Determinant

It's a function whose definition domain is the \(n\times n\) matrix \(A\).

Für eine nur aus einem Koeffizienten bestehende \(1\times1\)-Matrix \(A\) ist

\(\det A=\det(a_{11})=a_{11}\)

ist \(A\) eine \(2\times 2\)-Matrix, dann ist

\(\det A=\det\left(\begin{array}{cc}a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right)=a_{11}*a_{22}-a_{12}*a_{21}\)

für eine \(N\times N\)-Matrix \(A\) gilt die Formel:

\(\det A=\sum_{j_1\cdots j_n}(-1)^{\tau(j_1\cdots j_n)}a_{1j_1}a_{2j_2}\cdots a_{nj_n}\)

which \(\tau\) means number of reverse pairs of permutation \(j_1\cdots j_n\)

Matrix

A table of numbers in m rows and n columns arranged by m*n numbers \(a_{ij}\) is called a matrix of m rows and n columns.

Addition, Subtraction, Multiple with numbers, Multiple with another matrix, Transpose of the matrix

Invertible Matrix

\(A^*\) is adjugate matrix

define: \(A_{ij}\) is matrix determinant multiply \((-1)^{i+j}\) which delete i-th row and j-th column

\(A^*=\left(\begin{array}{cc} A_{11}&\cdots&A_{1n}\\ &\cdots&\\mathbb A_{n1}&\cdots& A_{nn} \end{array}\right)\)

\(A^{-1}=\frac{1}{\det A}A^*\)

Another way of calculating invert matrix

\((A,I)\) Elementary Line Transformation to \((I,B)\), then \(B=A^{-1}\)

Cramer's Rule

\(\left\{\begin{array}{rcl} a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n=b_1\\a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n=b_2\\ \cdots\cdots\cdots\\a_{n1}x_1+a_{n2}x_2+\cdots+a_{nn}x_n=b_n \end{array}\right.\)

\(D=\det\left(\begin{array}{cc}a_{11}&a_{12}&\cdots&a_{1n}\\a_{21}&a_{22}&\cdots&a_{2n}\\&\cdots&\cdots\\a_{n1}&a_{n2}&\cdots&a_{nn} \end{array}\right)\)

\(D_1=\det\left(\begin{array}{cc}b_1&a_{12}&\cdots&a_{1n}\\b_2&a_{22}&\cdots&a_{2n}\\&\cdots&\cdots\\b_n&a_{n2}&\cdots&a_{nn} \end{array}\right)\)

\(D_2=\det\left(\begin{array}{cc}a_{11}&b_1&\cdots&a_{1n}\\a_{21}&b_2&\cdots&a_{2n}\\&\cdots&\cdots\\a_{n1}&b_n&\cdots&a_{nn} \end{array}\right)\)

......

\(D_n=\det\left(\begin{array}{cc}a_{11}&a_{12}&\cdots&b_1\\a_{21}&a_{22}&\cdots&b_2\\&\cdots&\cdots\\a_{n1}&a_{n2}&\cdots&b_n \end{array}\right)\)

Then, \(x_i=\frac{D_i}{D}\). (if D=0, no solution.)

Eigen Value

矩阵的特征值

z.B. what's the eigen value & eigen vector of \(A=\left(\begin{array}{cc} -1&1&1\\1&-1&1\\1&1&-1 \end{array}\right)\)?

Eigen polynomial of \(A\) is: \(|\lambda I-A|=\det\left(\begin{array}{cc} \lambda+1&-1&-1\\-1&\lambda+1&-1\\-1&-1&\lambda+1 \end{array}\right)\)

\(=\det\left(\begin{array}{cc} \lambda+1-1-1&-1+\lambda+1-1&-1-1+\lambda+1\\-1&\lambda+1&-1\\-1&-1&\lambda+1 \end{array}\right)\)

\(=(\lambda-1)\times\det\left(\begin{array}{cc} 1&1&1\\-1&\lambda+1&-1\\-1&-1&\lambda+1 \end{array}\right)\)

\(=(\lambda-1)\times\det\left(\begin{array}{cc} 1&1&1\\0&\lambda+2&0\\0&0&\lambda+2 \end{array}\right)\)

\(=(\lambda-1)(\lambda+2)^2\)

\(|\lambda I-A|=0\Rightarrow\lambda=1,-2\)

So 1 and -2 are eigen values of matrix A.

For eigen vector \(x\), \(Ax=\lambda x\)

\((A-\lambda I)x=0\), solve the 2 equations with \(\lambda=1,-2\)

when \(\lambda=1\), \(x_1=x_2=x_3\), so \(k_1=(1,1,1)^T\), \(\xi k_1(\xi\neq0)\) is eigen vector of \(A\).

when \(\lambda=-2\), \(x_1+x_2+x_3=0\) so \(k_2=(a,b,-a-b)^T\) is eigen vector of \(A\).

trace \(tr(A)=\sum\lambda\)

determinant \(\det A=\Pi\lambda\)