Question Number 159123 by BHOOPENDRA last updated on 13/Nov/21
Answered by aleks041103 last updated on 14/Nov/21
![Def. Linear transform L: 1) L(a+b)=L(a)+L(b) 2)L(αa)=αL(a) T( [(a),(b),(c),(d) ]+ [(A),(B),(C),(D) ])=T( [((a+A)),((b+B)),((c+C)),((d+D)) ])= [((a+A+b+B)),((b+B−c−C)),((a+A+d+D)) ]= = [((a+b)),((b−c)),((a+d)) ]+ [((A+B)),((B−C)),((A+D)) ]=T( [(a),(b),(c),(d) ])+T( [(A),(B),(C),(D) ]) ⇒T(a+b)=T(a)+T(b) T(α [(a),(b),(c),(d) ])=T( [((αa)),((αb)),((αc)),((αd)) ])= [((αa+αb)),((αb−αc)),((αa+αd)) ]= =α [((a+b)),((b−c)),((a+d)) ]=αT( [(a),(b),(c),(d) ]) ⇒T(αa)=αT(a) ⇒T is linear](https://www.tinkutara.com/question/Q159206.png)
$${Def}.\:{Linear}\:{transform}\:{L}: \\ $$$$\left.\mathrm{1}\right)\:{L}\left({a}+{b}\right)={L}\left({a}\right)+{L}\left({b}\right) \\ $$$$\left.\mathrm{2}\right){L}\left(\alpha{a}\right)=\alpha{L}\left({a}\right) \\ $$$$ \\ $$$${T}\left(\begin{bmatrix}{{a}}\\{{b}}\\{{c}}\\{{d}}\end{bmatrix}+\begin{bmatrix}{{A}}\\{{B}}\\{{C}}\\{{D}}\end{bmatrix}\right)={T}\left(\begin{bmatrix}{{a}+{A}}\\{{b}+{B}}\\{{c}+{C}}\\{{d}+{D}}\end{bmatrix}\right)=\begin{bmatrix}{{a}+{A}+{b}+{B}}\\{{b}+{B}−{c}−{C}}\\{{a}+{A}+{d}+{D}}\end{bmatrix}= \\ $$$$=\begin{bmatrix}{{a}+{b}}\\{{b}−{c}}\\{{a}+{d}}\end{bmatrix}+\begin{bmatrix}{{A}+{B}}\\{{B}−{C}}\\{{A}+{D}}\end{bmatrix}={T}\left(\begin{bmatrix}{{a}}\\{{b}}\\{{c}}\\{{d}}\end{bmatrix}\right)+{T}\left(\begin{bmatrix}{{A}}\\{{B}}\\{{C}}\\{{D}}\end{bmatrix}\right) \\ $$$$\Rightarrow{T}\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)={T}\left(\boldsymbol{{a}}\right)+{T}\left(\boldsymbol{{b}}\right) \\ $$$$ \\ $$$${T}\left(\alpha\begin{bmatrix}{{a}}\\{{b}}\\{{c}}\\{{d}}\end{bmatrix}\right)={T}\left(\begin{bmatrix}{\alpha{a}}\\{\alpha{b}}\\{\alpha{c}}\\{\alpha{d}}\end{bmatrix}\right)=\begin{bmatrix}{\alpha{a}+\alpha{b}}\\{\alpha{b}−\alpha{c}}\\{\alpha{a}+\alpha{d}}\end{bmatrix}= \\ $$$$=\alpha\begin{bmatrix}{{a}+{b}}\\{{b}−{c}}\\{{a}+{d}}\end{bmatrix}=\alpha{T}\left(\begin{bmatrix}{{a}}\\{{b}}\\{{c}}\\{{d}}\end{bmatrix}\right) \\ $$$$\Rightarrow{T}\left(\alpha\boldsymbol{{a}}\right)=\alpha{T}\left(\boldsymbol{{a}}\right) \\ $$$$ \\ $$$$\Rightarrow{T}\:{is}\:{linear} \\ $$
Answered by aleks041103 last updated on 14/Nov/21
![c) T(a)=0, ∀a∈N, N? N={ [(a),(b),(c),(d) ]∣T( [(a),(b),(c),(d) ])= [((a−b)),((b−c)),((a+d)) ]= [(0),(0),(0) ]} [((a−b)),((b−c)),((a+d)) ]= [(0),(0),(0) ]⇒ { ((a−b=0)),((b−c=0)),((a+d=0)) :}⇒a=b=c=−d=p But: [(a),(b),(c),(d) ]= [(p),(p),(p),((−p)) ]=p [(1),(1),(1),((−1)) ] ⇒N={ [(a),(b),(c),(d) ]∣T( [(a),(b),(c),(d) ])= [((a−b)),((b−c)),((a+d)) ]= [(0),(0),(0) ]} N=Span( [(1),(1),(1),((−1)) ])⇒Dim(N)=1](https://www.tinkutara.com/question/Q159207.png)
$$\left.{c}\right) \\ $$$$ \\ $$$${T}\left(\boldsymbol{{a}}\right)=\mathrm{0},\:\forall\boldsymbol{{a}}\in{N},\:{N}? \\ $$$${N}=\left\{\begin{bmatrix}{{a}}\\{{b}}\\{{c}}\\{{d}}\end{bmatrix}\mid{T}\left(\begin{bmatrix}{{a}}\\{{b}}\\{{c}}\\{{d}}\end{bmatrix}\right)=\begin{bmatrix}{{a}−{b}}\\{{b}−{c}}\\{{a}+{d}}\end{bmatrix}=\begin{bmatrix}{\mathrm{0}}\\{\mathrm{0}}\\{\mathrm{0}}\end{bmatrix}\right\} \\ $$$$\begin{bmatrix}{{a}−{b}}\\{{b}−{c}}\\{{a}+{d}}\end{bmatrix}=\begin{bmatrix}{\mathrm{0}}\\{\mathrm{0}}\\{\mathrm{0}}\end{bmatrix}\Rightarrow\begin{cases}{{a}−{b}=\mathrm{0}}\\{{b}−{c}=\mathrm{0}}\\{{a}+{d}=\mathrm{0}}\end{cases}\Rightarrow{a}={b}={c}=−{d}={p} \\ $$$${But}: \\ $$$$\begin{bmatrix}{{a}}\\{{b}}\\{{c}}\\{{d}}\end{bmatrix}=\begin{bmatrix}{{p}}\\{{p}}\\{{p}}\\{−{p}}\end{bmatrix}={p}\begin{bmatrix}{\mathrm{1}}\\{\mathrm{1}}\\{\mathrm{1}}\\{−\mathrm{1}}\end{bmatrix} \\ $$$$\Rightarrow{N}=\left\{\begin{bmatrix}{{a}}\\{{b}}\\{{c}}\\{{d}}\end{bmatrix}\mid{T}\left(\begin{bmatrix}{{a}}\\{{b}}\\{{c}}\\{{d}}\end{bmatrix}\right)=\begin{bmatrix}{{a}−{b}}\\{{b}−{c}}\\{{a}+{d}}\end{bmatrix}=\begin{bmatrix}{\mathrm{0}}\\{\mathrm{0}}\\{\mathrm{0}}\end{bmatrix}\right\} \\ $$$${N}={Span}\left(\begin{bmatrix}{\mathrm{1}}\\{\mathrm{1}}\\{\mathrm{1}}\\{−\mathrm{1}}\end{bmatrix}\right)\Rightarrow{Dim}\left({N}\right)=\mathrm{1} \\ $$