矩阵秩的不等式证明

矩阵秩的不等式及其证明

$B)^{\mathrm{T}}=B^{\mathrm{T}} A^{\mathrm{T}} ( A B ) T = B T A T$

证明

：设

(

)

(

)

A=\left(a_{i j}\right)_{m \times s}, B=\left(b_{i j}\right)_{s \times n}

$A = (a_{i j})_{m \times s}, B = (b_{i j})_{s \times n}$

，记

(

)

(

)

A B=C=\left(c_{i j}\right)_{m \times n}, B^{T} A^{T}=D=\left(d_{i j}\right)_{n \times m}

$A B = C = (c_{i j})_{m \times n}, B^{T} A^{T} = D = (d_{i j})_{n \times m}$

,于是根据矩阵乘法公式，有

∑

c_{j i}=\sum_{k=1}^{s} a_{j k} b_{k i}

$c_{j i} = k = 1 \sum s a_{j k} b_{k i}$

而

\boldsymbol{B}^{\mathbf{T}}

$B^{T}$

的第

i

$i$

行为

⋯

)

\left(b_{1 i}, \cdots, b_{s i}\right), \boldsymbol{A}^{\mathrm{T}}

$(b_{1 i}, \dots, b_{s i}), A^{T}$

的第

j

$j$

列为

⋯

)

\left(a_{j 1}, \cdots, a_{j s}\right)^{T}

$(a_{j 1}, \dots, a_{j s})^{T}$

，因此

∑

d_{i j}=\sum_{k=1}^{s} b_{k i} a_{j k}=\sum_{k=1}^{s} a_{j k} b_{k i}

$d_{i j} = k = 1 \sum s b_{k i} a_{j k} = k = 1 \sum s a_{j k} b_{k i}$

所以

(

⋯

;

⋯

)

d_{i j}=c_{j i} \quad(i=1,2, \cdots, n ; j=1,2, \cdots, m)

$d_{i j} = c_{j i} (i = 1, 2, \dots, n; j = 1, 2, \dots, m)$

即

D=C^{\mathrm{T}}

$D = C^{T}$

$\leqslant r(A) \leqslant \min \{m, n\} 0 ⩽ r ( A ) ⩽ min { m , n }$

这个由定义就很容易理解，不再证明

$\leqslant \min \{r(\boldsymbol{A}), r(\boldsymbol{B})\} r ( A B ) ⩽ min { r ( A ) , r ( B ) }$

分析

：对于向量组

⋯

\alpha_{1}, \alpha_{2}, \cdots, \alpha_{s}

$α_{1}, α_{2}, \dots, α_{s}$

及

⋯

\beta_{1}, \beta_{2}, \cdots, \beta_{t}

$β_{1}, β_{2}, \dots, β_{t}$

，若任一

(

⋯

)

\boldsymbol{\beta}_{i}(i=1,2, \cdots, t)

$β_{i} (i = 1, 2, \dots, t)$

均可由

⋯

\alpha_{1}, \alpha_{2}, \cdots, \alpha_{s}

$α_{1}, α_{2}, \dots, α_{s}$

线性表出，则

(

⋯

)

⩽

(

⋯

)

r\left(\boldsymbol{\beta}_{1}, \boldsymbol{\beta}_{2}, \cdots, \boldsymbol{\beta}_{t}\right) \leqslant r\left(\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \cdots, \boldsymbol{\alpha}_{s}\right)

$r (β_{1}, β_{2}, \dots, β_{t}) ⩽ r (α_{1}, α_{2}, \dots, α_{s})$

，这是本题证明的关键依据

**证明：**不妨设

A_{m \times n}, B_{n \times s}

$A_{m \times n}, B_{n \times s}$

，将

B, A B

$B, A B$

按行分块为

[

⋮

]

[

⋮

]

B=\left[\begin{array}{c}\beta_{1} \\ \beta_{2} \\ \vdots \\ \beta_{n}\end{array}\right], A B=C=\left[\begin{array}{c}\gamma_{1} \\ \gamma_{2} \\ \vdots \\ \gamma_{m}\end{array}\right]

$B = ⎣ ⎢ ⎢ ⎢ ⎡ β_{1} β_{2} ⋮ β_{n} ⎦ ⎥ ⎥ ⎥ ⎤, A B = C = ⎣ ⎢ ⎢ ⎢ ⎡ γ_{1} γ_{2} ⋮ γ_{m} ⎦ ⎥ ⎥ ⎥ ⎤$

