線代CH6定義整理

# 線代CH6定義整理 ###### tags: `線代` > form Linear Algebra ,Stephen Friedberg. --- ## 6.1.1 inner product #### source: DP342 Let V be a vector space over F. An inner product on V is a function that assigns, to every ordered pair of vectors x and y in V. ascalar in F, denoted〈x,y〉, such that for all x, y and z in V and all c in F, the following hold: 1. 〈x + z,y〉= 〈x,y〉+〈z,y〉 2. 〈cx,y〉= c〈x,y〉 3. $\overline{〈x,y〉}$=〈y,x〉, where the bar denotes complex conjugation. $\overline{〈x,y〉}$為共軛複數 4. 〈x,x〉> 0 if x≠0. 純量需為正 - $〈\sum\limits_{i = 1}^n{a_iv_i,y}〉=\sum\limits_{i = 1}^n{a_i〈v_i,y〉}$ ## 6.1.2 conjugate transpose / adjiont #### source: DP344 Let A ∈$M_{m×n}(F)$. We dfine the conjugate transpose or adjoint of A to be the n×m matrix $A^*$such that$(A^*)_{ij} = \overline{A_{ij}}$ for all i,j. - 此為標準定義內積 ## 6.1.3 norm / lenth #### source: DP346 Let V be an inner product space. For x ∈V, we define the norm or lenth of x by $||x||$ = $\root \of{〈x,x〉}$. ## 6.1.4 orthogonal / perpendicular #### source: DP348 Let V be an inner product space. Vectors x and y in V are orthogonal ( perpendicular ) if 〈x,y〉= 0. A subset S of V is orthogonal if any two distinct vectors in S are orthogenal. A vector x in V is a unit vector if $||x||$ = 1. Finally, a subset S of V is orthonormal if S is orthogonal and consists entirely of unit vectors. ## 6.2.1 orthonormal basis #### source: DP354 Let V be an inner product space. A subset of V is an orthonormal basis for V if it is an ordered basis that is orthonormal. - 即互相垂直且長度為1 ## 6.2.2 orthonormal complement #### source: DP362 Let S be a nonempty subset of an inner product space V.We define $S^⊥$ ( read S perp) to be the set of all vectors in V that are orthogonal to every vector in S; that is , $S^⊥$ = {x∈V: 〈x,y〉= 0 for all y∈S }. The set $S^⊥$ is called th orthogonal complement of S. ## 6.4.1 normal #### source: DP383 Let V be an inner product space, Let T be a liner operator on V.we say that T is normal if $TT^* = T^*T$. An n×n real or $L^*_A(x) = A^*x$ complex martrix A is normal if $AA^* = A^*A$. - $TT^* = T^*T → 〈T(x),y〉= 〈x,T^*(y)〉$ > Q.這樣可以對角化嗎? Ans.幾乎可以了！ ## 6.4.2 real skew - symmetric matrix #### source: DP384 $A^t = -A$ ## 6.4.3 self-adjiont / Haermitaian #### source: DP386 Let T be a linear operator on an inner product space V. We say that T is self-adjiont ( Hermitaian ) if $T=T^*$. An n×n real or complex martrix A is self-adjiont ( Hermitaian ) if $A = A^*$. - self有自己幫助自己的意思 - 如果β是orthogonal base 且含有eigenavectors則可對角化 - 如果T是 self-adjiont 則矩陣代表A也是 self-adjiont ## 6.4.4 positive definite / postitive semidefinite #### source: DP390 A linear opear T on a finite-dimensional inner product space is called positive definite ( postitive semidefinite) if T is self-adjiont and 〈T(x),x > 0 ( 〈T(x),x >≧0〉) for all x≠0. An n×n matrix A with entries form R or C is call positive definite( positive semidefinite ) if L_A is positive definite( positive semidefinite ) ## 6.5.1 unitary operator & orthogonal operator #### source: DP392 Let T be a linear operator on a finite-dimensional inner product space V(over F). If $||T(x)||$ = $||x||$ for all x∈V, we call T a unnitary operator if F = C and an orthogonal operator if F = R. - $Q^{-1}AQ = D$ where $Q^{-1} = Q^* Q = I$ ## 6.5.2 reflection #### source: DP395 Let L be a one-dimensional subspace of $R^2$. We may view L as a line in the plane through the orgin A linear opeartor T on $R^2$ is called a reflection of $R^2$ about L if T(x) = x for all x ∈L and T(x) = -x for all x∈$L^⊥$. ## 6.5.3 orthogonal matrix & unitary #### source: DP395 A square martrix A is called an orthogonal matrix if $A^tA = AA^t = I$ and unitary if $A^*A = AA^*=I$. ## 6.5.4 rigid motion #### source: DP398 Let V be an real inner product space. A functionf : V→V