特殊矩阵

特殊矩阵

这篇博客篇幅较长,总结了很多特殊矩阵:

在实数域:

  1. 对称矩阵: 原矩阵 = 转置矩阵
  2. Gramian 矩阵: \(A \to A^TA\)
  3. 正交矩阵: 矩阵 \(A\) 对应的 Gramian 矩阵为 \(n\) 阶单位阵

在复数域:

  1. Hermitian matrix: 原矩阵 = 共轭转置矩阵
  2. unitary matrix: 逆矩阵 = 共轭转置矩阵
  3. normal martix

1. 对称矩阵

如果 \(A\) 满足 \(A^T = A\) ,则 \(A\) 为对称矩阵,(注: 先不讨论复数)

性质:

  1. 不同特征值对应的特征向量正交
    证明: 假设由两个特征值 \(\lambda\)\(\mu\) ,且 \(\lambda \neq \mu\) ,对应的特征向量为 \(\mathbf {x}\)\(\mathbf {y}\), 即 \(A \mathbf {x} = \lambda \mathbf {x}\)\(A \mathbf {y} = \mu \mathbf {y}\) 则: \[
    \begin {align}
    \mu \mathbf {y}^{T} \mathbf {x} &= (\mu \mathbf {y})^{T} \mathbf {x} = (A\mathbf {y})^{T} \mathbf {x} = \mathbf {y}^{T} A^T\mathbf {x} \\
    &= \mathbf {y}^{T} A\mathbf {x} =\mathbf {y}^{T} (A\mathbf {x}) = \lambda \mathbf {y}^{T} \mathbf {x}
    \end {align}
    \]
    \(\lambda \neq \mu\) ,故 \(\mathbf {y}^{T} \mathbf {x} = 0\) ,即 \(\mathbf {x}\)\(\mathbf {y}\) 正交

  2. 所有特征值均为实数,且对应特征向量是实向量
    证明: 假定 \(\mathbf {x} = \lambda \mathbf {x}\), \(\lambda \in \mathbb {C}\) , \(\mathbf {x} \in \mathbb {C}\)\(\mathbf {x} \neq \mathbf {0}\) , 等式两边同取共轭,就有 \(\overline {A} \overline {\mathbf {x}} = \overline {\lambda} \overline {\mathbf {x}}\), 再取转置, \[
    \overline {\mathbf {x}}^T \overline {A}^T = \overline {\lambda} \overline {\mathbf {x}}^T
    \]
    因为 \(A\) 为实对称矩阵,\(\overline {A}^T = \overline {A} = A\) , 于是 \[
    \overline { \mathbf {x}}^T A = \overline {\lambda} \overline {\mathbf {x}}^T
    \]
    上式左右同乘以 \(\mathbf {x}\), 得到: \[
    \overline { \mathbf {x}}^T A \mathbf {x} = \overline { \mathbf {x}}^T (\lambda \mathbf {x}) = \lambda \overline { \mathbf {x}}^T \mathbf {x}
    \]
    于是得到: \[
    (\lambda – \overline \lambda) \overline { \mathbf {x}}^T \mathbf {x} = 0
    \]
    \(\overline { \mathbf {x}}^T \mathbf {x} = \vert x_1 \vert ^2 + \cdots + \vert x_n \vert^2 \neq 0\), 故可推出 \(\lambda = \overline \lambda\) , 换句话说,特征值 \(\lambda\) 一定是实数,特征空间 \(N(A – \lambda I)\)\(\mathbb {R}^n\) 的子空间,故 \(\mathbf {x} \in N(A-\lambda I)\) 是非零实向量

  3. \(A\) 存在 \(n\) 个单位正交特征向量 \(\mathbf {q}_1, \ldots, \mathbf {q}_n\) , 它们可以组成正交矩阵 \(Q\) ,满足 \(QQ^T = I\)

