Matrix Inverses and Subspaces of $$\\mathbb{R}^n$$

Criteria and Formulas for Matrix Invertibility

• Recall of Invertibility for 2x2 Matrices:     * For a matrix $A = \begin{pmatrix} a & b \ c & d \end{pmatrix}$, the matrix is invertible if and only if the determinant is non-zero.     * Determinant Formula: $\det(A) = ad - bc \neq 0$.

• Fact: Calculation of the Inverse for a 2x2 Matrix:     * If $A = \begin{pmatrix} a & b \ c & d \end{pmatrix}$ is invertible, then the inverse matrix $A^{-1}$ is calculated as:     * $A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}$.

• Example Case Study:     * Let $A = \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}$.     * First, calculate the determinant: $\det(A) = (1 \times 4) - (2 \times 3) = 4 - 6 = -2$.     * Since $\det(A) = -2 \neq 0$, matrix $A$ is invertible.     * Applying the formula: $A^{-1} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \ \frac{3}{2} & -\frac{1}{2} \end{pmatrix}$.

• Verification (Check):     * Multiplying $A$ by its inverse: $\begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix} \begin{pmatrix} -2 & 1 \ \frac{3}{2} & -\frac{1}{2} \end{pmatrix} = \begin{pmatrix} -2 + 3 & 1 - 1 \ -6 + 6 & 3 - 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}$.

Properties and Computation of Matrix Inverses

• Fact on Matrice Multiplicity:     * For any $n \times n$ matrices $A$ and $B$, the following statement holds true: $AB = I_n \iff BA = I_n$.

• Inverse of a Product:     * The inverse of a product of two invertible matrices is the product of their inverses in reverse order:     * $(AB)^{-1} = B^{-1} A^{-1}$.

• Computing the Inverse using Gauss-Jordan Elimination:     * We compute $A^{-1}$ by executing the Gauss-Jordan algorithm on matrix $A$ and the identity matrix $I_n$ simultaneously.     * This is represented as performing row operations on the augmented matrix $[A | I_n]$ until it reaches the form $[I_n | A^{-1}]$.

• Step-by-Step Example of Inversion:     * Given matrix $A = \begin{pmatrix} 1 & 1 & 1 \ 0 & 1 & 1 \ -1 & 1 & -1 \end{pmatrix}$.     * Augmented Matrix: $\begin{pmatrix} 1 & 1 & 1 & | & 1 & 0 & 0 \ 0 & 1 & 1 & | & 0 & 1 & 0 \ -1 & 1 & -1 & | & 0 & 0 & 1 \end{pmatrix}$.     * Step 1: Add Row 1 to Row 3 ($R_3 + R_1 \rightarrow R_3$):         * $\begin{pmatrix} 1 & 1 & 1 & | & 1 & 0 & 0 \ 0 & 1 & 1 & | & 0 & 1 & 0 \ 0 & 2 & 0 & | & 1 & 0 & 1 \end{pmatrix}$.     * Step 2: Subtract $2 \times$, Row 2 from Row 3 ($R_3 - 2R_2 \rightarrow R_3$):         * $\begin{pmatrix} 1 & 1 & 1 & | & 1 & 0 & 0 \ 0 & 1 & 1 & | & 0 & 1 & 0 \ 0 & 0 & -2 & | & 1 & -2 & 1 \end{pmatrix}$.     * Step 3: Multiply Row 3 by $-\frac{1}{2}$ ($R_3 \times (-\frac{1}{2}) \rightarrow R_3$):         * $\begin{pmatrix} 1 & 1 & 1 & | & 1 & 0 & 0 \ 0 & 1 & 1 & | & 0 & 1 & 0 \ 0 & 0 & 1 & | & -\frac{1}{2} & 1 & -\frac{1}{2} \end{pmatrix}$.     * Step 4: Subtract Row 3 from Row 2 ($R_2 - R_3 \rightarrow R_2$) and subtract Row 3 from Row 1 ($R_1 - R_3 \rightarrow R_1$):         * $\begin{pmatrix} 1 & 1 & 0 & | & \frac{3}{2} & -1 & \frac{1}{2} \ 0 & 1 & 0 & | & \frac{1}{2} & 0 & \frac{1}{2} \ 0 & 0 & 1 & | & -\frac{1}{2} & 1 & -\frac{1}{2} \end{pmatrix}$.     * Step 5: Subtract Row 2 from Row 1 ($R_1 - R_2 \rightarrow R_1$):         * $\begin{pmatrix} 1 & 0 & 0 & | & 1 & -1 & 0 \ 0 & 1 & 0 & | & \frac{1}{2} & 0 & \frac{1}{2} \ 0 & 0 & 1 & | & -\frac{1}{2} & 1 & -\frac{1}{2} \end{pmatrix}$.     * Result: The inverse matrix is $A^{-1} = \begin{pmatrix} 1 & -1 & 0 \ \frac{1}{2} & 0 & \frac{1}{2} \ -\frac{1}{2} & 1 & -\frac{1}{2} \end{pmatrix}$.