is called a rigid motion if $||f(x) - f(y)||$ = $||x-y||$ for all x,y∈V. ## 6.6.1 orthogonal projection #### source: DP411 Let V be an inner product space, and let T : V →V be a projection. We say that T is an orthogonal projection if $R(T)^⫠ = N(T)$ and $N(T)^⫠ =R(T)$ ## 6.7.1 the singular values of A and linear transformation(SVD分解定義) #### source: DP423 Let A be a m×n matrix. We define th singular values of A to be the singular values of the linear transformation $L_A$. - Let A be a m×n matrix of rank r with the positive singular values $σ_1≧σ_2≧…≧σ_r$, and let $\sum$ be the m×n matrix difined by $\sum_{ij} = \cases{ σ_i & \text{if i = j≦r}\\ 0 & \text{O.W} }$ - 矩陣版本： $A_{v_i} = \cases{ \lambda_iu_i & \text{1≦i≦k}\\ 0u_i & \text{k≦i≦n} }$ >$v_i$為定義域的基底；$u_i$為值域的基底。 Then $A = U\sum V^*$, where $\sum$ is 對角矩陣(由eigen value組成) ## 6.7.2 singular value decomposition(SVD分解) #### source: DP423 Let A be a m×n matrix of rank r with postivesingular values $σ_1≧σ_2≧…≧σ_r$, A foctorize A = U $\sum V^*$ where U and V are unitary matrocies and $\sum$ is the m×n matrix defined as 6.7.1的註解 is called a singular value decomposition of A. ## SVD簡單說 $\exists \beta$ = {$v_1,v_2,...,v_n$} (定義域基底) $\exists \gamma$ = {$u_1,u_2,...,u_n$} (值域基底) is orthogonal bases, then: 1. $T(v_i) = \lambda_iu_i$ 2. $\lambda^2_i = T^*T$ > $\lambda_i$ = singular value ## 6.7.3 pseudoinverse #### source: DP426 Let V and W be finite-dimensional inner product spaces over the same field, and let T: V → W be a linear transformation. Let $L:N(T)^⊥ → R(T)$ be the linear transformation defined by L(x) = T(x) for all x$\in N(T)^⊥$. The pseudoinverse( or Moore-Penose generalized inverse) of T, denoted by $T^┼$, is defined as the unque linear transformation from W to V such that $T^┼(v_i) = \cases{ L^{-1}(y) & \text{for y ∈ R(T)}\\ 0 & \text{for y ∈ R(T)^⊥} }$ - $T: N(T)^⊥ -> Range(T)$ - L = $T_1 \in(T)^⊥$(1-1 & onto) 另一種寫法： $T^┼(v_i) = \cases{ \lambda_iu_i & \\ 0 & }$ > 適用非1-1的情況[$\because$限制L在Range(T)上，需要垂直(⊥)，即可變成唯一] ### Example A = $\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ 1 & 0 \\ 1 & 1 \\ \end{array} \right)$，求$A^┼$ $sol.$ $A^*A = A^TA =$ $\left( \begin{array}{cccc} 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ \end{array} \right)$$\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ 1 & 0 \\ 1 & 1 \\ \end{array} \right)$ = $\left( \begin{array}{cc} 3 & 2 \\ 2 & 3 \\ \end{array} \right)$ $rank(A^*A) =2$ $det\left( \begin{array}{cc} 3 - \lambda & 2 \\ 2 & 3 - \lambda \\ \end{array} \right)$ $=(\lambda - 5)(\lambda-2) = 0$ $\lambda = 5$代入: $A^*A - 5\lambda=$$\left( \begin{array}{cc} -2 & 2 \\ 2 & -2 \\ \end{array}\right)$ 得到： $u_1 = \left( \begin{array}{c} 1\over \sqrt2 \\ 1\over \sqrt2 \\ \end{array}\right)$ $u_2 = \left( \begin{array}{c} 1\over \sqrt2 \\ -1\over \sqrt2 \\ \end{array}\right)$ :::warning :warning: $u_1$、$u_2$需為orthonormal basis. ::: 如此可得U: $U = \left( \begin{array}{cC} 1\over \sqrt2 & 1\over \sqrt2 \\ 1\over \sqrt2 & -1\over \sqrt2 \\ \end{array}\right)$ Note $W=AU$ $W=AU=$$\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ 1 & 0 \\ 1 & 1 \\ \end{array} \right)$$\left( \begin{array}{cC} 1\over \sqrt2 & 1\over \sqrt2 \\ 1\over \sqrt2 & -1\over \sqrt2 \\ \end{array}\right)=$$\left( \begin{array}{cc} \sqrt 2 & 0 \\ 1\over \sqrt 2 & -1\over \sqrt 2 \\ 1\over \sqrt 2 & 1\over \sqrt 2 \\ \sqrt 2 & 0 \\ \end{array} \right)=(w_1,w_2)$ 求$v_1、v_2$ - $v_i =$$w_i \over \lambda_i$ $v_1=$$w_1 \over \sqrt 5$, $v_2=$$w_2 \over \sqrt 1$ :::info :information_source: Why? Recall: $Q^*A^*AQ =$$\left(\begin{matrix} \lambda^2_{1} & \cdots & 0\\ \vdots & \lambda^2_{k} & \ddots \\ 0 & \cdots & a_{mn} \end{matrix}\right)=W^*W$ $V^*W = V^*AU = \sum$ $<w_i,w_i> = \lambda^2_i<w_j,w_i> = 0$ $\therefore v_i =$$w_i \over \lambda_i$ :::