，于是

[

⋯

⋮

⋯

]

[

⋮

]

[

⋯

⋮

⋯

]

[

⋮

]

A B=\left[\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & & \vdots \\ a_{m 1} & a_{m 2} & \cdots & a_{n n} \end{array}\right]\left[\begin{array}{c} \beta_{1} \\ \beta_{2} \\ \vdots \\ \beta_{n} \end{array}\right]=\left[\begin{array}{c} a_{11} \beta_{1}+a_{12} \beta_{2}+\cdots+a_{1 n} \beta_{n} \\ a_{21} \beta_{1}+a_{22} \beta_{2}+\cdots+a_{2 n} \beta_{n} \\ \vdots \\ a_{m 1} \beta_{1}+a_{m 2} \beta_{2}+\cdots+a_{m} \beta_{n} \end{array}\right]=\left[\begin{array}{c} \gamma_{1} \\ \gamma_{2} \\ \vdots \\ \gamma_{m} \end{array}\right]

$A B = ⎣ ⎢ ⎢ ⎢ ⎡ a_{11} a_{21} ⋮ a_{m 1} a_{12} a_{22} ⋮ a_{m 2} \dots \dots \dots a_{1 n} a_{2 n} ⋮ a_{n n} ⎦ ⎥ ⎥ ⎥ ⎤ ⎣ ⎢ ⎢ ⎢ ⎡ β_{1} β_{2} ⋮ β_{n} ⎦ ⎥ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎢ ⎡ a_{11} β_{1} + a_{12} β_{2} + \dots + a_{1 n} β_{n} a_{21} β_{1} + a_{22} β_{2} + \dots + a_{2 n} β_{n} ⋮ a_{m 1} β_{1} + a_{m 2} β_{2} + \dots + a_{m} β_{n} ⎦ ⎥ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎢ ⎡ γ_{1} γ_{2} ⋮ γ_{m} ⎦ ⎥ ⎥ ⎥ ⎤$

所以

AB

$A B$

的行向量

(

⋯

)

\gamma_{i}(i=1,2, \cdots, m)

$γ_{i} (i = 1, 2, \dots, m)$

均可由

B

$B$

的行向量线性表出，故

(

)

⩽

(

)

r(A B) \leqslant r(B)

$r (A B) ⩽ r (B)$

同理可证

(

)

⩽

(

)

r(A B) \leqslant r(A)

$r (A B) ⩽ r (A)$

，故有

(

)

⩽

min

⁡

{

(

)

(

)

}

r(A B) \leqslant \min \{r(A), r(B)\}

$r (A B) ⩽ min {r (A), r (B)}$

$\leqslant r(A)+r(B) r ( A + B ) ⩽ r ( A ) + r ( B )$

证明

:设

(

)

(

)

r(A)=p, r(\boldsymbol{B})=q

$r (A) = p, r (B) = q$

，将

A,B

$A, B$

按列分块为

[

⋯

]

[

⋯

]

A=\left[\alpha_{1}, \alpha_{2}, \cdots, \alpha_{s}\right], B=\left[\beta_{1}, \beta_{2}, \cdots, \beta_{s}\right]

$A = [α_{1}, α_{2}, \dots, α_{s}], B = [β_{1}, β_{2}, \dots, β_{s}]$

于是

]

[

⋯

]

[

⋯

]

\begin{array}{c} {[A, B]=\left[\alpha_{1}, \alpha_{2}, \cdots, \alpha_{s}, \beta_{1}, \beta_{2}, \cdots, \beta_{s}\right]} \\ A+B=\left[\alpha_{1}+\beta_{1}, \alpha_{2}+\beta_{2}, \cdots, \alpha_{s}+\beta_{s}\right] \end{array}

$[A, B] = [α_{1}, α_{2}, \dots, α_{s}, β_{1}, β_{2}, \dots, β_{s}] A + B = [α_{1} + β_{1}, α_{2} + β_{2}, \dots, α_{s} + β_{s}]$

因

(

⋯

)

\boldsymbol{\alpha}_{i}+\boldsymbol{\beta}_{i}(i=1,2, \cdots, s)