  4. \(A\) 可正交对角化, 即 \(A = Q \Lambda Q^T\)
    根据性质 3, 可以得到: \[
    \begin {align}
    A = Q \Lambda Q^T &=
    \begin {bmatrix}
    \mathbf {q}_1 & \mathbf {q}_2 &\cdots& \mathbf {q}_n \\
    \end {bmatrix}
    \begin {bmatrix}
    \lambda_1 & & \\
    & \ddots & \\
    & & \lambda_n
    \end {bmatrix}
    \begin {bmatrix}
    \mathbf {q}^T_1 \\
    \mathbf {q}^T_2 \\
    \vdots \\
    \mathbf {q}^T_n
    \end {bmatrix} \\
    &= \lambda_1 \mathbf {q}_1 \mathbf {q}^T + \cdots + \lambda_n \mathbf {q}_n \mathbf {q}_n^T \\
    &= \lambda_1 P_1 + \cdots + \lambda_n P_n
    \end {align}
    \]
    其中 \(Q\) 为正交矩阵 ,矩阵 \(\Lambda\) 为对角阵,即 \(\Lambda = \mathrm {diag}(\lambda_1, \ldots, \lambda_n)\)
    \(P_i\) 是正交投影矩阵,满足以下性质:

    • \(P_i^2 = P_i = P_i^T, i = 1, \ldots, n\)
    • \(P_i P_j = 0, i \neq j\)
    • \(P_1 + \cdots + P_n = I_n\)

    这就是著名的 principal axis theorem 或者 spectral theorem

注: 如果 \(A\) 满足 \(A^T = -A\) ,则 \(A\) 为 反对称矩阵 (skew-symmetric matrix)

2. Gramian 矩阵

假设 \(A\) 为一个 \(m \times n\) 阶实矩阵,则 \(n\) 阶方阵 \(G = [g_{ij}] = A^T A\) 称为 Gramian 矩阵。

对于方阵, Gramian 矩阵非常有效,请牢记一句话:

With rectangular matrices, the key is almost always to consider \(A^TA\) and \(AA^T\)

考虑 \(A\) 的列向量表达式 \(A = \begin {bmatrix} \mathbf {a}_1 & \mathbf {a}_2 & \cdots & \mathbf {a}_n\end {bmatrix}\) , \(\mathbf {a}_i \in \mathbb {R}^m\) ,则 \[
\begin {align}
G &= A^T A =
\begin {bmatrix} \mathbf {a}_1^T \\ \mathbf {a}_2^T \\ \vdots \\ \mathbf {a}_n^T \end {bmatrix} \begin {bmatrix} \mathbf {a}_1 & \mathbf {a}_2 & \cdots & \mathbf {a}_n \end {bmatrix} \\
&= \begin {bmatrix} \mathbf {a}_1^T \mathbf {a}_1 & \mathbf {a}_1^T \mathbf {a}_2 & \cdots & \mathbf {a}_1^T\mathbf {a}_n \\ \mathbf {a}_2^T \mathbf {a}_1 & \mathbf {a}_2^T \mathbf {a}_2 & \cdots & \mathbf {a}_2^T\mathbf {a}_n \\ \vdots & \vdots & \ddots & \vdots \\ \mathbf {a}_n^T \mathbf {a}_1 & \mathbf {a}_n^T \mathbf {a}_2 & \cdots & \mathbf {a}_n^T\mathbf {a}_n\end {bmatrix}
\end {align}
\]
根据上面的表达式,Gramian matrix \(G\) 的每一项都是 \(A\) 的列向量两两之间的内积,因此 Gramian matrix 也被称为 cross-product matrix

性质:

  1. \(A^T A\)对称矩阵,于是它有对称矩阵的性质,其中最重要的一条:
    \(A^TA\) 的相异特征值对应的特征向量正交

  2. \(A^T A\)半正定矩阵
    证明: 对 \(\mathbf {x} \in \mathbb {R}^n\)\(\mathbf {x} \neq \mathbf {0}\) , 满足 \[
    \mathbf {x}^T A^T A \mathbf {x} = (A\mathbf {x})^T (A\mathbf {x}) = ||A \mathbf {x}||^2 \ge 0
    \]
    根据定义,\(A^TA\) 为半正定矩阵
    半正定矩阵的性质:

