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A linear transformation is a transformation \(T: \mathbb{R}^n \to \mathbb{R}^m\) satisfying
\[\begin{eqnarray} T(\vec{u}+\vec{v}) &=& T(\vec{u}) + T(\vec{v}) \tag{1} \\ T(c\vec{u}) &=& cT(\vec{u}) \tag{2} \end{eqnarray}\]for all vectors \(\vec{u}, \vec{v} \in \mathbb{R}^n\) and all scalars \(c\).
\(T(0) = 0\) 是 linear transformation 的結論，可以用來快速檢驗某 transformation 是否 linear。於是任何有常數的 transformation 都不 linear；如 \(T(x) = x + 1\) is non-linear for \(T(0) \not = 0\).
加上平移叫做 affine。
問題：找出符合 property of linearity \((1)\) 但不符合 \((2)\) 的 transformation。（目前無解）
Let \(T: \mathbb{R}^n \to \mathbb{R}^m\) be a linear transformation. Let \(A\) be the \(m \times n\) matrix
\[A = \Big(T(\vec{e_1})\ T(\vec{e_2})\ \cdots\ T(\vec{e_n})\Big).\]（\(\vec{e_i}\) 是單位向量（standard vector）；\(T(\vec{e_i})\) 是 column vector） Then \(T\) is the matrix transformation associated with \(A\), i.e. \(T(\vec{x}) = A\vec{x}\).
Proof
\[\begin{eqnarray} T\begin{pmatrix} x \\ y \\ z \end{pmatrix} &:=& T\Big(x\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix} + y\begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix} + z\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}\Big) \\ &=& xT(\vec{e_1}) + yT(\vec{e_2}) + zT(\vec{e_3}) \\ &=& \Big(T(\vec{e_1})\ T(\vec{e_2})\ \cdots\ T(\vec{e_n})\Big)\begin{pmatrix}x \\ y \\ z\end{pmatrix} \\ &=& A\begin{pmatrix}x \\ y \\ z\end{pmatrix} \tag*{$\blacksquare$} \end{eqnarray}\]此處 \(A\) 稱作 the standard matrix for \(T\)。從本 theorem 也可以看出，matrix transformation 和 linear transformation 是等價的。
Define \(T:\mathbb{R}^2 \to \mathbb{R}^3\) by the formula
\(T\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}3x-y \\ y \\ x\end{pmatrix}\).
已知 \(A = \Big(T(\vec{e_1})\ T(\vec{e_2})\ \cdots\ T(\vec{e_n})\Big)\)，因此我們只需依序對 \(T\) 帶入 \(\vec{e_1}, \vec{e_2}\)，就能求出相應的 matrix \(A\)：
\[T(\vec{e_1}) = \begin{pmatrix}3(1)-0 \\ 0 \\ 1\end{pmatrix}\] \[T(\vec{e_2}) = \begin{pmatrix}3(0)-1 \\ 1 \\ 0\end{pmatrix}\]所以 \(A = \begin{pmatrix}3 & -1 \\ 0 & 1 \\ 1 & 0\end{pmatrix}\)。
更多範例。
設 \(T: \mathbb{R}^n \to \mathbb{R}^m,\ T(x) = Ax,\ A \in \mathbb{R}^{m \times n}\)
若 \(A\) 符合以下等價條件：
則 \(T\) is injective。
\(\text{dim(N(A))} = 0\) 相當於 \(\text{ker}(T) = \{I\}\)。
進一步解釋：Full column rank 代表 \(A\) 的 columns 是 linearly independent，所以 \(A\vec{x} = A\vec{y} \iff \vec{x} = \vec{y}\)；這就是 one-to-one。然後，wide matrices (\(m < n\)) 不可能造就 one-to-one transformation，也就是「大打到小」不可能 one-to-one。
若 \(A\) 符合以下等價條件：
進一步解釋：tall matrices (\(m > n\)) 不可能造就 onto transformation，也就是「小打到大」不可能 onto。
若 \(A\) 是 square (\(m = n\))，則 \(T\) is one-to-one if only if it is onto.
“Homomorphism” comes from the greek homo (same) and morphus (form or shape).
So a “homomorphism” is a map that “preserves the shape” or “preserves the structure.” –StackExchange（很棒的參考！）
Linear transformation 本身就 preserve the vector space structure（參照 Vector Spaces and Subspaces 和 linear trans. 的定義）；homomorphisms of vector spaces 和 linear trans. 其實意義相同！
更多 homomorphism 見 StackExchange。
The word isomorphism derives from the Greek iso, meaning “equal,” and morphosis, meaning “to form” or “to shape.”
The term isomorphism is mainly used for algebraic structures. In this case, mappings are called homomorphisms, and a homomorphism is an isomorphism if and only if it is bijective. –Wiki
也就是說，Two algebraic structures are isomorphic if there exist:
between them.
因為是 bijective，所以存在 inverse mapping between two structures，這正是 isomorphism 和 homomorphism 的不同之處。
再次注意：homomorphism 是 isomorphism 的必要條件
就線性代數而言：
A linear map \(T\) is called an isomorphism if \(T\) is bijective.
為什麼這裡沒看到 homomorphism？因為這裡已經說明，linear map 本身就是 homomorphism！
見 Isomorphism。