MMT-006 Solved Assignment 2023

IGNOU MMT-006 Solved Assignment 2023 | M.Sc. MACS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU MMT-006 Assignment Question Paper 2023

mmt-006-qp-f7c5c3cd-c8c6-4a67-bc59-92f7593d91ab
  1. State whether the following statements are true or false. Justify with a short proof or a counter example.
    a) The space l 3 l 3 l^(3)l^{3}l3 is a Hilbert space
    b) Any non zero bounded linear functional on a Banach space is an open map.
    c) Every bounded linear map on a complex Banach space has an eigen value.
    d) The image of a Cauchy sequence under a bounded linear map is also a Couchy sequence.
    e) If A A A\mathrm{A}A is a bounded linear operator on a Hilbert space such that A A = I A A = I AA^(**)=I\mathrm{AA}^{*}=\mathrm{I}AA=I, then A A = I A A = I A^(**)A=I\mathrm{A}^{*} \mathrm{~A}=\mathrm{I}A A=I.
  2. a) Characterise all bounded linear functionals on a Hilbert space.
    b) Show that the map T : R 3 R 2 T : R 3 R 2 T:R^(3)rarrR^(2)\mathrm{T}: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}T:R3R2 given by T ( x 1 , x 2 , x 3 ) = ( x 1 + x 2 , x 3 ) T x 1 , x 2 , x 3 = x 1 + x 2 , x 3 T(x_(1),x_(2),x_(3))=(x_(1)+x_(2),x_(3))\mathrm{T}\left(\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}\right)=\left(\mathrm{x}_{1}+\mathrm{x}_{2}, \mathrm{x}_{3}\right)T(x1,x2,x3)=(x1+x2,x3) is an open map.
c) Check whether a finite dimensional normed linear space is reflexive? Justify your answer.
  1. a) Show how a real linear functional u u uuu on a complex linear normed space gives rise to a complex linear functional f f f\mathrm{f}f. What is the relation between the boundedness of u u u\mathrm{u}u and that of f f f\mathrm{f}f ?
b) In a Hilbert space. Prove that x n x x n x x_(n)rarrx\mathrm{x}_{\mathrm{n}} \rightarrow \mathrm{x}xnx provided x n x x n x ||x_(n)||rarr||x||\left\|\mathrm{x}_{\mathrm{n}}\right\| \rightarrow\|\mathrm{x}\|xnx and x n , x x , x x n , x x , x (:x_(n),x:)rarr(:x,x:)\left\langle\mathrm{x}_{\mathrm{n}}, \mathrm{x}\right\rangle \rightarrow\langle\mathrm{x}, \mathrm{x}\ranglexn,xx,x.
c) Are Hahn-Banach extensions always unique? Justify.
  1. a) State the principle of uniform boundedness. Use it to show that a set E E EEE in a normed space X X XXX is bounded if f ( E ) f ( E ) f(E)f(E)f(E) is bounded in K K KKK for every f X f X f inX^(‘)f \in X^{\prime}fX.
b) If H H H\mathrm{H}H is a Hilbert space and S C H S C H SCH\mathrm{SCH}SCH, show that S = S ⊥⊥⊥ S = S ⊥⊥⊥ S^(_|_)=S^(⊥⊥⊥)\mathrm{S}^{\perp}=\mathrm{S}^{\perp \perp \perp}S=S⊥⊥⊥. When S S S\mathrm{S}S is the same as S ⊥⊥ S ⊥⊥ S^(⊥⊥)\mathrm{S}^{\perp \perp}S⊥⊥ ? Justify.