    • 所有特征值 \(\lambda_i\) 满足 \(\lambda_i \ge 0\)
    • 所有主子式 (principal submatrices) 对应的行列式非负
    • 所有主轴 (pivots) 非负
  3. \(N(A^TA) = N(A)\) ; \(\mathrm {rank} (A^T A) = \mathrm {rank} A\)
    先证明: \(A\mathbf {x} = \mathbf {0} \to A^T A \mathbf {x} = \mathbf {0}\) , 这是显然的。
    然后证明: \(A^T A \mathbf {x} = \mathbf {0} \to A\mathbf {x} = \mathbf {0}\)
    两种证明思路:

    1. 利用 \(\mathbf {x}^T A^T A \mathbf {x} = ||A \mathbf {x}||^2 = \mathbf {0}\) ,推出 \(A \mathbf {x} = \mathbf {0}\)

    2. 利用 \(A\mathbf {x} \in C(A)\)\(A^T(A\mathbf {x}) = \mathbf {0} \to A\mathbf {x} \in N(A^T)\)
      \(C(A) \cap N(A^T) = \{ \mathbf {0} \}\) ,于是 \(A \mathbf {x} = \mathbf {0}\)

    于是: \(N(A) = N(A^T A)\) , \(\dim N(A) = \dim N(A^TA)\)
    根据 秩-零度定理,得到: \[
    \mathrm {rank} A = n – \dim N(A) \\
    \mathrm {rank} (A^TA) = n – \dim N(A^TA)
    \]
    于是: \(\mathrm {rank} (A^T A) = \mathrm {rank} A\)

  4. 当且仅当 \(A\) 的列向量 \(\mathbf {a}_1, \ldots, \mathbf {a}_n\) 线性独立时,\(A^T A\) 可逆并且是正定矩阵
    根据定义,\(\dim C(A)\) 其实就是 \(A\) 的线性独立列向量个数,于是 \(\dim C(A) = n\) , 或者说 \(\mathrm{rank} A = n\) :
    根据性质 3,得出 \(\text {rank}(A^TA) = \text {rank}(A) = n\) ,说明 \(A^T A\) 可逆
    然后证明正定矩阵:
    根据 \(A^TA\) 可逆,推出 \(\det (A^TA) \neq 0\) ,得到:
    \(\prod_{i=1}^n \lambda_i = \det(A^TA) \neq 0\) ,又由于 \(\lambda_i \ge 0\) ,于是 \(\lambda_i \gt 0\)
    于是 \(A^TA\) 为正定矩阵

  5. 实对称半正定矩阵 \(M\) 均可表示为 Gramian 矩阵, 即存在一个矩阵 \(A\) 使得 \(M = A^T A\)
    证明: 对称矩阵 \(M\) 可以正交对角化: \(M = Q\Lambda Q^T\), 其中 \(QQ^T =I\), \(\Lambda = \text{diag}(\lambda_1, \ldots, \lambda_n)\)。 又因为 \(M\) 半正定,满足 \(\lambda_i \ge 0\), 故可令 \(\Lambda^{1/2} = \text{diag}(\sqrt{\lambda_1}, \ldots, \sqrt {\lambda_n})\), 且 \(A = \Lambda^{1/2} Q^T\) ,马上得到: \[
    M = Q\Lambda Q^T = (Q\Lambda^{1/2})(\Lambda^{1/2}Q^T) = (\Lambda^{1/2} Q^T)^T (\Lambda^{1/2} Q^T) = A^TA
    \]

3. 正交矩阵 (orthogonal matrix)

根据前一小节的定义,显然 \(Q^TQ\) 就是 \(Q\) 的 Gramian matrix,当 \(Q^TQ\) 满足: \[
Q^T Q = I
\]
其中 \(I\) 为单位矩阵,我们将这种特殊的矩阵 \(Q\) 称为正交矩阵 (orthogonal matrix)