• Final Check:     * Multiplying $A$ and the resulting $A^{-1}$ yields $\begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}$.

Subspaces of $\mathbb{R}^n$ and the Concept of Images

• Definition of the Image of a Function:     * For a function $f: X \rightarrow Y$, the image, denoted as $Im(f)$, is defined as the set of all outputs resulting from inputs in the domain $X$.     * $Im(f) = { f(x) : x \in X }$.     * Alternatively expressed as: $Im(f) = { y \in Y : y = f(x) \text{ for some } x \in X }$.

• Remark on Terminology:     * The image of a function is also commonly referred to as the range of $f$.

• The Image of a Matrix:     * The image of an $n \times m$ matrix $A$ is defined as the image of the linear transformation $T$ defined by that matrix.     * $Im(A) = Im(T)$.     * Given an $n \times m$ matrix $A = \begin{pmatrix} \vec{v}_1 & \vec{v}_2 & \dots & \vec{v}_m \end{pmatrix}$, where $\vec{v}_i$ are the columns and $\vec{x} \in \mathbb{R}^m$:     * $Im(A) = { A\vec{x} : \vec{x} \in \mathbb{R}^n }$.     * This is the set of all linear combinations of the column vectors: $Im(A) = { x_1 \vec{v}_1 + \dots + x_m \vec{v}_m : x_1, \dots, x_m \in \mathbb{R} }$.

• Definition of Span:     * The set of all linear combinations of vectors $\vec{v}_1, \dots, \vec{v}_m$ is called the span of these vectors.

Questions & Discussion

• Question regarding Projections:     * Q: What is the image of the projection $proj_L: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ onto the line $L$?     * A: The image is the line $L$ itself!

• Example Analysis of Matrix Images:     * Consider matrix $A = \begin{pmatrix} 1 & 0 & 1 \ 0 & 1 & 0 \end{pmatrix}$.     * $Im(A) = { x_1 \begin{pmatrix} 1 \ 0 \end{pmatrix} + x_2 \begin{pmatrix} 0 \ 1 \end{pmatrix} + x_3 \begin{pmatrix} 1 \ 0 \end{pmatrix} : x_1, x_2, x_3 \in \mathbb{R} }$.     * This simplifies to $\begin{pmatrix} x_1 + x_3 \ x_2 \end{pmatrix}$.     * In the plane, this represents the entirety of the $x_1-x_2$ plane (the entirety of $\mathbb{R}^2$) because any vector $\begin{pmatrix} a \ b \end{pmatrix}$ can be achieved by choosing appropriate weights for the columns.

• Problem-Solving - Membership in the Image:     * Q: Is the vector $\begin{pmatrix} 1 \ 2 \end{pmatrix}$ in the image of the matrix $A = \begin{pmatrix} 0 & 1 \ 2 & 3 \end{pmatrix}$?     * A: To determine this, we check if there exists a vector $\vec{x} = \begin{pmatrix} x_1 \ x_2 \end{pmatrix}$ such that $A\vec{x} = \vec{b}$, i.e., $\begin{pmatrix} 0 & 1 \ 2 & 3 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 1 \ 2 \end{pmatrix}$.     * This yields the linear system:         1. $0 \cdot x_1 + 1 \cdot x_2 = 1 \implies x_2 = 1$.         2. $2 \cdot x_1 + 3 \cdot x_2 = 2$.     * Substituting $x_2 = 1$ into the second equation: $2x_1 + 3(1) = 2 \implies 2x_1 = -1 \implies x_1 = -\frac{1}{2}$.     * Conclusion: YES, the vector is in the image because it is the image of the vector $\vec{x} = \begin{pmatrix} -1/2 \ 1 \end{pmatrix}$.