5 a) Show that Q Q Q\mathrm{Q}Q defined on ( C [ 0 , 1 ] , ) C [ 0 , 1 ] , (C[0,1],||*||_(oo))\left(\mathrm{C}[0,1],\|\cdot\|_{\infty}\right)(C[0,1],) by Q ( x ) = 0 1 t x ( t ) d t Q ( x ) = 0 1 t x ( t ) d t Q(x)=int_(0)^(1)tx(t)dt\mathrm{Q}(\mathrm{x})=\int_{0}^{1} \mathrm{t} \mathrm{x}(\mathrm{t}) \mathrm{dt}Q(x)=01tx(t)dt is a bounded linear functional. Calculate Q Q ||Q||\|\mathrm{Q}\|Q. b) Let A A AAA be an operator on a Hilbert space H H HHH. Show if A x = A x A x = A x ||Ax||=||A^(**)x||\|A x\|=\left\|A^{*} x\right\|Ax=Ax for every x H x H x in Hx \in HxH, then A A A\mathrm{A}A is normal. Is it converse true? Justify.
c) Let { u n } u n {u_(n)}\left\{\mathrm{u}_{\mathrm{n}}\right\}{un} be the sequence in l 2 l 2 l^(2)l^{2}l2 with 1 in the n th n th  n^(“th “)\mathrm{n}^{\text {th }}nth  place and zeroes else where prove that the set { u n } u n {u_(n)}\left\{\mathrm{u}_{\mathrm{n}}\right\}{un} is an orthonormal basis for l 2 l 2 l^(2)l^{2}l2.
6 a) Let X = C [ 0 , 1 ] X = C [ 0 , 1 ] X=C[0,1]X=C[0,1]X=C[0,1] with Sup norm defined by f = Sup { | f ( x ) | } f = Sup { | f ( x ) | } ||f||=Sup{|f(x)|}\|f\|=\operatorname{Sup}\{|f(x)|\}f=Sup{|f(x)|}.
x [ 0 , 1 ] x [ 0 , 1 ] xin[0,1]\mathrm{x} \in[0,1]x[0,1]
Let T T T\mathrm{T}T be a linear map defined on X X X\mathrm{X}X by T ( f ) = f ( 1 2 ) T ( f ) = f 1 2 T(f)=f((1)/(2))\mathrm{T}(\mathrm{f})=\mathrm{f}\left(\frac{1}{2}\right)T(f)=f(12).
Show that T T T\mathrm{T}T is a bounded linear map such that T = 1 T = 1 ||T||=1\|\mathrm{T}\|=1T=1.
b) Define Eigen Spectrum of a bounded linear operator on a Banach space. Show that the eigen spectrum of the operator T T T\mathrm{T}T on l 2 l 2 l^(2)l^{2}l2 given by T ( α 1 , α 2 ) = ( 0 , α 1 , α 2 ) T α 1 , α 2 = 0 , α 1 , α 2 T(alpha_(1),alpha_(2)dots)=(0,alpha_(1),alpha_(2)dots)\mathrm{T}\left(\alpha_{1}, \alpha_{2} \ldots\right)=\left(0, \alpha_{1}, \alpha_{2} \ldots\right)T(α1,α2)=(0,α1,α2) is empty.
c) Let ||*||\|\cdot\| be a norm on a linear space X X XXX. if x , y X x , y X x,y in Xx, y \in Xx,yX and x + y = x + y x + y = x + y ||x+y||=||x||+||y||\|x+y\|=\|x\|+\|y\|x+y=x+y, then show that s x + t y = s x + t y s x + t y = s x + t y ||sx+ty||=s||x||+t||y||\|s x+t y\|=s\|x\|+t\|y\|sx+ty=sx+ty for all s 0 , t 0 s 0 , t 0 s >= 0,t >= 0s \geq 0, t \geq 0s0,t0.
