Matrix

Conjugate transpose: sign as \(A^H\) or \(A^*\)
Suppose \( A= \begin{bmatrix} 1-i & 2+3i & 3-2i \\ 4+3i & 5-6i & 6+7i \end{bmatrix} \)
\( A^H=\overline{A^T}= \begin{bmatrix} 1+i & 4-3i \\ 2-3i & 5+6i \\ 3+2i & 6-7i \end{bmatrix} \)
Suppose \(A\in F^{n\times n}\)
Symmetric matrix: \(A^T=A\)
ex:
\( A= \begin{bmatrix} 5 & 2 & 1 \\ 2 & 7 & 5 \\ 1 & 5 & 8 \end{bmatrix} \)
 \(A\) is any square matrix.
\(A+A^T\) is a symmetric.

Proof:

Suppose \(B=A+A^T\)
\( (A+A^T)^T= A^T+(A^T)^T= A^T+A= A+A^T \)
\(B^T=B\)
Skew-symmetric matrix: \(A^T=-A\)
ex:
\( A= \begin{bmatrix} 0 & 3 & 2 \\ -3 & 0 & 4 \\ -2 & -4 & 0 \end{bmatrix} \)
 \(A\) is any square matrix.
\(A-A^T\) is a skew-symmetric matrix.
** \(tr(A-A^T)=0\)

Proof:

Suppose \(B=A-A^T\)
\( (A-A^T)^T= A^T-(A^T)^T= A^T-A= -(A-A^T) \)
\(B^T=-B\)
Hermitian matrix: \(A^H=A=\overline{A^T}\)
The diagonal are all real number.
ex:
\( A= \begin{bmatrix} 1 & 3+2i & 5-4i \\ 3-2i & 4 & 5+5i \\ 5+4i & 5-5i & 7 \end{bmatrix} \)
Skew-hermitian matrix: \(A^H=-A=-\overline{A^T}\)
The diagonal are either 0 or complex number(no real part).
ex:
\( A= \begin{bmatrix} i & 3+i & 5-i \\ -3+i & 0 & 1+3i \\ -5-i & -1+3i & 6i \end{bmatrix} \)
Suppose matrix in \(F^{n\times n}\)
\(AB\neq BA\), except \(B=A^{-1}\)
\( A= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, B= \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \)
\( AB= \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} , BA= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
\( A^n = O \),
\(A=O\) may not be right.
\( A= \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, A^2= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}, \)
\( A^2=A,\)
\( A=I \text{ or } A=O \) may not be right.
\( A= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, A^2= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \)
\(A\neq O\), \(B\neq O\),
\(AB\neq O\) may not be right.
\( A= \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \), \( B= \begin{bmatrix} 0 & 0 \\ 2 & 0 \end{bmatrix} \)
\( AB= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
\( (AB=AC)\;\wedge\;(A\neq O) \)
\( B=C \) may not be right.
\( A= \begin{bmatrix} 1 & 0 \\ 2 & 0 \end{bmatrix} \), \( B= \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix} \), \( C= \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix} \)
\( AB= \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \), \( AC= \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \)
\( X^n = A \)
It has n solutions in \(X^n\) may not be right.
\( A= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}^2= \) \( \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}^2= \) \( \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}^2= \) \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}^2= \) \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}^2 \)
\( X \) is at least 4 solutions.
\( (A+B)^2 = A^2+AB+BA+B^2 \) is right.
\((A+B)^2=A^2+2AB+B^2\) may not be right. Except \(AB=BA\)
\( A= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \), \( B= \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \)
\( (A+B)^2 = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \)
\( A^2+2AB+B^2= \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix} \)


rref

reduced row echelon form
notice: all pivots = 1
pivot column 上除了 pivot為1 其他為0。
兩相同三角矩陣相乘還是相同的三角矩陣

線性系統

解線性系統的題目(通解(無限多解:(特解, 齊次解), 無解, 唯一解))

Reference

黃子嘉-線性代數