$α_{i} + β_{i} (i = 1, 2, \dots, s)$

均可由向量组

⋯

\alpha_{1}, \alpha_{2}, \cdots, \alpha_{s}, \beta_{1}, \beta_{2}, \cdots, \beta_{s}

$α_{1}, α_{2}, \dots, α_{s}, β_{1}, β_{2}, \dots, β_{s}$

线性表出，故

(

)

⩽

(

[

]

)

r(A+B) \leqslant r([A, B])

$r (A + B) ⩽ r ([A, B])$

又设

A,B

$A, B$

的列向量的极大线性无关组分别为

⋯

\alpha_{1}, \alpha_{2}, \cdots, \alpha_{p}

$α_{1}, α_{2}, \dots, α_{p}$

和

⋯

\boldsymbol{\beta}_{1}, \boldsymbol{\beta}_{2}, \cdots, \boldsymbol{\beta}_{q}

$β_{1}, β_{2}, \dots, β_{q}$

。将

A

$A$

的极大线性无关组

⋯

\alpha_{1}, \alpha_{2}, \cdots, \alpha_{p}

$α_{1}, α_{2}, \dots, α_{p}$

扩充成

]

[A,B]

$[A, B]$

的极大线性无关组，设为

⋯

\alpha_{1}, \alpha_{2}, \cdots, \alpha_{p}, \beta_{1}, \beta_{2}, \cdots, \beta_{w}

$α_{1}, α_{2}, \dots, α_{p}, β_{1}, β_{2}, \dots, β_{w}$

，显然

⩽

w \leqslant q

$w ⩽ q$

，故有

(

[

]

)

⩽

(

)

(

)

r([A, B])=p+w \leqslant p+q=r(A)+r(B)

$r ([A, B]) = p + w ⩽ p + q = r (A) + r (B)$

故

(

)

⩽

(

[

]

)

⩽

(

)

(

)

r(A+B) \leqslant r([A, B]) \leqslant r(A)+r(B)

$r (A + B) ⩽ r ([A, B]) ⩽ r (A) + r (B)$

(

[

A

O

O

B

]

)

=

r

(

A

)

+

r

(

B

)

r\left(\left[\begin{array}{ll}A & O \\ O & B\end{array}\right]\right)=r(A)+r(B)

$r ([A O O B]) = r (A) + r (B)$
(

A

)

+

r

(

B

)

⩽

r

(

[

A

O

C

B

]

)

⩽

r

(

A

)

+

r

(

B

)

+

r

(

C

)

r(\boldsymbol{A})+r(\boldsymbol{B}) \leqslant r\left(\left[\begin{array}{cc}\boldsymbol{A} & \boldsymbol{O} \\ \boldsymbol{C} & \boldsymbol{B}\end{array}\right]\right) \leqslant r(\boldsymbol{A})+r(\boldsymbol{B})+r(\boldsymbol{C})

$r (A) + r (B) ⩽ r ([A C O B]) ⩽ r (A) + r (B) + r (C)$

证明：

(

)

(

)

(

[

]

)

(

[

]

)

(

[

]

)

⩽

(

[

]

)

⩽

(

[

]

)

(

[

]

)

⩽

(

)

(

)

(

)

\begin{aligned} r(A)+r(B) &=r([A, O])+r([O, B]) \\ &=r\left(\left[\begin{array}{cc} A & O \\ O & B \end{array}\right]\right) \leqslant r\left(\left[\begin{array}{cc} A & O \\ C & B \end{array}\right]\right) \\ & \leqslant r([A, O])+r([C, B]) \\ & \leqslant r(A)+r(B)+r(C) \end{aligned}

$r (A) + r (B) = r ([A, O]) + r ([O, B]) = r ([A O O B]) ⩽ r ([A C O B]) ⩽ r ([A, O]) + r ([C, B]) ⩽ r (A) + r (B) + r (C)$

（因

(

)

⩽

(

[

]

)

⩽

(

)

(

)

r(\boldsymbol{B}) \leqslant r([\boldsymbol{C}, \boldsymbol{B}]) \leqslant r(\boldsymbol{B})+r(\boldsymbol{C})

