linear algebra代写| 线性代数代写| dimension argument代写

证明 . (a) We have that \(T\left(O_{v}\right)=T\left(O_{v}+U_{v}\right)=T\left(O_{v}\right)+T\left(0_{v}\right)\). Adding \(-T\left(0_{v}\right)\) to both sider we get \(-\underbrace{T\left(0_{v}\right)+T\left(0_{v}\right)}_{O_{w}}=-\underbrace{T\left(0_{v}\right)+T\left(0_{v}\right)}_{O_{w}}+T\left(0_{v}\right)\) So, \(\quad O_{W}=T\left(U_{v}\right)\) (b) This follows from (c) with \(n=2\), (c) \(\Rightarrow\) Suppose \(T\) is linear. We prove the resvit by induction, If \(n=1\), then \(T\left(\alpha_{1} x_{1}\right)=\alpha_{1} T\left(x_{1}\right)\) since \(T\) is linear. If \(n=2\), Then \(T\left(x_{1} x_{1}+\alpha_{2} x_{2}\right)=T\left(\alpha_{1} x_{1}\right)+T\left(\alpha_{2} x_{2}\right)\) \(=\alpha_{1} T\left(x_{1}\right)+\alpha_{2} T\left(x_{2}\right)\), Let \(k \geqslant 2\) in formula. Suppose the formula is true for \(n=k\), Then $$ T\left(\alpha_{1} x_{1}+\alpha_{2} x_{2}+\ldots+\alpha_{k} x_{k}+\alpha_{k+1} x_{k+1}\right)=T\left(\alpha_{1} x_{1}+\ldots+\alpha_{k} x_{k}\right)+T\left(\alpha_{k+1} x_{k+1}\right)=…$$ $$=\alpha_{1} T\left(x_{1}\right)+\alpha_{2} T\left(x_{2}\right)+\alpha_{k} T\left(x_{k}\right)+\alpha_{k+1} T\left(x_{k+1}\right)$$ \((\Leftarrow)\) Suppose \(T\left(\sum_{i=1}^{n} \alpha_{i} x_{i}\right)=\sum_{i=1}^{n} \alpha_{i} T\left(x_{i}\right)\) for all \(\alpha_{i} \in F\) and \(x_{i} \in V_{1}\) Setting \(n=2\) and \(\alpha_{1}=\alpha_{2}=1\) we get \(T\left(x_{1}+x_{2}\right)=T\left(x_{1}\right)+T\left(x_{2}\right)\) for all \(x_{1}, x_{2} \in V\). Setting \(n=1\), we get \(T\left(\alpha_{1} x_{1}\right)=\alpha_{1} T\left(x_{1}\right)\) for all \(\alpha_{1} \in F\) and \(x_{1} \in V\), Thus, \(T\) is linear,

证明 . (a)This linear, Let \(x=(a, b, c)\) and \(y=(d, e, f)\), Let \(\alpha, \beta \in \mathbb{R}\), Then \(\begin{aligned} T(\alpha x+\beta y) &=T(\alpha a+\beta d, \alpha b+\beta e, \alpha c+\beta f) \\ &=(\alpha a+\beta d-\alpha b-\beta e, 2 \alpha c+2 \beta f) \\ &=(\alpha a-\alpha b, 2 \alpha c)+(\beta d-\beta e, 2 \beta f) \\ &=\alpha(a-b, 2 c)+\beta(d-e, 2 f) \\ &=\alpha T(x)+\beta T(y) . \end{aligned}\) (i)(ii) are easy to check, (iii) A basis for \(N(T)\) is \(\left\{\left(\begin{array}{l}1 \\ 0\end{array}\right)\right\} .\) So, Nullity \((T)=1\). (iv) \(R(T)=\mathbb{R}^{2}\), So, \(\operatorname{rank}(T)=2\), (v) \(T\) is not one-to one. For example, \(T(0,0,0)=T(1,1,0)=(0,0)\) (vi) T is onto since \(R(t)=\mathbb{R}^{2}\). (b) T is not linear, For example, \(T(1,1)+T(2,2)=\left(1-1,1^{2}\right)+\left(2-2,2^{2}\right)\) \(=(0,1)+(0,4)=(0,5)\) but \(T((1,1)+(2,2))=T(3,3)=\left(3-3,3^{2}\right)=(0,9)\) (c) \(T\) is linear. Let \(x=\left(\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ d_{1} & e_{1} & f_{1}\end{array}\right)\) and \(y=\left(\begin{array}{ll}a_{2} & b_{2} c_{2} \\ d_{2} & e_{2} f_{2}\end{array}\right)\) and \(\alpha, \beta \in \mathbb{R},\) Then \(\begin{aligned}  &T(\alpha x+\beta y)=T\left(\begin{array}{cc}\alpha a_{1}+\beta a_{2} & \alpha b_{1}+\beta b_{2} & \alpha c_{1}+p_{2} z_{2} \\ \alpha d_{1}+\beta d_{2} & \alpha e_{1}+\beta e_{2} & \alpha f_{1}+\beta f_{2}\end{array}\right) \\ &=\left(\begin{array}{cc}2\left(\alpha a_{1}+\beta a_{2}\right)-\left(\alpha_{1}+\beta b_{2}\right) & \alpha c_{1}+\beta c_{2}+2\left(\alpha d_{1}+\beta d_{2}\right) \\ 0 & 0\end{array}\right) \\ &=\alpha\left(\begin{array}{cc}2 a_{1}-b_{1} & c_{1}+2 d_{1} \\ 0 & 0\end{array}\right)+\beta\left(\begin{array}{cc}2 a_{2}-b_{2} & c_{2}+2 d_{2} \\ 0 & 0\end{array}\right) \\ &=\alpha T(x)+\beta T(y) \end{aligned}\)

