1∗1))\begin{pmatrix} 1 & 3 & -4\\ 0 & 1 & -1\\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 6 & 2 \\ -3 & 1 \end{pmatrix} = (A\begin{pmatrix}1\\6\\-3\end{pmatrix} A\begin{pmatrix}-1\\2\\1\end{pmatrix}) = ( \begin{pmatrix} 1*1 + 3*6 + (-4)(-3) \\ 0*1 + 1*6 + (-1)*(-3) \\ 0*1 + 0*6 + 1*(-3) \end{pmatrix} \begin{pmatrix} 1*(-1) + 3*2 + (-4)*1 \\ 0*(-1) + 1*2 + (-1)*1 \\ 0*(-1) + 0*2 + 1*1 \end{pmatrix} )