Matrix

• Prerequisites
• 1. Matrix
• 2. Matrix Addition
• 3. Scalar Multiplication
• 4. Matrix Multiplication
  • 4.1. Properties of Matrix Multiplication
    • 4.1.1. Associativity
    • 4.1.2. Distributivity
    • 4.1.3. Square Matrix
• 5. Transpose
• 6. Conjugate
• 7. Hermitian
• 8. Block Matrix Multiplication
• 9. Trace
• 10. Inner Product
• 11. Determinant
  • 11.1. Alternating multilinear map
  • 11.2. Laplace expansion
• 12. Adjugate matrix

Prerequisites

• 环论(一)

Let $F$ be a field(e.g. $F=\mathbb{Q},F=\mathbb{R}$ or $F=\mathbb{C}$)

$0_F$ is the addtive unit element of $F$.

$1_F$ is the multiply unit element of $F$.

We will use $0$ and $1$ to denote $0_F$ and $1_F$ respectively if there is no confusion about the field $F$.

1. Matrix

A matrix is a rectangular array of elements from a field $F$.

$$A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{m1}& a_{m2}& \cdots & a_{mn}\end{array}\right]$$

The element $a_{ij}$ is called the $(i,j)$-th entry of $A$.

$F^{m\times n}$ is the set of all $m\times n$ matrices with entries in $F$.

2. Matrix Addition

Let $A=[a_{ij}]$ and $B=[b_{ij}]$ be two matrices of the same size. The sum of $A$ and $B$ is the matrix $C=[c_{ij}]$ defined by $c_{ij}=a_{ij}+b_{ij}$.

It is easy to see that $⟨F^{m\times n},+⟩$ is an commutative group.

3. Scalar Multiplication

Let $A=[a_{ij}]$ be a matrix and $c\in F$ be a scalar. The product of $A$ and $c$ is the matrix $C=[c_{ij}]$ defined by $c_{ij}=c{a}_{ij}$.

It is easy to see that $⟨F^{m\times n},+,\cdot ⟩$ is a vector space over $F$.

4. Matrix Multiplication

Let $A=[a_{ij}]\in F^{m\times n}$ and $B=[b_{ij}]\in F^{n\times p}$ be two matrices. The product of $A$ and $B$ is the matrix $C=[c_{ij}]\in F^{m\times p}$ defined by $c_{ij}={\sum }_{k=1}^n{a}_{ik}{b}_{kj}$.

4.1. Properties of Matrix Multiplication

4.1.1. Associativity

Let $A=[a_{ij}]\in F^{m\times n}$, $B=[b_{ij}]\in F^{n\times p}$, $C=[c_{ij}]\in F^{p\times q}$ be three matrices.

$$(AB)C=A(BC)$$

proof

$$\begin{array}{rl}((AB)C{)}_{ij}& =\sum _{k=1}^p(AB{)}_{ik}{c}_{kj}\\ & =\sum _{k=1}^p(\sum _{l=1}^n{a}_{il}{b}_{lk})c_{kj}\\ & =\sum _{k=1}^p\sum _{l=1}^n{a}_{il}{b}_{lk}{c}_{kj}\end{array}$$ $$\begin{array}{rl}A(BC{)}_{ij}& =\sum _{l=1}^n{a}_{il}(BC{)}_{lj}\\ & =\sum _{l=1}^n{a}_{il}\sum _{k=1}^p{b}_{lk}{c}_{kj}\\ & =\sum _{l=1}^n\sum _{k=1}^p{a}_{il}{b}_{lk}{c}_{kj}\\ & =\sum _{k=1}^p\sum _{l=1}^n{a}_{il}{b}_{lk}{c}_{kj}\\ & =((AB)C{)}_{ij}\end{array}$$

4.1.2. Distributivity

$$A(B+C)=AB+AC$$

proof

$$\begin{array}{rl}(A(B+C){)}_{ij}& =\sum _{k=1}^n{a}_{ik}(B+C{)}_{kj}\\ & =\sum _{k=1}^n{a}_{ik}(b_{kj}+c_{kj})\\ & =\sum _{k=1}^n{a}_{ik}{b}_{kj}+\sum _{k=1}^n{a}_{ik}{c}_{kj}\\ & =(AB{)}_{ij}+(AC{)}_{ij}\\ & =(AB+AC{)}_{ij}\end{array}$$

4.1.3. Square Matrix

A matrix $A$ is called a square matrix if $m=n$.