证明 . Let \(\alpha, \beta \in \mathbb{R}\) and \(f, g \in C(\mathbb{R})\). Then \(\begin{aligned} T(\alpha f+\beta g)=\int_{a}^{b} \alpha f(t)+\beta g(t) d t &=\alpha \int_{a}^{b} f(t) d t+\beta \int_{a}^{b} g(x) d t \\ &=\alpha T(f)+\beta T(g) \end{aligned}\)

证明 . (a) \(L_{A}(x)=\left(\begin{array}{cc}1 & \pi \\ y / 2 & -10\end{array}\right)\left(\begin{array}{c}17 \\ -5\end{array}\right)=\left(\begin{array}{cc}17 & -5 \pi \\ \frac{17}{2} & +50\end{array}\right)=\left(\begin{array}{c}17-5 \pi \\ 117 / 2\end{array}\right)\) (b) \(L_{A}(x)=\left(\begin{array}{ccc}-i & 1 & 0 \\ 1+i & 0 & -1\end{array}\right)\left(\begin{array}{c}-2 i \\ 4 \\ 1,57\end{array}\right)=\left(\begin{array}{c}2 i^{2}+4+0 \\ -2 i-2 i^{2}+0-1,57\end{array}\right)\) \(=\left(\begin{array}{c}2 \\ -2 i+0.43\end{array}\right)\)

证明 . Let \(y \in R(T)\). Then there exists \(x \in V\) with \(T(x)=y\) by def of \(R(T)\). Since \(x \in V\) and \(V=\operatorname{span}\left(\left\{v_{1}, \ldots, v_{n}\right\}\right)\) we have \(x=c_{1} v_{1}+c_{2} v_{2} t_{14}+c_{1} v_{1}\) for some \(c_{i} \in F\), So, \(y=T(x)=T\left(c_{1} v_{1}+c_{2} v_{2}+u_{1}+c_{1} v_{n}\right)=c_{1} T\left(v_{1}\right)+c_{2} T\left(v_{2}\right)+m+c_{n} T\left(v_{n}\right)\) co \(y \in \operatorname{span}\left(\left\{T\left(v_{1}\right), T\left(v_{2}\right), \ldots, T\left(v_{n}\right)\right\}\right)\) Hence \(R(T) \subseteq \operatorname{span}\left(\left\{T\left(v_{1}\right), T\left(v_{2}\right)_{1}, \ldots, T\left(v_{n}\right)\right\}\right)\) Thus, \(\left.R(T)=\operatorname{spn}\left(\left\{T\left(V_{1}\right)_{1}, \mu_{1}\right) T\left(V_{n}\right)\right\}\right)=R(T)\)

证明 . \(\Rightarrow\) is easy. Suppuse that T is one-to-one. We know that (T\left(0_{v}\right)=0_{w}\) So, \(O_{v} \in N(T)\). Since \(T\) is one-to-one, \(0_{V}\) is the only solution Thus, \(N(T)=\{x \in V \mid T(x)=0 w\}=\left\{0_{v}\right\}\) Let \(x, y \in V\) with \(T(x)=T(y)\) Then \(T(x)-T(y)=0 w\). So, \(T(x-y)=O_{w}\), since \(T\) ir linear. Thus, \(x-y \in N(T)\). so, \(x-y=0_{V}\). Thus, \(x=y\). (b) Suppoce that \(\operatorname{dim}(V)=\operatorname{dim}(W)\) Rocale that nullity \((T)+\operatorname{can} k(T)=\operatorname{dim}(V)\) \(\Rightarrow\) Suppore that \(T\) is one-tu-one. By pact \((a)\), \(N(T)=\{0,\}\}\) so nullity \((T)=0\), So, mea \(0+\operatorname{cank}(T)=\operatorname{dim}(V)\) Since \(\operatorname{dim}(V)=\operatorname{dim}(\omega)\) we get 1. Since \(R\) (T) is a subspace of w and since dim (R(T) Hence T is onto. \(\Leftrightarrow\) Suppose that \(T\) is onto. So, \(R(T)=W\), So, \(R(T)=W_{C}=\operatorname{dim}(w)\), Thus, rank \((T)=\operatorname{dim}(v)\) since \(dim(w)=dim(v)\)we get that \(null(T)+dim(T)=dim(v)\) because \(null(T)+dim(w)=dim(w)\), so \(null(T)=0\), thus \(N(T)=\{0_v\}\). (c) Supeose that \(V\) and \(W\) ane finite dimen ritinal cand T is \(1-1\) and onto. basis for \(V\), Let \(w_{i}=T\left(v_{i}\right)\) for \(i=1,2, \ldots, n\), By thm in the notes since \(T\) By thm in the notes \(\left.w_{1}, w_{2}, \ldots, w_{n}\right\}\) is a basis fur \(w_{1}\) s a basis \((V)=n=\operatorname{dim}(\omega)\),

证明 . (a) is easy. (b) We have that \(nullity (T)+rank(T)=\operatorname{dim}(V)\). Since \(\operatorname{din}(V)>\operatorname{dim}(\omega)\) we have that $$nullity \((T)+ranK(T)>\operatorname{dim}(W)$$ if \(nullity(T)=0\), then \(rank(T)>dim(W)\). But that would imply that \(R(T)\), which is a subspace of \(W\) but has bigger dimension than \(W\)\), this couyld not happen, so \(nullity (T) >0\), So, \(\operatorname{dim}(N(T))>0\). so \(N(T) \neq\left\{0_{v}\right\}\)so, \(T\) is not one to one b problem 6.