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$\underline{Def}$ (Left Product Map) Let $A \in M_{m \times n}(\mathbb{F})$. We define a map $L_{A} \colon \mathbb{F}^{n} \rightarrow \mathbb{F}^{m}$ by $$L_{A}(x)=Ax \;\; \forall a \in \mathbb{F}^{n},$$ called the left-product map by $A$. $\underline{Ex}$ $$A = \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 2 \end{bmatrix} \in M_{2 \times 3}(\mathbb{R})$$ $$L_{A} \colon \mathbb{R}^{3} \rightarrow \mathbb..

$\underline{Thm}$ $A \in M_{m \times n},\, B \in M_{n \times p}$ Let $$AB=\begin{bmatrix} u_{1} & u_{2} & \cdots & u_{p} \end{bmatrix},\, B=\begin{bmatrix} v_{1} & v_{2} & \cdots & v_{p} \end{bmatrix}$$ Then $$u_{j}=Av_{j}, v_{j}=Be_{j}\;\;\;(B=BI_{p})$$ $\underline{Thm}$ Let $V,W$ be finite dimensional vector spaces with ordered bases $\beta, \gamma$ respectively. Let $T$ be linear. Then $$[T(u..

$\underline{Def}$ Let $A \in M_{n \times n}$. Define $A^{1}=A,\, A^{2}= A^{1} \cdot A.$ Define recursively that $A^{k}=A^{k-1} \cdot A$ $\forall \geq 2$. Define $A^{0}=I_{n}$. $\underline{Ex}$ $A=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix},\; A^{2}=0$. Cancellation law does not hold for matrix product.

$\underline{Thm}$ Let $A \in M_{m \times n},\, B,C \in M_{n \times p},\, D,E \in M_{q \times m}$. Then $(a)$ $A(B+C)=AB+AC,\, (D+E)A=DA+EA$ $(b)$ $a(AB)=(aA)B=A(aB)$ $(a \in \mathbb{F})$ $(c)$ $I_{m}A=A=AI_{n}$ $(d)$ If $V$ is an n-dimensional vector space with an ordered basis, then $[I_{V}]_{\beta}=I_{n}$. $\underline{Proof \, of \, (c)}$ $$\displaystyle \begin{equation} \begin{split} (I_{m}A)..

$\underline{Thm}$ Let $V,W,Z$ be vector spaces over $\mathbb{F}$, and let $T \colon V \rightarrow W$ & $U \colon W \rightarrow Z$ be linear. Then $UT=U \circ T \colon V \rightarrow Z$ is linear. $\underline{Thm}$ Let $V$ be a vector space. Let $T, U_{1}, U_{2} \in \mathcal{L}(V)=\mathcal{L}(V,V)$. Then $(a)$ $T(U_{1}+U_{2})=TU_{1}+TU_{2}, (U_{1}+U_{2})T=U_{1}T+U_{2}T$ $(b)$ $T(U_{1}U_{2})=(TU_{1..

$\underline{Thm}$ Let $V,W$ be finite dimensional vector space over $\mathbb{F}$ with ordered basis $\beta,\gamma$, respectively. Let $T,U \colon V \rightarrow W$ be linear & $a \in \mathbb{F}$. Then $$[T+U]_{\beta}^{\gamma}=[T]_{\beta}^{\gamma}+[U]_{\beta}^{\gamma},$$ $$[aT]_{\beta}^{\gamma}=a[T]_{\beta}^{\gamma}.$$ That is, the map $\mathcal{J} \colon \mathcal{L}(V,W) \rightarrow M_{m \times n..

$\underline{Def}$ Let $V,W$ be vector space over $\mathbb{F}$ and let $a \in \mathbb{F}$. Let $T,U \colon V \rightarrow W$ be any maps. We define two maps $T+U,aT \colon V \rightarrow W$ by $$(T+U)(v)=T(v)+U(v)$$ $$(aT)(v)=aT(v)\,for\, v \in V$$ $\underline{Rmk}$ Let $\mathcal{F}(V,W)$ be the set of all maps from $V$ into $W$. With the operations of additaion & scalar multiplication above, $\mat..

$\underline{Def}$ Let $V,W$ be finite-dimensional vector spaces with ordered basis $\beta=\{v_{1},\cdots,v_{n}\},\,\gamma=\{w_{1},\cdots,w_{m}\}$, respectively. Let $T \colon V \rightarrow W$ be linear. For each $v_{j}(1\leq j\leq n)$, since $T(v_{j}) \in W$, $\exists ! a_{1j},\cdots,a_{mj} \in \mathbb{F}$ such that $\displaystyle T(v_{j})=\sum_{i=1}^{m} a_{ij}w_{i}$. We write $[T]_{\beta}^{\gam..

$\underline{Def}$ Let $\beta=\{u_{1},\cdots,u_{n}\}$ be an ordered basis for an n-dimensional vector space $V$. $\forall x \in V$, let $a_{1},\cdots,a_{n} \in \mathbb{F}$ be the unique scalars such that $\displaystyle x=\sum_{i=1}^{n} a_{i}u_{i}$; then we define the coordinate vector of $x$ by $\beta$ (relative to $\beta$), denoted by $[x]_{\beta}$, as $$\displaystyle [x]_{\beta}=\begin{bmatrix}..

$\underline{Def}$ Let $V$ be a finite dimensional vector space. An ordered basis for $V$ is a finite sequence of linear independent vectors in $V$ that spans $V$. $\underline{Ex}$ In $\mathbb{F}^{3},\, \{ e_{1},e_{2},e_{3} \}$ is an ordered basis for $\mathbb{F}^{3}$. Also, $\{e_{2},e_{1},e_{3} \}$ is another ordered basis for $\mathbb{F}^{3}$. $\underline{Rmk}$ For a given basis $\beta=\{v_{1},..