正交矩阵的逆矩阵也是它的转置矩阵

如果熟悉矩阵乘法,可以知道:
\[
\begin {align}
Q^T Q &= \begin {bmatrix}
\mathbf {q}^T_1 \\
\mathbf {q}^T_2 \\
\vdots \\
\mathbf {q}^T_n
\end {bmatrix}
\begin {bmatrix}
\mathbf {q}_1 & \mathbf {q}_2 &\cdots& \mathbf {q}_n \\
\end {bmatrix} \\
&= \begin {bmatrix}
\mathbf {q}_1^T \mathbf {q}_1 & \mathbf {q}_1^T\mathbf {q}_2 &\cdots& \mathbf {q}_1^T\mathbf {q}_n \\
\mathbf {q}_2^T \mathbf {q}_1 & \mathbf {q}_2^T\mathbf {q}_2 &\cdots& \mathbf {q}_2^T\mathbf {q}_n \\
\vdots & \vdots & \ddots & \vdots \\
\mathbf {q}_n^T \mathbf {q}_1 & \mathbf {q}_n^T\mathbf {q}_2 &\cdots& \mathbf {q}_n^T\mathbf {q}_n \\
\end {bmatrix}
\end{align}
\]
\(I\) 对比,可以得出结论: \(Q\) 的列向量两两之间规范正交 ,即 \[
\mathbf {q}_i^T\mathbf {q}_j = \begin{cases}
0, &\text{whenever } i \neq j ,\quad \text {giving the orthogonality}\\
1, &\text{whenever } i = j, \quad \text {giving the normalization}
\end{cases}
\]
note:

  1. orthogonal matrix 严格来说应该叫 orthonormal matrix ,这是历史遗留问题
  2. \(Q\) 不一定为方阵 (square matrix),它可以是普通矩阵。但如果 \(Q\) 满足 \(QQ^T = I\) ,则 \(Q\) 的 row columns 也满足两两正交的条件,且 \(Q\) 为方阵

实数域的正交矩阵对应到复数域就是 unitary matrix, 因此正交矩阵具有 unitary matrix 的所有性质

4. Hermitian matrix

实对称矩阵 \(A\) 对应在复数域就是 Hermitian matrix

Hermitian matrix 的共轭转置矩阵等于它本身 \[
A^{H} = A
\]
如果 \(A\) 是 Hermitian matrix,则它有如下性质:

  1. 对于任意向量 \(\mathbf {x} \in \mathbb {C}^n\), \(\mathbf {x}^{H}A\mathbf {x}\) 是实数,反之,亦成立
    请记住: \(\mathbf {x}^{H}A\mathbf {x}\) 是一个数,即 \(1\times 1\) 矩阵
    证明: \((\mathbf {x}^{H}A\mathbf {x})^{H} = \mathbf {x}^{H}A (\mathbf {x}^H)^H = \mathbf {x}^{H}A\mathbf {x}\) ,于是 \(\mathbf {x}^{H}A\mathbf {x}\) 是实数
    反向证明,利用 \((\mathbf {x}+\mathbf {y})^{H}A(\mathbf {x}+\mathbf {y})\)

  2. \(A\) 的特征值均为实数
    利用 \(A\mathbf {x} = \lambda \mathbf {x}\) ,可得 \(\mathbf {x}^{H}A\mathbf {x} = \lambda \mathbf {x}^{H}\mathbf {x}\)
    于是: \(\lambda = \mathbf {x}^{H}A\mathbf {x}/\mathbf {x}^{H}\mathbf {x}\)
    然后由性质1可得,\(\lambda\) 为实数

  3. 相异特征值的特征向量互为正交

注: 如果 \(K^H = -K\) ,那么 \(K\) 称为 skew-Hermitian 矩阵 (反共轭对称矩阵)

