Section05 Inverse Matrices

$$\boldsymbol{A}^{-1}\boldsymbol{A} = \boldsymbol{A}\boldsymbol{A}^{-1} = \boldsymbol{I}\\ $$

Notes

1. The inverse exsits if and only if elimination produces $n$ pivots.

2. The matrix $\boldsymbol{A}$ can't have two different inverses. $$\begin{array}{c} \text{IF } \boldsymbol{AB}= \boldsymbol{AC} = \boldsymbol{I} \text{ then } \boldsymbol{B} = \boldsymbol{C} \\ \end{array}$$

   • proof $$\begin{array}{ll} \because & \boldsymbol{B} = \boldsymbol{BI} = \boldsymbol{B(AC)} = \boldsymbol{(BA)C} = \boldsymbol{IC} = \boldsymbol{C} \\ \therefore & \boldsymbol{B} = \boldsymbol{C} \\ \end{array}$$

3. If $\boldsymbol{A}$ is invertible, the one and only one solution to $\boldsymbol{Ax} = \boldsymbol{b}$ is $\boldsymbol{x} = \boldsymbol{A}^{-1}\boldsymbol{b}$

   • proof $$\boldsymbol{x} = \boldsymbol{Ix} = \boldsymbol{(A^{-1}A)x} = \boldsymbol{A^{-1}(Ax)} = \boldsymbol{A^{-1}b}$$

4. Suppose there is a nonezero vector $\boldsymbol{x}$ such that $\boldsymbol{Ax} = \boldsymbol{0}$. Then $\boldsymbol{A}$ cann't have an inverse. No matrix can bring $\boldsymbol{0}$ to $\boldsymbol{x}$. $$\text{If } \boldsymbol{A} \text{ is invertible, then } \boldsymbol{Ax} = \boldsymbol{0} \text{ can only have the zero solution } \boldsymbol{x} = \boldsymbol{A}^{-1}\boldsymbol{0} = \boldsymbol{0}$$

5. A 2 by 2 matrix is invertible if and only if $ad - bc \ne 0$ $$\left[ \begin{matrix} a & b \\ c & d \\ \end{matrix} \right] ^ {-1} = \frac{1}{ad - bc} \left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right]$$

   • $ad - bc$ is the determinant of this 2 by 2 matrix.

6. A diagnonal matrix has an inverse provided no diagnonal entries are zero. $$\text{If } \boldsymbol{A} = \left[ \begin{matrix} d_{1} && \\ & \ddots & \\ & & & d_{n} \end{matrix} \right] \text{ then } \boldsymbol{A}^{-1} = \left[ \begin{matrix} \frac{1}{d_{1}} && \\ & \ddots & \\ & & & \frac{1}{d_{n}} \end{matrix} \right]$$

The Inverse of a Product $\boldsymbol{AB}$

$$(\boldsymbol{AB})^{-1} = \boldsymbol{B}^{-1}\boldsymbol{A}^{-1}$$

• proof $$\begin{array}{ll} \because & (\boldsymbol{AB})(\boldsymbol{AB})^{-1} = \boldsymbol{I} \\ & \text{and } \boldsymbol{AB}\boldsymbol{B^{-1}}\boldsymbol{A^{-1}} = \boldsymbol{AA}^{-1} = \boldsymbol{I} \\ \therefore & (\boldsymbol{AB})^{-1} = \boldsymbol{B}^{-1}\boldsymbol{A}^{-1} \\ \end{array}$$

• 从含义出发

  $\boldsymbol{A_{1}A_{2}X} = \boldsymbol{B}$

  从$\boldsymbol{X}$变化到$\boldsymbol{B}$分为两个步骤，

  1. 对矩阵$\boldsymbol{X}$进行$\boldsymbol{A}_{2}$的行变化，
  2. 然后再进行$\boldsymbol{A}_{1}$的行变化。

  $\boldsymbol{X} = (\boldsymbol{A_{1}A_{2}})^{-1} \boldsymbol{B}$

  从$\boldsymbol{B}$ 变化到 $\boldsymbol{X}$ 则需要进行和之前相反的变化，

  1. 首先，对矩阵$\boldsymbol{B}$ 进行与 $\boldsymbol{A}_{1}$相反的行变化$\boldsymbol{A_{1}}^{-1}$
  2. 然后进行与 $\boldsymbol{A}_{2}$相反的行变化$\boldsymbol{A_{2}}^{-1}$

  所以，$(\boldsymbol{A_{1}A_{2}})^{-1} = \boldsymbol{A_{2}}^{-1} \boldsymbol{A_{1}}^{-1}$

Calculating $\boldsymbol{A}^{-1}$ by Gauss-Jordan Elimination

$$\left[ \begin{matrix} \boldsymbol{A} & \boldsymbol{I} \end{matrix} \right] \boldsymbol{A}^{-1} \Rightarrow \left[ \begin{matrix} \boldsymbol{I} & \boldsymbol{A}^{-1} \end{matrix} \right]$$

steps

1. Construct a augment matrix $\left[ \begin{matrix} \boldsymbol{A} & \boldsymbol{I} \end{matrix} \right]$
2. Elminating the $\boldsymbol{A}$ block of augment matrix to $\boldsymbol{I}$. Then the $\boldsymbol{I}$ block of augment matrix is eliminated to $\boldsymbol{A}^{-1}$

Properties of inverse matrix

1. If $\boldsymbol{A}$ is a symmetric across its main diagnonal. Then $\boldsymbol{A}^{-1}$ is also symmetric.
2. If $\boldsymbol{A}$ is tridiagonal. But $\boldsymbol{A}^{-1}$ is a dense matrix with no zeros. ( The inverse of band matrix is always a dese matrix. )
3. When $\boldsymbol{A}$ is already eliminated. The product of pivots($a_{11} \cdot a_{22}\cdots a_{nn} = \prod_{i=1}^{n}a_{ii}$) is the determinant of matrix $\boldsymbol{A}$.