$r (B) ⩽ r ([C, B]) ⩽ r (B) + r (C)$

）

注解

:第五题也为本题的一个特例

$\geqslant r(A)+r(B)-n r ( A B ) ⩾ r ( A ) + r ( B ) − n (当 A B = O A B=O A B = O 时， r ( A ) + r ( B ) ⩽ n r(A)+r(B) \leqslant n r ( A ) + r ( B ) ⩽ n ， n n n 是 A A A 的列数或 B B B 的行数)$

证明：

：对

]

\left[\begin{array}{ll}\mathbf{A} & \boldsymbol{O} \\ \boldsymbol{E}_{n} & \boldsymbol{B}\end{array}\right]

$[A E_{n} O B]$

做初等变换，得：

]

→

[

−

]

→

[

−

]

→

[

]

\left[\begin{array}{cc} A & O \\ E_{n} & B \end{array}\right] \rightarrow\left[\begin{array}{cc} O & -A B \\ E_{n} & B \end{array}\right] \rightarrow\left[\begin{array}{cc} O & -A B \\ E_{n} & O \end{array}\right] \rightarrow\left[\begin{array}{cc} E_{n} & O \\ O & A B \end{array}\right]

$[A E_{n} O B] \to [O E_{n} - A B B] \to [O E_{n} - A B O] \to [E_{n} O O A B]$

又显然

(

)

(

)

(

[

]

)

⩽

(

[

]

)

r(A)+r(B)=r\left(\left[\begin{array}{ll}A & O \\ O & B\end{array}\right]\right) \leqslant r\left(\left[\begin{array}{ll}A & O \\ E_{n} & B\end{array}\right]\right)

$r (A) + r (B) = r ([A O O B]) ⩽ r ([A E_{n} O B])$

，则

(

)

(

)

⩽

(

[

]

)

(

)

(

)

(

)

r(\boldsymbol{A})+r(\boldsymbol{B}) \leqslant r\left(\left[\begin{array}{cc} \boldsymbol{E}_{n} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{A} \boldsymbol{B} \end{array}\right]\right)=r\left(\boldsymbol{E}_{n}\right)+r(\boldsymbol{A} \boldsymbol{B})=n+r(\boldsymbol{A} \boldsymbol{B})

$r (A) + r (B) ⩽ r ([E_{n} O O A B]) = r (E_{n}) + r (A B) = n + r (A B)$

所以

(

)

⩾

(

)

(

)

−

r(A B) \geqslant r(A)+r(B)-n

$r (A B) ⩾ r (A) + r (B) - n$

(

A

)

=

r

(

A

T

)

=

r

(

A

A

T

)

=

r

(

A

T

A

)

r(A)=r\left(A^{\mathrm{T}}\right)=r\left(A A^{\mathrm{T}}\right)=r\left(A^{\mathrm{T}} A\right)

$r (A) = r (A^{T}) = r (A A^{T}) = r (A^{T} A)$
(

A

∗

)

=

{

n

,

r

(

A

)

=

n

1

,

r

(

A

)

=

n

−

1

0

,

r

(

A

)

<

n

−

1

r\left(A^{*}\right)=\left\{\begin{array}{ll}n, & r(A)=n \\ 1, & r(A)=n-1 \\ 0, & r(A)<n-1\end{array}\right.

$r (A^{*}) = ⎩ ⎨ ⎧ n, 1, 0, r (A) = n r (A) = n - 1 r (A) < n - 1$

，其中

A

$A$

为

n

$n$

阶方阵

证明：

：当

(

)

r(\boldsymbol{A})=n

$r (A) = n$

时，

A

$A$

可逆，

∣

≠

|A| \neq 0

$∣ A ∣ \neq = 0$

，由

∗

∣

A A^{*}=|A| E

$A A^{*} = ∣ A ∣ E$

可知，

A

$A$

和

∗

A^{*}

$A^{*}$

均是可逆矩阵，故

(

∗

)

r\left(A^{*}\right)=n

$r (A^{*}) = n$

当

(

)

−

r(\boldsymbol{A})=n-1

$r (A) = n - 1$

时，由矩阵的秩的定义知，

∣

|A|

$∣ A ∣$

中存在

−

n-1

$n - 1$

阶子式不等于零，而

∗

由

A^{*}由

$A^{*} 由$

∣

|A|

$∣ A ∣$

的元素的

a_{i j}

$a_{i j}$

的代数余子式

A_{i j}

$A_{i j}$

组成，故

(

∗

)