如果 \(A\) 是 Hermitian matrix, 那么 \(K = iA\) 满足: \[
K^H = (iA)^H = -i A^H = – iA = -K
\]
于是我们根据 Hermitian matrix \(A\) 构造出 skew-Hermitian 矩阵 \(K\)

重要的对称矩阵或者 Hermitian matrix 例子:

  1. Hessian 矩阵 \(H\) – 二阶偏导矩阵, 用于梯度下降
    矩阵每个元素的值如下: \[
    h_{ij}(\mathbf {x}) = \frac {\partial ^2 f(x)}{ \partial x_i \partial x_j}
    \]
    根据导数法则,交换偏导次序,导数值不变,即 \[
    \frac {\partial ^2 f(x)}{ \partial x_i \partial x_j} = \frac {\partial ^2 f(x)}{ \partial x_j \partial x_i}
    \]
    因此 \(H^T = H\) , 即任何二次可导的实函数的 Hessian 矩阵都是实对称矩阵

  2. 协方差矩阵 \(\Sigma\) , 用于度量两个随机变量的相似程度
    假设有 \(n\) 个随机变量 \(X_1, \ldots, X_n \in \mathbb {R}^n\) \[
    (\Sigma)_{ij} = \mathrm {cov} (X_i , X_j) = E[(X_i – \mu_i)(X_j – \mu_j)]
    \]
    其中 \(\mu_i = E[X_i]\) 代表 \(X_i\) 的期望值,明显地, \[
    E[(X_i – \mu_i)(X_j – \mu_j)] = E[(X_j – \mu_j)(X_i – \mu_i)]
    \]
    因此 \(\Sigma_{ij} = \Sigma_{ji}\)

  3. 二次型
    假设 \(A = [a_{ij}]\)\(n \times n\) 阶实矩阵 (或复矩阵),\(\mathbf {x}\)\(n\) 维实向量 (或复向量),考虑二次型 \(f(\mathbf {x}) = \mathbf {x}^T A \mathbf {x}\) \[
    \begin {align}
    f(\mathbf {x}) &= \mathbf {x}^T A \mathbf {x} = \sum_{i=1}^n \sum_{j=1}^n a_{ij} x_i x_j \\
    &= \sum_{i=1}^n \sum_{j=1}^n (\frac {a_{ij} + a_{ji}} {2}) x_i x_j \\
    &= \mathbf {x}^T (\frac {A + A^T}{2}) \mathbf {x}
    \end {align}
    \]
    上式指出 \(A\)\((A + A^T)/2\) 有相同的二次型,且 \((A + A^T)/2\) 是对称矩阵

5. unitary matrix

如果把正交矩阵 \(Q\) 推广到复数域,就得到下面的 unitary matrix \(U\) (中文名很奇怪,就不说了)

其中转置操作变成了共轭转置运算

a complex square matrix \(U\) is unitary if its conjugate transpose (共轭转置) \(U^{H}\) is also its inverse. That is. \[
U^{H} U = I
\]
where \(I\) is the identity matrix

Note:

  • 共轭转置(conjugate transpose) 符号有两种表示方式: \(U^H\) 或者 \(U^{\ast}\)
  • 矩阵 \(U\) 和上面的 \(Q\) 类似,不一定为方阵

A complex matrix with orthonormal columns is called a unitary matrix. \[
\mathbf {u}_i^{H}\mathbf {u}_j = \begin{cases}
0, &\text{whenever } i \neq j ,\quad \text {giving the orthogonality}\\
1, &\text{whenever } i = j, \quad \text {giving the normalization}
\end{cases}
\]
性质:

  1. \((U\mathbf {x})^H (U\mathbf {y}) = \mathbf {x}^H U^H Uy = \mathbf {x}^H \mathbf {y}\)
    Length unchanged: \(||U\mathbf {x}||^2 = (U\mathbf {x})^H (U\mathbf {x}) = \mathbf {x}^H\mathbf {x} = ||\mathbf {x}||^2\)

