MAT223: 12_Inverse Functions and Matrices

Inverse Functions and Inverse Matrices §

Invertible functions §

• Identity Function - Let $X$ be a set the identity function with domain and codomain $X$, notated $id:X\to X$, is the function satisfying, for all $x\in X$

  $$\text{id}(x)=x$$
  • Identity function is the function that nodes nothing to tis input

• A function is invertible if it can be undone. (If there exists an inverse function that when composed with the original function produces the identity function and vice versa)

• Inverse function - Let $f:X\to Y$ be a function, $f$ is invertible if there exist a function $g:Y\to X$ so $f\circ g=id$ and $g\circ f=id$. In this case, call $g$ an inverse of $f$ and write

  $$f^{-1}=g$$

• Every invertible function is both one-to-one and onto and every one-to-one and onto functions is invertible

  • One-to-one - Let $f:x\to Y$ be a fiction, say $f$ is one-to-one (or injective) if distinct inputs to $f$ produces distinct outputs.. That is $f(x)=f(y)$ implies $x=y$
  • Onto - Let $f:X\to Y$ be a function. We say $f$ is onto (or surjective) if every point in the codomain of $f$ gets mapped to. That is $range(f)=Y$.

Invertibility and Linear Transformations §

• $\mathcal{T}:{\mathbb{R}}^n\to {\mathbb{R}}^m$ Is invertible if and only if $nullity(\mathcal{T})=0$ and $rank(\mathcal{T})=m$

• $\mathcal{T}:{\mathbb{R}}^n\to {\mathbb{R}}^m$ Is invertible if and only if $m=n$ and $nullity(\mathcal{T})=0$

• $\mathcal{T}:{\mathbb{R}}^n\to {\mathbb{R}}^m$ Is invertible if and only if $m=n$ and $rank(\mathcal{T})=m$

  Let $\mathcal{P}:{\mathbb{R}}^2\to {\mathbb{R}}^2$ be projection onto the x-axis and let $\mathcal{R}:{\mathbb{R}}^2\to {\mathbb{R}}^2$ be rotation counter-clockwise by $15^{\circ }$. Classifty each of $\mathcal{P}$ and $\mathcal{R}$ as invertible or not §

  Notice that $\mathcal{P}({\vec{e}}_2)=P(2{\vec{e}}_2)=\vec{0}$ , therefore $\mathcal{P}$ is not one-to-one and so it is not invertible

  Let $\mathcal{Q}:{\mathbb{R}}^2\to {\mathbb{R}}^2$ be rotation clockwise by $15^{\circ }$. $\mathcal{R}$ And $\mathcal{Q}$ will undo each other. Thus:

  $$\mathcal{R}\circ \mathcal{Q}=id\text{and}\mathcal{Q}\circ \mathcal{R}=id$$

  Therefore, $\mathcal{Q}$ is an inverse of $\mathcal{R}$ and so $\mathcal{R}$ is invertible

• Let $\mathcal{T}$ be an invertible linear transformation, then ${\mathcal{T}}^{-1}$ is also a linear transformation.

Investability and Matrices §

• Identity matrix - an identity matrix is a square matrix with ones on the diagonal and zeros everywhere else. Then $n\times n$ identity matrix is denoted $I_{n\times n}$ or just $I$ when its size is implied
• Matrix inverse - the inverse of a matrix $A$ is a matrix $B$ such that $AB=I$ and $BA=I$. In this case, $B$ is called the inverse of $A$ and is notated $A^{-1}$
• An $n\times m$ matrix $A$ is invertible if and only if $nullity(A)=0$ and $rank(A)=n$
• An $n\times m$ matrix $A$ is invertible if and only if $nullity(A)=0$
• An $n\times m$ matrix $A$ is invertible if and only if $rank(A)=n$

Determine whether the matrices $A=\left[\begin{array}{cc}2& 5\\ -3& -7\end{array}\right]$ and $B=\left[\begin{array}{cc}-7& -5\\ 3& 2\end{array}\right]$ are inverse of each other §

AB = \begin{bmatrix}2 & 5 \\ -3 & -7\end{bmatrix} \begin{bmatrix}-7 & -5 \\ 3 & 2\end{bmatrix} = 

\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = I \\
BA = \begin{bmatrix}-7 & -5 \\ 3 & 2\end{bmatrix} \begin{bmatrix}2 & 5 \\ -3 & -7\end{bmatrix} = 
\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = I

Therefore, $A, B$ are inverses of each other

Matrix Algebra §

• Suppose $A$ is invertible, then

  $$A\vec{x}=\vec{b}\Rightarrow A^{-1}A\vec{x}=A^{-1}\vec{b}\Rightarrow \vec{x}=A^{-1}\vec{b}$$

  Thus, if having a inverse of a matrix handy, can use it to solve system of equations.

  Use th efact that ${\left[\begin{array}{cc}2& 5\\ -3& 7\end{array}\right]}^{-1}=\left[\begin{array}{cc}-7& -5\\ 3& 2\end{array}\right]$ to solve the system $\{\begin{array}{cc}2x+5y& =2\\ -3x-7y& =1\end{array}$ §

  The system can be rewritten as

  \begin{bmatrix}2 & 5 \ -3 & -7\end{bmatrix} \begin{bmatrix}x \ y\end{bmatrix} = \begin{bmatrix}2 \ 1\end{bmatrix}

  Multiplying both sides by $\begin{bmatrix}2 & 5 \\ -3 & -7 \end{bmatrix}^{-1}$ gives

  \begin{bmatrix}x \ y\end{bmatrix} = \begin{bmatrix}2 & 5 \ -3 & -7 \end{bmatrix}^{-1} \begin{bmatrix}2 \ 1\end{bmatrix} = \begin{bmatrix}-7 & -5 \ 3 & 2\end{bmatrix}\begin{bmatrix}2 \ 1\end{bmatrix} = \begin{bmatrix}-19 \ 8\end{bmatrix}

  Therefore $x = -19$ and $y=8$

• Finding a matrix inverse (write example here)

Elementary Matrices §

• Elementary matrix - is an identity matrix with a single elementary row operation applies
• Elementary row operations are
  • Multiply a row by a non-zero constant
  • Add a multiple of one row to another
  • Swap two rows
• A matrix $M$ is invertible if and only if there are elementary matrices $E_1,\dots E_k$ so that
  $$E_k\dots E_2{E}_{2}M=I$$
  • Let $Q=(E_k\dots E_2{E}_1)$, then $MQ=I$, then $Q$ is the inverse of $M$
• If $A$ is a square matrix and $AB=I$ for some matrices $B$, then $BA=I$
  • Write example of using this method to find inverse of a matrix

Decomposition int elementary matrices §

• A matrix $M$ is invertible if and only if it can be written as the product of elementary matrices