⩾

r\left(\boldsymbol{A}^{*}\right) \geqslant 1

$r (A^{*}) ⩾ 1$

，又

(

)

−

∣

∗

∣

r(\boldsymbol{A})=n-1,|\boldsymbol{A}|=0, \boldsymbol{A} \boldsymbol{A}^{*}=|\boldsymbol{A}| \boldsymbol{E}=\boldsymbol{O}

$r (A) = n - 1, ∣ A ∣ = 0, A A^{*} = ∣ A ∣ E = O$

，得

(

)

(

∗

)

⩽

r(A)+r\left(A^{*}\right) \leqslant n

$r (A) + r (A^{*}) ⩽ n$

，而

(

)

−

r(A)=n-1

$r (A) = n - 1$

，故

(

∗

)

⩽

r\left(A^{*}\right) \leqslant 1

$r (A^{*}) ⩽ 1$

（或

∗

A^{*}

$A^{*}$

的每一列均是

Ax=0

$A x = 0$

的解向量，故

(

∗

)

⩽

r\left(\boldsymbol{A}^{*}\right) \leqslant 1

$r (A^{*}) ⩽ 1$

），所以

(

∗

)

r\left(\boldsymbol{A}^{*}\right)=1

$r (A^{*}) = 1$

当

(

)

−

r(\boldsymbol{A})<n-1

$r (A) < n - 1$

时，由

A

$A$

的秩的定义知，

∣

|A|

$∣ A ∣$

的代数余子式全部为零，故

(

∗

)

r\left(A^{*}\right)=0

$r (A^{*}) = 0$

注意伴随矩阵、可逆矩阵一定是方阵

若
$A^{2}=A A 2 = A ,则 r ( A ) + r ( E − A ) = n r(\boldsymbol{A})+r(\boldsymbol{E}-\boldsymbol{A})=n r ( A ) + r ( E − A ) = n$

证明

：由

A^{2}=A

$A^{2} = A$

，得

(

−

)

\boldsymbol{A}(\boldsymbol{A}-\boldsymbol{E})=\boldsymbol{O}

$A (A - E) = O$

，故

(

)

(

−

)

⩽

r(\boldsymbol{A})+r(\boldsymbol{A}-\boldsymbol{E}) \leqslant n

$r (A) + r (A - E) ⩽ n$

又

(

)

(

−

)

(

)

(

−

)

⩾

(

−

)

(

)

r(A)+r(A-E)=r(A)+r(E-A) \geqslant r(A+E-A)=r(E)=n

$r (A) + r (A - E) = r (A) + r (E - A) ⩾ r (A + E - A) = r (E) = n$

，得证

(

)

(

−

)

r(\boldsymbol{A})+r(\boldsymbol{A}-\boldsymbol{E})=n

$r (A) + r (A - E) = n$

若
$A^{2}=E A 2 = E ,则 r ( A + E ) + r ( A − E ) = n r(A+E)+r(A-E)=n r ( A + E ) + r ( A − E ) = n$

解析

：由题设

A^{2}=E

$A^{2} = E$

知

)

(

−

)

−

(A+E)(A-E)=A^{2}-E=O

$(A + E) (A - E) = A^{2} - E = O$

于是有

(

)

(

−

)

⩽

r(A+E)+r(A-E) \leqslant n

$r (A + E) + r (A - E) ⩽ n$

注意到

(

−

)

(

−

)

r(\boldsymbol{A}-\boldsymbol{E})=r(-\boldsymbol{A}+\boldsymbol{E})

$r (A - E) = r (- A + E)$

，则

(

)

(

−

)

(

)

(

−

)

⩾

(

−

)

(

)

\begin{aligned} r(A+E)+r(A-E) &=r(A+E)+r(-A+E) \\ & \geqslant r(A+E-A+E) \\ &=r(2 E)=n \end{aligned}

$r (A + E) + r (A - E) = r (A + E) + r (- A + E) ⩾ r (A + E - A + E) = r (2 E) = n$

综上所述

(

)

(

−

)

r(A+E)+r(A-E)=n

$r (A + E) + r (A - E) = n$

得证

原文链接：https://blog.csdn.net/baidu_41922918/article/details/116453437

矩阵秩的不等式及其证明

你可能也喜欢