> For the complete documentation index, see [llms.txt](https://sejkai.gitbook.io/academic/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://sejkai.gitbook.io/academic/linear-algebra/matrix-transformations/transformations-and-matrix-multiplication.md). # Transformations and matrix multiplication ## Compositions of linear transformations 1 * 現在要來討論的是， linear transformation 的組合要怎麼表示？他還算是 linear transformation 嗎？ ![](/files/-LoxNU_O7PbmsEycg0v8) 我們現在有 Rn, Rm, Rl 空間，以及 S 和 T 兩個 linear transformation function $$ \begin{aligned} \&S: X \to Y \mid X \subseteq \mathbb{R}^n, Y \subseteq \mathbb{R}^m \\ \&S(\vec{x}) = \mathbf{A}\vec{x} \mid \mathbf{A} \text{ is } m \times n \\\\ \&T: Y \to Z \mid Y \subseteq \mathbb{R}^m, Z \subseteq \mathbb{R}^l \\ \&T(\vec{x}) = \mathbf{B}\vec{x} \mid \mathbf{B} \text{ is } l \times m \end{aligned} $$ 那我們要怎麼一次定義 x 到 z ，也就是先對 vector 做 S 再做 T 的 transformation 這段變化呢 $$ T \circ S: X \to Z \text{ ( The composition of T with S )} \\ T \circ S = T(S(\vec{x})) $$ 那這樣子定義還是不是 linear transformation 呢 * 條件一 OK $$ \begin{aligned} T \circ S(\vec{x} + \vec{y}) &= T(S(\vec{x}+\vec{y}))\\ &= T(S(\vec{x})+S(\vec{y}))\\ &= T(S(\vec{x})) + T(S(\vec{y}))\\ &= T \circ S(\vec{x}) + T \circ S(\vec{y}) \end{aligned} $$ * 條件二 OK $$ \begin{aligned} T \circ S(c\vec{x} ) &= T(S(c\vec{x})) \\ &= T(cS(\vec{x})) \\ &= cT(S(\vec{x})) \\ &= c T \circ S(\vec{x}) \end{aligned} $$ 所以我們的確可以把 S 和 T 兩個 transformation 合併，用一個 matrix 來表達 $$ T \circ S (\vec{x}) = \mathbf{C}\vec{x} \mid C \text{ is } l \times n $$ ## Compositions of linear transformations 2 * 現在將更進一步探討 S 和 T 兩個 transformation 如何轉為 matrix 表示 $$ T(\vec{x})= \mathbf{B}\vec{x} \mid \mathbf{B} \text{ is } l \times m\\ S(\vec{x})= \mathbf{A}\vec{x} \mid \mathbf{A} \text{ is } m \times n\\ T \circ S = \mathbf{B}(\mathbf{A}(\vec{x})) = \mathbf{C}\vec{x} $$ 我們要如何找到這個 C ，首先這個 C 將會是多大 size 因為 vector 從 Rn 開始被轉換，所以先定義 identity matrix 為 In $$ \mathbf{I\_n} = \begin{bmatrix} 1&0&\cdots& 0\ 0&1&\cdots&0 \0&0&\ddots&\vdots\0&0&\cdots&1\end{bmatrix} $$ 而我們的 C 就會等於 $$ \mathbf{C} = \begin{bmatrix} \mathbf{B}(\mathbf{A}\left(\begin{bmatrix} 1\0\\\vdots\0\end{bmatrix}\right))& \mathbf{B}(\mathbf{A}\left(\begin{bmatrix} 0\1\\\vdots\0\end{bmatrix}\right))& \cdots& \mathbf{B}(\mathbf{A}\left(\begin{bmatrix} 0\0\\\vdots\1\end{bmatrix}\right)) \end{bmatrix} $$ 雖然看起來很複雜，但我們將 A 拆開來看，A 的運算應該如下 $$ \mathbf{A} = \begin{bmatrix} \vec{a\_1} &\vec{a\_2}&\cdots &\vec{a\_n}\end{bmatrix} \begin{bmatrix} x\_1\x\_2\\\vdots\x\_n\end{bmatrix} = x\_1\vec{a\_1}+x\_2\vec{a\_2}+\cdots+x\_n\vec{a\_n} $$ 所以 C 的 A 只有乘到每一項對應的 vector 而已，而這個 matrix 也就是 BA 相乘 ! $$ \mathbf{C} = \begin{bmatrix} \mathbf{B}(\vec{a\_1})& \mathbf{B}(\vec{a\_2})& \cdots& \mathbf{B}(\vec{a\_n}) \end{bmatrix} = \mathbf{BA} $$ ## Matrix product examples * 我們現在知道 B 和 A 相乘的意義其實就是 S 和 T 這兩個 transformation 