  1. a) Let X X XXX be an inner product space with the inner product given by < < <<<, > > >>>. For x X x X x in Xx \in XxX, define the function : X K : X K ||*||:X rarr K\|\cdot\|: X \rightarrow K:XK given by x =< x , x > 1 / 2 x =< x , x > 1 / 2 ||x||=<x,-x > 1//2\|x\|=<x,-x>1 / 2x=<x,x>1/2, the non negative square root of < x , x > < x , x > < x,x ><\mathrm{x}, \mathrm{x}><x,x>. Show that : X K : X K ||*||:XrarrK\|\cdot\|: \mathrm{X} \rightarrow \mathrm{K}:XK defines a norm on X X X\mathrm{X}X and I < ( x , y ) > I x y I < ( x , y ) > I x y I < (x,y) > I <= ||x||||y||\mathrm{I}<(\mathrm{x}, \mathrm{y})>\mathrm{I} \leq\|\mathrm{x}\|\|\mathrm{y}\|I<(x,y)>Ixy for all x , y X x , y X x,yinX\mathrm{x}, \mathrm{y} \in \mathrm{X}x,yX. Also show that for all x , y X x , y X x,yinX\mathrm{x}, \mathrm{y} \in \mathrm{X}x,yX, x + y 2 + x y 2 = 2 ( x 2 + y 2 ) x + y 2 + x y 2 = 2 x 2 + y 2 ||x+y||^(2)+||x-y||^(2)=2(||x||^(2)+||y||^(2))\|x+y\|^{2}+\|x-y\|^{2}=2\left(\|x\|^{2}+\|y\|^{2}\right)x+y2+xy2=2(x2+y2).
b) Let X X X\mathrm{X}X be a vector space. Let 1 1 ||*||^(1)\|\cdot\|^{1}1 and 2 2 ||*||^(2)\|\cdot\|^{2}2 be two norms on X X X\mathrm{X}X. When are these norms said to be equivalent? Justify your answer.
Let X = R 3 X = R 3 X=R^(3)\mathrm{X}=\mathbb{R}^{3}X=R3. For x = ( x 1 , x 2 , x 3 ) x = x 1 , x 2 , x 3 x=(x_(1),x_(2),x_(3))\mathrm{x}=\left(\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}\right)x=(x1,x2,x3).
Let x 1 = | x 1 | + | x 2 | + | x 3 | x 1 = x 1 + x 2 + x 3 ||x||^(1)=|x_(1)|+|x_(2)|+|x_(3)|\|x\|^{1}=\left|x_{1}\right|+\left|x_{2}\right|+\left|x_{3}\right|x1=|x1|+|x2|+|x3|
x 2 = | x 1 | 2 + | x 2 | 2 + | x 3 | 2 x 2 = x 1 2 + x 2 2 + x 3 2 ||x||^(2)=sqrt(|x_(1)|^(2)+|x_(2)|^(2)+|x_(3)|^(2))\|x\|^{2}=\sqrt{\left|x_{1}\right|^{2}+\left|x_{2}\right|^{2}+\left|x_{3}\right|^{2}}x2=|x1|2+|x2|2+|x3|2
Show that 1 1 ||*||^(1)\|\cdot\|^{1}1 and 1 1 ||*||^(1)\|\cdot\|^{1}1 are equivalent.
  1. a) Let X = L 2 [ 0 , 2 π ] X = L 2 [ 0 , 2 π ] X=L^(2)[0,2pi]X=L^{2}[0,2 \pi]X=L2[0,2π] and u n ( t ) = e int 2 π , t { π , π } , n Z u n ( t ) = e int  2 π , t { π , π } , n Z u_(n)(t)=(e^(“int “))/(sqrt(2pi)),t in{-pi,pi},n in Zu_{n}(t)=\frac{e^{\text {int }}}{\sqrt{2 \pi}}, t \in\{-\pi, \pi\}, n \in Zun(t)=eint 2π,t{π,π},nZ. Show that the set E = { u 1 , u 2 } E = u 1 , u 2 E={u_(1),u_(2)dots}\mathrm{E}=\left\{\mathrm{u}_{1}, \mathrm{u}_{2} \ldots\right\}E={u1,u2} is an orthonormal set in X X X\mathrm{X}X.
b) Let H H H\mathrm{H}H be a Hilbert space. For any subset A A A\mathrm{A}A of H H H\mathrm{H}H, define A A A^(_|_)\mathrm{A}^{\perp}A. If A B H A B H AsubeBsubeH\mathrm{A} \subseteq \mathrm{B} \subseteq \mathrm{H}ABH, then show that:
i) B A B A quadB^(_|_)subeA^(_|_)\quad \mathrm{B}^{\perp} \subseteq \mathrm{A}^{\perp}BA ii) A A ⊥⊥ A A ⊥⊥ quadAsubeA^(⊥⊥)\quad \mathrm{A} \subseteq \mathrm{A}^{\perp \perp}AA⊥⊥
State conditions on A A A\mathrm{A}A so that A ⊥⊥ = A A ⊥⊥ = A A^(⊥⊥)=A\mathrm{A}^{\perp \perp}=\mathrm{A}A⊥⊥=A.