  2. Every eigenvalue of \(U\) has absolute value \(|\lambda| = 1\) ,即所有的特征值都在复数域的单位圆上。
    证明: 由于 \(U\mathbf {x} = \lambda \mathbf {x}\) ,根据性质1,\(||U\mathbf {x}||^2 = ||\mathbf {x}||^2\)
    于是 \(||U\mathbf {x}||^2 = ||\lambda \mathbf {x}||^2 = |\lambda|^2 \cdot||\mathbf {x}||^2 = ||\mathbf {x}||^2\) ,而 $||||^2 0 $, 于是 \(|\lambda| = 1\)

  3. 相异特征值对应的特征向量正交
    证明: 假设有 \(U\mathbf {x} = \lambda_1 \mathbf {x}\)\(U\mathbf {y} = \lambda_2 \mathbf {y}\) ,则根据性质1: \[
    \mathbf {x}^H \mathbf {y} = (U\mathbf {x})^H (U\mathbf {y}) = (\lambda_1 \mathbf {x})^H (\lambda_2 \mathbf {y}) = \overline \lambda_1 \lambda_2 \mathbf {x}^H \mathbf {y}
    \]
    根据性质2,所有特征值都有 \(|\lambda| = 1\), 如果 \(\lambda_1 \neq \lambda_2\), 则 \(\overline \lambda_1 \lambda_2 \neq 1\) ,于是 \(\mathbf {x}^H \mathbf {y} = 0\)

Unitary matrix 最重要的例子就是 Fast Fourier matrix \[
U = \cfrac {1}{\sqrt {n}}
\begin {bmatrix}
1 & 1 & \cdots & 1 \\
1 & w & \cdots & w^{n-1} \\
\vdots & \vdots & \ddots & \vdots \\
1 & w^{n-1} & \cdots & w^{(n-1)^2}
\end {bmatrix}
= \cfrac {\text {Fourier matrix}} {\sqrt {n}}
\]
其中 复数 \(w\) 是单位圆 (unit circle) 上 \(\theta = 2\pi i/n\) 对应的点,即 \(w = e^{2\pi i/n}\)

由于 \(U\) 的所有元素都是纯虚数,因此 \(U^{-1} = U^H = \overline U\)

6. 正规矩阵 (normal matrix)

\(A\)\(n \times n\) 的复数矩阵,如果 \(A\)\(A^H\) 是可交换矩阵,即 \[
A^H A = A A^H
\]
\(A\) 为 normal matrix

normal matrix 性质:

  1. unitarily diagonalizable (可么正对角化 ), 即:
    \[
    A = U D U^H
    \]
    其中 \(U\) 为 unitary matrix, 满足 \(U^{-1} = U^H\) , \(D\) 为对角矩阵

  2. \(N(A^H) = N(A)\)\(C(A^H) = C(A)\)
    证明: \[
    \vert \vert A^H \mathbf {x} \vert \vert^2 = (A^H \mathbf {x})^H (A^H \mathbf {x}) = \mathbf {x}^H AA^H \mathbf {x} = \mathbf {x}^H A^HA \mathbf {x} = (A \mathbf {x})^H (A \mathbf {x}) = \vert \vert A \mathbf {x} \vert \vert^2
    \]
    可知 \(A^H \mathbf {x} = \mathbf {0}\) 同义于 \(A \mathbf {x} = \mathbf {0}\) , 即 \(N(A^H) = N(A)\)
    另一方面,如果 \(A^H \mathbf {x} = \mathbf {0}\), 对于任一 $ $, \(\mathbf {x}^H (A \mathbf {y}) = (A^H \mathbf {x})^H \mathbf {y} = \mathbf {0}\) ,由于 $ N(A^H)$, 而 \((A \mathbf {y}) \in C(A)\), 说明 \(C(A) = N(A^H)^{\perp}\) , 将 \(A\) 替换为 \(A^H\), 又得到 \(C(A^H) = N(A)^{\perp}\) ,由于 \(N(A^H) = N(A)\), 有 \(C(A^H) = C(A)\)

