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$\underline{Lemma}$ Let $T \colon V \rightarrow W$ be linear & invertible. Then $dim(V) < \infty \Leftrightarrow dim(W) < \infty.$ In this case, $dim(V)=dim(W).$ $\underline{Proof}$ $(\Rightarrow)$ Let $\beta= \{ x_{1}, \cdots, x_{n} \}$ be a basis for $V$. By thm, $T(\beta)$ spans $R(T)=W$. By thm, $dim(W) < \infty$. $(\Leftarrow)$ Since $T^{-1} \colon W \rightarrow V$ is linear & invertible, i..

$\underline{Def}$ Let $T \colon V \rightarrow W$ be linear, where $V,W$ are vector spaces over $\mathbb{F}$. $(i)$ A function $U \colon W \rightarrow V$ is an inverse $T$ if $TU=I_{W},\,UT=I_{V}$. $(ii)$ $T$ is invertible if it has an inverse. $(iii)$ Such an inverse $U$ of $T$ is unique if ie exists. In this case, we write $U=T^{-1}.$ $\underline{Facts}$ Let $T \colon V \rightarrow W,\, U \colo..

$\underline{Def}$ (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..