c) Let X = C [ 0 , 1 ] X = C [ 0 , 1 ] X=C^(‘)[0,1]\mathrm{X}=\mathrm{C}^{\prime}[0,1]X=C[0,1] and Y = C [ 0 , 11 ] Y = C [ 0 , 11 ] Y=C[0,11]\mathrm{Y}=\mathrm{C}[0,11]Y=C[0,11] and let T : X Y T : X Y T:XrarrY\mathrm{T}: \mathrm{X} \rightarrow \mathrm{Y}T:XY be the linear operator from X X X\mathrm{X}X to Y Y YYY given by T ( f ) = f T ( f ) = f T(f)=f^(‘)T(f)=f^{\prime}T(f)=f, the derivative of f f fff on [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1]. Show that T T TTT is not continuous.
  1. a) Let X X X\mathrm{X}X and Y Y Y\mathrm{Y}Y be Banach spaces and F : X Y F : X Y F:XrarrY\mathrm{F}: \mathrm{X} \rightarrow \mathrm{Y}F:XY be a linear map which is continuous and open. Will F F F\mathrm{F}F always be closed? Will F F F\mathrm{F}F be always surjective? Give reasons for your answer.
b) Check whether the identity map on an infinite dimensional space is compact.
c) Define A : C 3 C 3 A : C 3 C 3 A:C^(3)rarrC^(3)\mathrm{A}: \mathbb{C}^{3} \rightarrow \mathbb{C}^{3}A:C3C3 by A ( z 1 , z 2 , z 3 ) = ( i z z 1 , e 2 i z 2 , z 3 ) A z 1 , z 2 , z 3 = i z z 1 , e 2 i z 2 , z 3 A(z_(1),z_(2),z_(3))=(izz_(1),e^(2i)z_(2),z_(3))\mathrm{A}\left(\mathrm{z}_{1}, \mathrm{z}_{2}, \mathrm{z}_{3}\right)=\left(\mathrm{iz} \mathrm{z}_{1}, \mathrm{e}^{2 \mathrm{i}} \mathrm{z}_{2}, \mathrm{z}_{3}\right)A(z1,z2,z3)=(izz1,e2iz2,z3).
Check whether A A AAA is i) self adjoint, ii) unitary.
  1. a) Let A : X 0 X Y A : X 0 X Y A:X_(0)subeXrarrY\mathrm{A}: \mathrm{X}_{0} \subseteq \mathrm{X} \rightarrow \mathrm{Y}A:X0XY be a closed operator where X X X\mathrm{X}X and Y Y Y\mathrm{Y}Y are Banach spaces.
Define x A = x + A x , x X 0 x A = x + A x , x X 0 ||x||_(A)=||x||+||Ax||,xinX_(0)\|\mathrm{x}\|_{\mathrm{A}}=\|\mathrm{x}\|+\|\mathrm{Ax}\|, \mathrm{x} \in \mathrm{X}_{0}xA=x+Ax,xX0.
Then show that the norm A A ||*||_(A)\|\cdot\|_{\mathrm{A}}A is complete.
b) Find a bounded linear functional f f f\mathrm{f}f on 3 3 ℓ^(3)\ell^{3}3 such that f ( e 3 ) = 3 f e 3 = 3 f(e_(3))=3\mathrm{f}\left(\mathrm{e}_{3}\right)=3f(e3)=3 and f = 3 f = 3 ||f||=3\|\mathrm{f}\|=3f=3.
c) Prove that l 1 l 2 l 1 l 2 l^(1)subl^(2)l^{1} \subset l^{2}l1l2. If: T : ( l 2 , 2 ) ( l 1 , 2 ) T : l 2 , 2 l 1 , 2 T:(l^(2),||*||_(2))rarr(l^(1),||*||_(2))\mathrm{T}:\left(l^{2},\|\cdot\|_{2}\right) \rightarrow\left(l^{1},\|\cdot\|_{2}\right)T:(l2,2)(l1,2) is a compact operator, show that: T : ( l 2 , 2 ) ( l 2 , 2 ) T : l 2 , 2 l 2 , 2 T:(l^(2),||*||_(2))rarr(l^(2),||*||_(2))\mathrm{T}:\left(l^{2},\|\cdot\|_{2}\right) \rightarrow\left(l^{2},\|\cdot\|_{2}\right)T:(l2,2)(l2,2) is also compact.
\(sin\:3\theta =3\:sin\:\theta -4\:sin^3\:\theta \)

