Gram 矩阵及其主要性质
文章目录
Gram 矩阵6 大性质
Gram 矩阵
假设
A
A
A 是一个
m
×
n
m\times n
m×n 阶矩阵,
列向量 Gram 矩阵
A
A
A 由列向量
α
i
\mathbf{\alpha}_i
αi 表示, 即
A
=
[
α
1
α
2
⋯
α
n
]
A=\begin{bmatrix}\mathbf{\alpha}_1 & \mathbf{\alpha}_2 &\cdots & \mathbf{\alpha}_n \end{bmatrix}
A=[α1α2⋯αn] 则
G
=
A
T
A
=
[
α
1
T
α
2
T
⋮
α
n
T
]
[
α
1
α
2
⋯
α
n
]
=
[
α
1
T
α
1
α
1
T
α
2
⋯
α
1
T
α
n
α
2
T
α
1
α
2
T
α
2
⋯
α
2
T
α
n
⋮
⋮
⋮
α
n
T
α
1
α
n
T
α
2
⋯
α
n
T
α
n
]
\begin{aligned} G &= \, A^{\mathsf T}A \\[3pt] &= \begin{bmatrix} \mathbf{\alpha}_1^{\mathsf T} \\ \mathbf{\alpha}_2^{\mathsf T} \\ \vdots \\ \mathbf{\alpha}_n^{\mathsf T} \end{bmatrix} \begin{bmatrix} \mathbf{\alpha}_1 & \mathbf{\alpha}_2 & \cdots & \mathbf{\alpha}_n \end{bmatrix} \\[3pt] & = \begin{bmatrix} \mathbf{\alpha}_1^{\mathsf T}\mathbf{\alpha}_1 & \mathbf{\alpha}_1^{\mathsf T}\mathbf{\alpha}_2 & \cdots & \mathbf{\alpha}_1^{\mathsf T}\mathbf{\alpha}_n \\ \mathbf{\alpha}_2^{\mathsf T}\mathbf{\alpha}_1 & \mathbf{\alpha}_2^{\mathsf T}\mathbf{\alpha}_2 & \cdots &\mathbf{\alpha}_2^{\mathsf T}\mathbf{\alpha}_n \\ \vdots & \vdots & & \vdots \\ \mathbf{\alpha}_n^{\mathsf T}\mathbf{\alpha}_1 & \mathbf{\alpha}_n^{\mathsf T}\mathbf{\alpha}_2 & \cdots & \mathbf{\alpha}_n^{\mathsf T}\mathbf{\alpha}_n \end{bmatrix} \end{aligned}
G=ATA=⎣⎢⎢⎢⎡α1Tα2T⋮αnT⎦⎥⎥⎥⎤[α1α2⋯αn]=⎣⎢⎢⎢⎡α1Tα1α2Tα1⋮αnTα1α1Tα2α2Tα2⋮αnTα2⋯⋯⋯α1Tαnα2Tαn⋮αnTαn⎦⎥⎥⎥⎤
行向量 Gram 矩阵
A
A
A 由行向量
β
i
T
\mathbf{\beta}_i^{\mathsf T}
βiT 表示, 即
A
=
[
β
1
T
β
2
T
⋮
β
m
T
]
A=\begin{bmatrix}\mathbf{\beta}_1^{\mathsf T} \\ \mathbf{\beta}_2^{\mathsf T} \\ \vdots \\ \mathbf{\beta}_m^{\mathsf T} \end{bmatrix}
A=⎣⎢⎢⎢⎡β1Tβ2T⋮βmT⎦⎥⎥⎥⎤ 则
G
=
A
A
T
=
[
β
1
T
β
2
T
⋮
β
m
T
]
[
β
1
β
2
⋯
β
m
]
=
[
β
1
T
β
1
β
1
T
β
2
⋯
β
1
T
β
m
β
2
T
β
1
β
2
T
β
2
⋯
β
2
T
β
m
⋮
⋮
⋮
β
m
T
β
1
β
m
T
β
2
⋯
β
m
T
β
m
]
\begin{aligned} G &= \, AA^{\mathsf T} \\[3pt] &= \begin{bmatrix}\mathbf{\beta}_1^{\mathsf T} \\ \mathbf{\beta}_2^{\mathsf T} \\ \vdots \\ \mathbf{\beta}_m^{\mathsf T} \end{bmatrix} \begin{bmatrix} \mathbf{\beta}_1 & \mathbf{\beta}_2 & \cdots & \mathbf{\beta}_m \end{bmatrix} \\[3pt] & = \begin{bmatrix} \mathbf{\beta}_1^{\mathsf T}\mathbf{\beta}_1 & \mathbf{\beta}_1^{\mathsf T}\mathbf{\beta}_2 & \cdots & \mathbf{\beta}_1^{\mathsf T}\mathbf{\beta}_m \\ \mathbf{\beta}_2^{\mathsf T}\mathbf{\beta}_1 & \mathbf{\beta}_2^{\mathsf T}\mathbf{\beta}_2 & \cdots &\mathbf{\beta}_2^{\mathsf T}\mathbf{\beta}_m \\ \vdots & \vdots & & \vdots \\ \mathbf{\beta}_m^{\mathsf T}\mathbf{\beta}_1 & \mathbf{\beta}_m^{\mathsf T}\mathbf{\beta}_2 & \cdots & \mathbf{\beta}_m^{\mathsf T}\mathbf{\beta}_m \end{bmatrix} \end{aligned}
G=AAT=⎣⎢⎢⎢⎡β1Tβ2T⋮βmT⎦⎥⎥⎥⎤[β1β2⋯βm]=⎣⎢⎢⎢⎡β1Tβ1β2Tβ1⋮βmTβ1β1Tβ2β2Tβ2⋮βmTβ2⋯⋯⋯β1Tβmβ2Tβm⋮βmTβm⎦⎥⎥⎥⎤
6 大性质
下面只考虑列向量 Gram 矩阵
(1)
G
=
A
T
A
G = \, A^{\mathsf T}A
G=ATA 是对称矩阵
G
T
=
(
A
T
A
)
T
=
A
T
A
=
G
G^{\mathsf T } = \, (A^{\mathsf T}A)^{\mathsf T} = \, A^{\mathsf T}A = G
GT=(ATA)T=ATA=G
(2) 对于实矩阵
A
A
A
r
a
n
k
(
A
T
A
)
=
r
a
n
k
(
A
)
\mathrm{rank} (A^{\mathsf T}A) = \mathrm{rank} (A)
rank(ATA)=rank(A)
证明
{
A
x
=
0
A
T
A
x
=
0
\begin{cases} A\mathsf{x} = 0 \\ A^{\mathsf T}A\mathbf{x} = 0 \end{cases}
{Ax=0ATAx=0 同解即可.