Identity Matrix

The $n\times n$ identity matrix $I_n$ is the matrix with $1_F$ on the diagonal and $0_F$ elsewhere.

$$I_n=\left[\begin{array}{cccc}{1}_F& 0_F& \cdots & 0_F\\ 0_F& 1_F& \cdots & 0_F\\ ⋮& ⋮& \ddots & ⋮\\ 0_F& 0_F& \cdots & 1_F\end{array}\right]$$

$I$ is the identity element of $⟨F^{n\times n},\cdot ⟩$.

$$AI=IA=A$$

proof

$$\begin{array}{rl}(AI{)}_{ij}& =\sum _{k=1}^n{a}_{ik}(I{)}_{kj}\\ & =\sum _{k=1}^n{a}_{ik}{\delta }_{kj}\\ & =a_{ij}\end{array}$$

similar, we have $IA=A$.

$⟨F^{n\times n},+,\cdot ⟩$ is a ring with identity.

because all the units in a ring form a group under multiplication(this group is called the group of units of the ring), the group of units of $⟨F^{n\times n},+,\cdot ⟩$ is denoted by $⟨GL(n,F),\cdot ⟩$ or $GL(n,F)$ or $G{L}_n(F)$.

$GL(n,F)$ is called the general linear group of degree $n$ over $F$.

5. Transpose

Let $A=[a_{ij}]\in F^{m\times n}$ be a matrix. The transpose of $A$ is the matrix $A^T=[a_{ji}]\in F^{n\times m}$ defined by $a_{ji}=a_{ij}$.

$A\in F^{m\times n},B\in F^{n\times p}$, we have:
$(AB{)}^T=B^T{A}^T$

proof

$$\begin{array}{rl}(AB{)}_{ij}^T& =(AB{)}_{ji}\\ & =\sum _{k=1}^n{a}_{jk}{b}_{ki}\\ & =\sum _{k=1}^n(A^T{)}_{kj}(B^T{)}_{ik}\\ & =(B^T{A}^T{)}_{ij}\end{array}$$

6. Conjugate

If $F$ is a subfield of $\mathbb{C}$, the conjugate of $a\in F$ is $\overline{a}\in F$.

Let $A=[a_{ij}]\in F^{m\times n}$ be a matrix. The conjugate of $A$ is the matrix $\overline{A}=[\overline{a_{ij}}]\in F^{m\times n}$

• $A,B\in F^{m\times n}$, $\overline{A+B}=\overline{A}+\overline{B}$

• $A\in F^{m\times n},B\in F^{n\times p}$, $\overline{A\cdot B}=\overline{A}\cdot \overline{B}$

7. Hermitian

If $F$ is a subfield of $\mathbb{C}$

Let $A=[a_{ij}]\in F^{m\times n}$ be a matrix. The Hermitian of $A$ is the matrix $A^H=[\overline{a_{ji}}]\in F^{n\times m}$ defined by $a_{ji}=\overline{a_{ij}}$.

8. Block Matrix Multiplication

let $A$ be a $m\times n$ matrix, $B$ be a $n\times p$ matrix, divide $A$ and $B$ into blocks:

$$A=\left[\begin{array}{cccc}{A}_{11}& A_{12}& \cdots & A_{1s}\\ A_{21}& A_{22}& \cdots & A_{2s}\\ ⋮& ⋮& \ddots & ⋮\\ A_{r1}& A_{r2}& \cdots & A_{rs}\end{array}\right],B=\left[\begin{array}{cccc}{B}_{11}& B_{12}& \cdots & B_{1t}\\ B_{21}& B_{22}& \cdots & B_{2t}\\ ⋮& ⋮& \ddots & ⋮\\ B_{s1}& B_{s2}& \cdots & B_{st}\end{array}\right]$$

where $A_{ij}$ is a $m_i\times n_j$ matrix, $B_{jk}$ is a $n_j\times p_k$ matrix, and satisfies:

• ${\sum }_{i=1}^r{m}_i=m$
• ${\sum }_{j=1}^s{n}_j=n$
• ${\sum }_{k=1}^t{p}_k=p$

then $C=AB$ can be represented as:

$$C=\left[\begin{array}{cccc}{C}_{11}& C_{12}& \cdots & C_{1t}\\ C_{21}& C_{22}& \cdots & C_{2t}\\ ⋮& ⋮& \ddots & ⋮\\ C_{r1}& C_{r2}& \cdots & C_{rt}\end{array}\right]$$

where $C_{ik}={\sum }_{j=1}^s{A}_{ij}{B}_{jk}$, and $C_{ik}$ is a $m_i\times p_k$ matrix.

9. Trace

Let $A=[a_{ij}]\in F^{n\times n}$ be a square matrix. The trace of $A$ is the sum of the diagonal elements of $A$, denoted by $tr(A)$.

$$tr(A)=\sum _{i=1}^n{a}_{ii}$$

Properties of Trace

• $tr(A+B)=tr(A)+tr(B)$
• $tr(cA)=ctr(A)$
• $tr(AB)=tr(BA)$
• $\overline{tr(A)}=tr(\overline{A})$
• $tr(A^T)=tr(A)$

proof : $tr(AB)=tr(BA)$

$$\begin{array}{rl}tr(AB)& =\sum _{i=1}^n(AB{)}_{ii}\\ & =\sum _{i=1}^n\sum _{j=1}^n{a}_{ij}{b}_{ji}\\ & =\sum _{j=1}^n\sum _{i=1}^n{b}_{ji}{a}_{ij}\\ & =tr(BA)\end{array}$$

10. Inner Product

Recall that $⟨F^{m\times n},+,\cdot ⟩$ is a vector space over $F$.

If $F$ is a subfield of $\mathbb{C}$, we can define an inner product on $F^{m\times n}$ as follows:

$$⟨A,B⟩=tr(A^{H}B)$$

It is easy to verify that $⟨A,B⟩$ satisfies the following properties:

• conjugate symmetry: $⟨A,B⟩=\overline{⟨B,A⟩}$
• right linearity: $⟨A,bB+cC⟩=b⟨A,B⟩+c⟨A,C⟩$
• positive-definiteness: $⟨A,A⟩\ge 0$, $⟨A,A⟩=0$ if and only if $A=0$

proof: $⟨A,B⟩=\overline{⟨B,A⟩}$

$$\begin{array}{rl}⟨A,B⟩& =tr(A^{H}B)\\ & =\overline{\overline{tr(A^{H}B)}}\\ & =\overline{tr(A^T\overline{B})}\\ & =\overline{tr(B^{H}A)}\\ & =\overline{⟨B,A⟩}\end{array}$$

proof: $⟨A,bB+cC⟩=b⟨A,B⟩+c⟨A,C⟩$

$$\begin{array}{rl}⟨A,bB+cC⟩& =tr(A^H(bB+cC))\\ & =tr(b{B}^{H}A+c{C}^{H}A)\\ & =btr(B^{H}A)+ctr(C^{H}A)\\ & =b⟨A,B⟩+c⟨A,C⟩\end{array}$$

proof: $⟨A,A⟩\ge 0$

$$\begin{array}{rl}⟨A,A⟩& =tr(A^{H}A)\\ & =\sum _{i=1}^n\sum _{j=1}^n{a}_{ij}\overline{a_{ij}}\\ & =\sum _{i=1}^n|a_{ii}{|}^2\\ & \ge 0\end{array}$$

It is easy to see that $⟨A,A⟩=0$ if and only if $A=0$.