  3. \(\mathbb {C}^n = N(A) \oplus C(A)\)\(\mathbb {C}^n = N(A^H) \oplus C(A^H)\)
    证明利用秩-零度定理和四个子空间的关系

  4. \(A\)\(A^H\) 有相同的特征向量
    \(A\) 为 normal matrix, 则 \(A – \lambda I\) 也是 normal matrix, 根据性质 2, \(N(A – \lambda I) = N(A^H – \overline \lambda I)\) , 这意味着若 \(A\) 有特征值 \(\lambda\), 则 \(A^H\) 有特征值 \(\overline \lambda\) ,且 \(A\) 的特征向量等于 \(A^H\) 的特征向量

重要的 normal matrix :

  1. 实对称矩阵, 因为如果 \(A = A^T\), 则 \(A^T A = A A^T = A^2\)
  2. Hermitian matrix
  3. orthogonal matrix and unitary matrix
  4. 下面提到的 skew-Hermitian matrix \(K^H K = (-K) (-K^H) = K K^H\)

normal matrix \(A = [a_{ij}]\) 的等价条件:

  1. \(A\) 可以对角化, \(A = U\Lambda U^H\) , 其中 \(\Lambda = \mathrm{diag} (\lambda_1, \ldots, \lambda_n)\) 是对角阵,\(U^H = U^{-1}\)\(U\) 为 unitary matrix
  2. \(A = B + iC\) , 其中 \(B\)\(C\) 是可交换 (\(BC = CB\)) Hermitian matrix
  3. \(\sum_{i=1}^n \vert \lambda_i \vert^2 = \sum_{i=1}^n \sum_{j=1}^n \vert a_{ij} \vert^2\)

7. 实对称矩阵 和 复对称矩阵 的对比

Real Complex
\(\mathbb {R}^n\) (n real components) \(\mathbb {C}^n\) (n complex components)
length: \(||\mathbf {x}||^2 = x_1^2 + \cdots + x_n^2\) length: \(||\mathbf {x}||^2 = |x_1|^2 + \cdots + |x_2|^2\)
transpose: \(A_{ij}^T = A_{ij}\) Hermitian transpose: \(A_{ij}^H = A_{ij}\)
inner product: \(\mathbf {x}^T \mathbf {y} = x_1 y_1 + \cdots + x_n y_n\) inner product: \(\mathbf {x}^H \mathbf {y} = \overline x_1 y_1 + \cdots + \overline x_n y_n\)
\((A\mathbf {x})^T \mathbf {y} = \mathbf {x}^T (A^T \mathbf {y})\) \((A\mathbf {x})^H \mathbf {y} = \mathbf {x}^H (A^H \mathbf {y})\)
orthogonality: \(\mathbf {x}^T \mathbf {y} = 0\) orthogonality: \(\mathbf {x}^H \mathbf {y} = 0\)
symmetric matrices: \(A^T = A\) Hermitian matrices: \(A^H = A\)
\(A = Q \Lambda Q^{-1} = Q \Lambda Q^T\)(real \(\Lambda\)) \(A = U \Lambda U^{-1} = U \Lambda U^H\)(real \(\Lambda\))
skew-symmetric \(K^T = -K\) skew-Hermitian \(K^H = -K\)
orthogonal \(Q^TQ = I\) or \(Q^T = Q^{-1}\) unitary \(U^H U = I\) or \(U^{-1} = U^H\)
\((Q\mathbf {x})^T (Q\mathbf {y}) = \mathbf {x}^T \mathbf {y}\) and \(||Q\mathbf {x}|| = ||\mathbf {x}||\) \((U\mathbf {x})^H (U\mathbf {y}) = \mathbf {x}^H \mathbf {y}\) and \(||U\mathbf {x}|| = ||\mathbf {x}||\)

The column, rows and eigenvectors of \(Q​\) and \(U​\) are orthonormal, and every \(|\lambda| = 1​\)

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