Inverse Matrix

Question

反矩陣可用來證明矩陣可逆
一些反矩陣的題目-反矩陣存在的證明

可逆矩陣相乘的結果還是可逆矩陣

齊次解恆有解, 當矩陣為可逆時，齊次解只有唯一零解。
在線性系統組裡，特解多於一解或無解時，齊次解為無限多解。
齊次解定義(homogenouse solution)

2-by-2 matrix

\( A= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \)
\( A^{-1}= \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)
\( \text{The Determinant of }A= det(A) = ad-bc \)
Adjugate matrix of \(A= adj(A)= \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)
\( A^{-1}= \frac{1}{det(A)}adj(A) \)
\( = \frac{1}{det(A)}\text{cofactor}(A)^T \)

cofactor matrix(餘因矩陣\( \rightarrow \)由多個餘因子構成)轉置之後，即為adjugate matrix(古典伴隨矩陣)。

Argument matrix

Using Gauss-Jordan form eliminate.
\( \left[ \begin{matrix} A \end{matrix} \left| \, \begin{matrix} I \end{matrix} \right. \right] = \left[ \begin{matrix} I \end{matrix} \left| \, \begin{matrix} A^{-1} \end{matrix} \right. \right] \)

Unique Inverse

Suppose \(A\) is \(n\) by \(n\) matrix and \(A\) is nonsingular matrix.
Its inverse matrix is unique.

Proof

Suppose \(B\) and \(C\) are both inverse of matrix of an \(n\) by \(n\) matrix \(A\)
let \( \begin{cases} BA=AB=I \\ CA=AC=I \end{cases} \)
\( B=B\cdot I = B(AC) = (BA)C = I\cdot C = C \)
\(\therefore\)the inverse of \(A\) matrix is unique.

\(A,B\in F^{n\times n}\) are both nonsingular matrix.
\(AB\) is also nonsingular matrix. \((AB)^{-1}=B^{-1}A^{-1}\)

proof

\( (AB)(AB)^{-1} =(AB)(B^{-1}A^{-1})=AIA^{-1}=I \)
\( (AB)^{-1}(AB)= (B^{-1}A^{-1})(AB)= B^{-1}IB=I \)

hint: \( 1-x^n = (1-x)(x^0+x^1+x^2+\dots+x^{n-1}) \)
\( =(1-x)(1+x^1+x^2+x^3+\dots+x^{n-1}) \)

Exercise

Suppose \(A\in F^{n\times n}\) and \(A^n=O\), show that \(I-A\) is nonsingular matrix and value of \((I-A)^{-1}\).

Solution

\( (I-A^n)= (I-A)(I+A+A^2+\dots+A^{n-1}) \)
\( \Rightarrow I-A^n = I-O=I \)
\( \Rightarrow I= (I-A)(I+A+A^2+\dots+A^{n-1}) \)
\( \therefore (I-A) \) is the inverse of an matrix to \( (I+A+A^2+\dots+A^{n-1}), \)
\( (I-A)^{-1}= (I+A+A^2+\dots+A^{n-1}) \)

Transpose and Hermintian

Suppose \( A\in F^{n\times n} \) is the nonsingular matrix.

\( A^T \) is a nonsingular matrix and if only if \( (A^T)^{-1}=(A^{-1})^T \).
Proof
- \( (A^T)(A^{-1})^T= (A^{-1}A)^T= (I)^T= I \)
- \( (A^{-1})^T(A^T)= (AA^{-1})^T= (I)^T= I \)
\( A^H \) is a nonsingular matrix and if only if \( (A^H)^{-1}=(A^{-1})^H \).
Proof
proof as same as \( (A^T)^{-1}=(A^{-1})^T \)
- \( I=(A^H)(A^{-1})^H= (A^{-1}A)^H=(I)^H=\overline{I^T}=I \)
- \( I=(A^{-1})^H(A^H)= (AA^{-1})^H=(I)^H= \overline{I^T}=I \)

Reference

黃子嘉-線性代數