11. Determinant

11.1. Alternating multilinear map

let $⟨F,V⟩$ be a vector space over $F$. A map $f:V^n\to F$ is called an alternating multilinear map if:

• multilinear: $f$ is linear in each variable
• alternating: $f(v_1,\cdots ,v_i,\cdots ,v_j,\cdots ,v_n)=-f(v_1,\cdots ,v_j,\cdots ,v_i,\cdots ,v_n)$ for any $v_i,v_j\in V$

Let $A=[a_{ij}]\in F^{n\times n}$ be a square matrix. The determinant of $A$ is the scalar $\det (A)$ defined by:

$$|A|:=\det (A)=\sum _{\sigma \in S_n}\text{sgn}(\sigma )\prod _{i=1}^n{a}_{i,\sigma (i)}$$

or equivalently,

$$|A|=\sum _{j_1,j_2,\cdots ,j_n\in [n]}\tau (j_1,j_2,\cdots ,j_n)\prod _{i=1}^n{a}_{i,j_i}$$

where $S_n$ is the symmetric group of degree $n$, $\text{sgn}(\sigma )$ is the sign of the permutation $\sigma$, and $a_{i,\sigma (i)}$ is the element of $A$ at the $i$-th row and $\sigma (i)$-th column.

$$\text{sgn}(\sigma )=\{\begin{array}{ll}{1}_F& \text{if }\sigma \text{ is an even permutation}\\ -1_F& \text{if }\sigma \text{ is an odd permutation}\end{array}$$

$\tau (j_1,j_2,\cdots ,j_n)$ is the number of inverse pairs of the sequence $(j_1,j_2,\cdots ,j_n)$.

It is not difficult to see that:

$$\text{sgn}(\sigma )=(-1{)}^{\tau (j_1,j_2,\cdots ,j_n)}$$

• Determinant is a multilinear map for each row of $A$
• Determinant does not change under transposition, i.e. $\det (A)=\det (A^T)$
• Determinant is a multilinear map for each column of $A$
• Determinant is an alternating multilinear map for both each row of $A$ and each column of $A$
• $\det (A)={\sum }_{j_1,j_2,\cdots ,j_n\in [n]}(-1{)}^{\tau (i_1,i_2,\cdots ,i_n)+\tau (j_1,j_2,\cdots ,j_n)}{\prod }_{k=1}^n{a}_{i_k,j_k}$

11.2. Laplace expansion

$I=i_1,i_2,\cdots ,i_k$
$I^{\prime }=[n]\setminus I$

$J=j_1,j_2,\cdots ,j_k$
$J^{\prime }=[n]\setminus J$

$\sum (X):={\sum }_{x\in X}x$

$A_{I,J}$ is the submatrix of $A$ obtained by deleting the $i$-th $\in I^{\prime }$ row and $j$-th $\in J^{\prime }$ column.

$$\det (A)=\sum _{J\subseteq [n]\wedge |J|=k}(-1{)}^{\sum (I)+\sum (J)}\det (A_{I,J})\det (A_{I^{\prime },J^{\prime }})$$

• $\det (AB)=\det (A)\det (B)$

12. Adjugate matrix

$A_{ij}$ is the submatrix of $A$ obtained by deleting the $i$-th row and $j$-th column.

Let $A=[a_{ij}]\in F^{n\times n}$ be a square matrix. The adjugate of $A$ is the matrix $A^{\#}:=adj(A)=[a_{ij}^{\#}]\in F^{n\times n}$ defined by $a_{ij}^{\#}=(-1{)}^{i+j}\det (A_{ji})$.

$$A{A}^{\#}=A^{\#}A=\det (A)I$$

proof

$$\begin{array}{rl}(A{A}^{\#}{)}_{ij}& =\sum _{k=1}^n{a}_{ik}{a}_{kj}^{\#}\\ & =\sum _{k=1}^n{a}_{ik}(-1{)}^{k+j}\det (A_{jk})\\ & =\det (A){\delta }_{ij}\end{array}$$

• $A$ is invertible if and only if $\det (A)\ne 0$
• if $A$ is invertible, then $A^{-1}=\frac{1_F}{\det (A)}{A}^{\#}$