的 composition 我們實際舉個例子來運算看看 $$ \underset{2\times3}{\mathbf{B}} = \begin{bmatrix}1&-1&2\0&-2&1\end{bmatrix},,, \underset{3\times4}{\mathbf{A}} = \begin{bmatrix}1&0&1&1\2&0&1&-1\3&1&0&2\end{bmatrix} $$ 所以 BA 會等於 $$ \begin{aligned} \mathbf{BA} &= \begin{bmatrix} \mathbf{B}(\vec{a\_1})& \mathbf{B}(\vec{a\_2})& \cdots& \mathbf{B}(\vec{a\_n}) \end{bmatrix} \\ &= \begin{bmatrix} \mathbf{B}\begin{bmatrix}1\2\3\end{bmatrix} & \mathbf{B}\begin{bmatrix}0\0\1\end{bmatrix} & \mathbf{B}\begin{bmatrix}1\1\0\end{bmatrix} & \mathbf{B}\begin{bmatrix}1\\-1\2\end{bmatrix} \end{bmatrix}\\ &= \begin{bmatrix} 5&2&0&6\\-1&1&-2&4 \end{bmatrix} \end{aligned} $$ > 更詳細的運算過程可以去看影片 我們可以發現，能夠運算如此順利，是因為 B 和 A 已經 well defined 因為 2 x 3 和 3 x 4 的矩陣相乘，我們可以得到一個 2 x 4 的矩陣 今天若是 B 和 A 交換，要求 AB 是求不出來的 所以這邊又點出一個觀念，就是 **矩陣相乘是沒有交換律的 (NO commutative)** $$ \mathbf{AB \neq BA} $$ ## Matrix product associativity * 那們矩陣相乘有沒有結合律呢？我們從更多的 transformation 來看 $$ H(\vec{x}) = \mathbf{A}\vec{x}\\ G(\vec{x}) = \mathbf{B}\vec{x}\\ F(\vec{x}) = \mathbf{C}\vec{x}\\ $$ 我們將轉移到 F 後，再轉移到 G ，最終轉移到 H，可以這樣寫 $$ \begin{aligned} ((H\circ G) \circ F)(\vec{x}) &= (H\circ G)(F(\vec{x})) \\&= H(G(F(\vec{x})))\\&= H(G\circ F(\vec{x})) \\&= (H\circ (G \circ F))(\vec{x}) \end{aligned} $$ 結果我們發現 $$ ((H\circ G) \circ F)(\vec{x}) = (H\circ (G \circ F))(\vec{x}) = (H\circ G \circ F)(\vec{x}) $$ 也就是說 **矩陣相乘是有結合律的 (HAS Associative)** $$ (\mathbf{AB})\mathbf{C} = \mathbf{A}(\mathbf{BC}) = \mathbf{ABC} $$ > 簡單來說就是可以忽略掉括號啦 ## Distributive property of matrix products * 最後我們來看矩陣相乘有沒有分配律 ! $$ \mathbf{A} = k \times m\\ \mathbf{B} = m \times n\\ \mathbf{C} = m \times n\\ $$ 試著乘乘看 $$ \begin{aligned} \mathbf{A}(\mathbf{B+C}) &= \mathbf{A} \begin{bmatrix}\vec{b\_1}+\vec{c\_1}& \vec{b\_2}+\vec{c\_2} & \cdots & \vec{b\_n}+\vec{c\_n}\end{bmatrix}\\ &= \begin{bmatrix}\mathbf{A}(\vec{b\_1}+\vec{c\_1})& \mathbf{A}(\vec{b\_2}+\vec{c\_2}) & \cdots & \mathbf{A}(\vec{b\_n}+\vec{c\_n})\end{bmatrix} \\ &= \begin{bmatrix}\mathbf{A}\vec{b\_1}+\mathbf{A}\vec{c\_1}& \mathbf{A}\vec{b\_2}+\mathbf{A}\vec{c\_2} & \cdots & \mathbf{A}\vec{b\_n}+\mathbf{A}\vec{c\_n}\end{bmatrix} \\ &= \begin{bmatrix}\mathbf{A}\vec{b\_1}& \mathbf{A}\vec{b\_2} & \cdots & \mathbf{A}\vec{b\_n}\end{bmatrix} + \begin{bmatrix}\mathbf{A}\vec{c\_1}& \mathbf{A}\vec{c\_2} & \cdots & \mathbf{A}\vec{c\_n}\end{bmatrix} \\ &= \mathbf{AB} + \mathbf{AC} \end{aligned} $$ 所以**矩陣相乘是有分配律的 (HAS Distributive)** 統整一下： * matrix multiplication has no commutative $$ \mathbf{AB} \neq \mathbf{BA} $$ * matrix multiplication has associative $$ \mathbf{A}(\mathbf{BC}) = (\mathbf{AB})\mathbf{C} = \mathbf{ABC} $$ * matrix multiplication has distributive $$ \mathbf{A}(\mathbf{B+C}) = \mathbf{AB}+\mathbf{AC} $$ --- # Agent Instructions This documentation is published with GitBook. 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