Computing in Programs

Matlab

%% Compute Directly
A_inverse = inv(A);

%% Compute in Gauss-Jordan Method
I = eye(size(A,1));
R = rref([A I]);
A_inverse = R(:, size(A,1)+1:end);

Julia

# Compute Directly (need to using LinearAlgebra Package)
using LinearAlgebra
A_inverse = inv(A);

# Compute in Gauss-Jordan Method
# In Julia we need to add RowEchelon Package
using RowEchelon
I = Matrix(I, size(A,1), size(A,2));
R = rref([A I]);
A_inverse = R[:,size(A,1):end];

Singular versus Invertible

• $\boldsymbol{A}^{-1}$ exsits exactly when $\boldsymbol{A}$ has a full set of $n$ pivots.

  1. if $\boldsymbol{A}$ doesn't have $n$ pivots, elimination will lead to a zero row.(Left-inverse)
  2. those elimination steps are taken by invertible matrix $\boldsymbol{M}$. So a row of $\boldsymbol{MA}$ is zero.
  3. If $\boldsymbol{AC} = \boldsymbol{I}$ had been possible, then $\boldsymbol{MAC} = \boldsymbol{M}$. The zero row of $\boldsymbol{MA}$ times $\boldsymbol{C}$, gives a zero row of $\boldsymbol{M}$ itself.
  4. An invertible matrix $\boldsymbol{M}$ can't have a zero row! $\boldsymbol{A}$ must have $n$ pivots if $\boldsymbol{AC} = \boldsymbol{I}$

• A trigular matrix is invertible if and only if no diagnonal entiries are zero.

Recognizing an Invertible Matrix

• diagnonally dominant matrices
  • Defination $$\boldsymbol{A} \text{ is a diagnonally dominant matrix, if } \vert a_{ii} \vert > \sum_{j \ne i}{\vert a_{ij} \vert}$$
  • diagnonally dominant matrices are invertible
  • proof $$\begin{array}{ll} \because & \text{Suppose any nonzero vector } \boldsymbol{x} \text{ whose largest component is } \vert x_{i} \vert \\ \therefore & i^{\text{th}} \text{ row of } \boldsymbol{Ax} \text{ is } \sum_{j=1}^{n}{a_{ij}x_{j}} \\ \because & \vert a_{ii} \vert > \sum_{j \ne i}{\vert a_{ij} \vert} \text{ and } \vert x_{i} \vert \text{ is the largest component of } \boldsymbol{x} \\ \therefore & \vert a_{ii} \vert \vert x_{i} \vert > \sum_{i \ne j}{\vert a_{ij} \vert \vert x_{i} \vert} > \sum_{i \ne j}{\vert a_{ij} \vert \vert x_{j} \vert} \\ \therefore & i ^{\text{th}} \text{ row of } \boldsymbol{Ax} \ne 0 \\ \therefore & \boldsymbol{Ax} \ne 0 \text{ until } \boldsymbol{x} = \boldsymbol{0}\\ \therefore & \boldsymbol{A} \text{ is invertible.} \\ \end{array}$$

Properties from Exercise

1. Exchange the row 1 and row 2 of an invertible matrix $\boldsymbol{A}$ to reach a new matrix $\boldsymbol{B}$, $\boldsymbol{B}$ is also invetible, and $\boldsymbol{B}^{-1}$ is reached by exhange the col 1 and col 2 of $\boldsymbol{A}^{-1}$.

   • proo $$\begin{array}{ll} \because & \boldsymbol{B} \text{ is reached by exchanging row 1 and row 2 of matrix } \boldsymbol{A} \\ \therefore & \text{Supppose a matrix } \boldsymbol{E} \text{, } \boldsymbol{EA} = \boldsymbol{B} \\ \therefore & \boldsymbol{E} = \left[ \begin{matrix} 0 & 1 &\cdots & 0 \\ 1 & 0 &\cdots & 0 \\ \vdots & \vdots &\ddots & \vdots \\ 0 & 0 &\cdots & 1 \end{matrix} \right] = \left[ \begin{array}{@c{}|@c{}} \begin{matrix} 0 & 1\\ 1 & 0 \end{matrix} & \boldsymbol{0} \\ \hline \boldsymbol{0} & \boldsymbol{I} \end{array} \right]\\ \therefore & \boldsymbol{B}^{-1} = (\boldsymbol{EA})^{-1} = \boldsymbol{A}^{-1}\boldsymbol{E}^{-1} = \boldsymbol{A}^{-1}\boldsymbol{E} \\ \therefore & \boldsymbol{B}^{-1} \text{ is reached by exchanging the col 1 and col 2 of matrix } \boldsymbol{A}^{-1} \\ \end{array}$$

2. Alternating Matrix $$\boldsymbol{A} = \left[ \begin{matrix} 1 & -1 & 1 & -1 & \cdots \\ 0 & 1 & -1 & 1 & \cdots \\ 0 & 0 & 1 & -1 & \cdots \\ 0 & 0 & 0 & 1 & \cdots \\ \vdots& \vdots & \vdots & \vdots & \ddots \end{matrix} \right] \boldsymbol{A}^{-1} = \left[ \begin{matrix} 1 & 1 & 0 & 0 & 0 & \cdots\\ 0 & 1 & 1 & 0 & 0 & \cdots\\ 0 & 0 & 1 & 1 & 0 & \cdots\\ 0 & 0 & 0 & 1 & 1 & \cdots\\ \vdots& \vdots & \vdots & \vdots & \vdots & \ddots \end{matrix} \right]$$