证明过程详见经典例题(第３小问)
(3) 若
A
T
A
=
0
A^{\mathsf T}A=0
ATA=0, 则
A
=
0
A = 0
A=0
由上面性质
r
a
n
k
(
A
T
A
)
=
r
a
n
k
(
A
)
=
r
a
n
k
(
0
)
=
0
\begin{aligned} \mathrm{rank} (A^{\mathsf T}A) &= \mathrm{rank} (A) \\ &= \mathrm{rank} \ (0) = 0 \end{aligned}
rank(ATA)=rank(A)=rank (0)=0
(4) 对于实矩阵
A
A
A, 则
A
T
A
A^{\mathsf T}A
ATA 是半正定矩阵
x
T
A
T
A
x
=
(
A
x
)
T
A
x
≥
0
\mathbf{x}^{\mathsf T}A^{\mathsf T}A\mathbf{x} = (A\mathbf{x})^{\mathsf T}A\mathbf{x} \geq 0
xTATAx=(Ax)TAx≥0
(5) 对于任意
n
n
n 阶实对称半正定矩阵
M
M
M, 存在矩阵
A
A
A 使得
M
=
A
T
A
M=A^{\mathsf T}A
M=ATA 成立.
因为矩阵
M
M
M 实对称, 所以
M
M
M 可以正交对角化, 即
M
=
Q
Λ
Q
T
M = Q\Lambda Q^{\mathsf T}
M=QΛQT 又因为矩阵
M
M
M 半正定, 所以其特征值 $\lambda_i \geq 0 $, 所以可记
Λ
1
/
2
=
d
i
a
g
(
λ
1
,
…
,
λ
n
)
\Lambda^{1/2} = \mathrm{diag} ( \sqrt{\lambda_1}, \dots, \sqrt{\lambda_n})
Λ1/2=diag(λ1
,…,λn
) 且 KaTeX parse error: Expected 'EOF', got '}' at position 29: …2}Q^\{\mathsf T}̲ 则可得
M
=
Q
Λ
Q
T
=
(
Λ
1
/
2
Q
T
)
T
Λ
1
/
2
Q
T
=
A
T
A
\begin{aligned} M &= Q\Lambda Q^{\mathsf T} \\ &= (\Lambda^{1/2}Q^{\mathsf T})^{\mathsf T}\Lambda^{1/2}Q^{\mathsf T} \\ &= A^{\mathsf T}A \end{aligned}
M=QΛQT=(Λ1/2QT)TΛ1/2QT=ATA
(6) 若
A
=
[
α
1
α
2
⋯
α
n
]
A=\begin{bmatrix}\mathbf{\alpha}_1 & \mathbf{\alpha}_2 &\cdots & \mathbf{\alpha}_n \end{bmatrix}
A=[α1α2⋯αn] 列满秩, 则
A
T
A
A^{\mathsf T}A
ATA 正定
由性质 (2), 知
r
a
n
k
(
A
T
A
)
=
r
a
n
k
(
A
)
=
n
\mathrm{rank} (A^{\mathsf T}A) = \mathrm{rank} (A) = n
rank(ATA)=rank(A)=n因为
A
x
=
0
A\mathbf{x}=0
Ax=0 只有零解, 结合性质 (4), 对于非零
x
∈
R
n
\mathbf{x}\in \mathbb{R}^n
x∈Rn
x
T
A
T
A
x
=
(
A
x
)
T
A
x
>
0
\mathbf{x}^{\mathsf T}A^{\mathsf T}A\mathbf{x} = (A\mathbf{x})^{\mathsf T}A\mathbf{x} > 0
xTATAx=(Ax)TAx>0
原文链接 [1] matnoble.me/posts/gram [2] 关注我吧