MMT-006 Sample Solution 2023

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Question:-01

  1. State whether the following statements are true or false. Justify with a short proof or a counter example.
a) The space l 3 l 3 l^(3)l^3l3 is a Hilbert space
Answer:

Statement: The space l 3 l 3 l^(3)l^3l3 is a Hilbert space.

Justification:

A Hilbert space is a complete inner product space. That is, it is a vector space equipped with an inner product that is complete in the metric induced by the inner product.
The space l 3 l 3 l^(3)l^3l3 consists of all sequences ( x 1 , x 2 , x 3 , ) ( x 1 , x 2 , x 3 , ) (x_(1),x_(2),x_(3),dots)(x_1, x_2, x_3, \ldots)(x1,x2,x3,) such that n = 1 | x n | 3 < n = 1 | x n | 3 < sum_(n=1)^(oo)|x_(n)|^(3) < oo\sum_{n=1}^{\infty} |x_n|^3 < \inftyn=1|xn|3<.
  1. Inner Product: We can define an inner product on l 3 l 3 l^(3)l^3l3 as follows:
x , y = n = 1 x n y n ¯ x , y = n = 1 x n y n ¯ (:x,y:)=sum_(n=1)^(oo)x_(n) bar(y_(n))\langle x, y \rangle = \sum_{n=1}^{\infty} x_n \overline{y_n}x,y=n=1xnyn¯
This definition satisfies the properties of an inner product (conjugate symmetry, linearity, and positive-definiteness).
  1. Completeness: The space l 3 l 3 l^(3)l^3l3 is not complete with respect to this inner product. To see this, consider the sequence of sequences x ( k ) = ( 1 , 1 2 , , 1 k , 0 , 0 , ) x ( k ) = ( 1 , 1 2 , , 1 k , 0 , 0 , ) x^((k))=(1,(1)/(2),dots,(1)/(k),0,0,dots)x^{(k)} = (1, \frac{1}{2}, \ldots, \frac{1}{k}, 0, 0, \ldots)x(k)=(1,12,,1k,0,0,). Each x ( k ) x ( k ) x^((k))x^{(k)}x(k) is in l 3 l 3 l^(3)l^3l3, but its limit as k k k rarr ook \to \inftyk, which is the sequence ( 1 , 1 2 , 1 3 , ) ( 1 , 1 2 , 1 3 , ) (1,(1)/(2),(1)/(3),dots)(1, \frac{1}{2}, \frac{1}{3}, \ldots)(1,12,13,), is not in l 3 l 3 l^(3)l^3l3 because n = 1 ( 1 n ) 3 = n = 1 1 n 3 = sum_(n=1)^(oo)((1)/(n))^(3)=oo\sum_{n=1}^{\infty} \left(\frac{1}{n}\right)^3 = \inftyn=1(1n)3=.
Therefore, l 3 l 3 l^(3)l^3l3 is not a Hilbert space because it is not complete with respect to the inner product.

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b) Any non zero bounded linear functional on a Banach space is an open map.
Answer:

Statement: Any non-zero bounded linear functional on a Banach space is an open map.

Justification:

A map T : X Y T : X Y T:X rarr YT: X \to YT:XY between topological spaces is said to be open if the image of every open set in X X XXX under T T TTT is open in Y Y YYY.
Let X X XXX be a Banach space and f : X R f : X R f:X rarrRf: X \to \mathbb{R}f:XR (or C C C\mathbb{C}C) be a non-zero bounded linear functional. We want to show that f f fff is an open map.
  1. Boundedness: Since f f fff is bounded, there exists a constant C > 0 C > 0 C > 0C > 0C>0 such that | f ( x ) | C x | f ( x ) | C x |f(x)| <= C||x|||f(x)| \leq C \|x\||f(x)|Cx for all x X x X x in Xx \in XxX.
  2. Non-Zero: f f fff is non-zero, which means there exists some x 0 X x 0 X x_(0)in Xx_0 \in Xx0X such that f ( x 0 ) 0 f ( x 0 ) 0 f(x_(0))!=0f(x_0) \neq 0f(x0)0.
  3. Open Map: Consider an open ball B ( x , ϵ ) B ( x , ϵ ) B(x,epsilon)B(x, \epsilon)B(x,ϵ) in X X XXX centered at x x xxx with radius ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0. We want to show that f ( B ( x , ϵ ) ) f ( B ( x , ϵ ) ) f(B(x,epsilon))f(B(x, \epsilon))f(B(x,ϵ)) is an open set in R R R\mathbb{R}R (or C C C\mathbb{C}C).
    • Take y f ( B ( x , ϵ ) ) y f ( B ( x , ϵ ) ) y in f(B(x,epsilon))y \in f(B(x, \epsilon))yf(B(x,ϵ)). Then there exists z B ( x , ϵ ) z B ( x , ϵ ) z in B(x,epsilon)z \in B(x, \epsilon)zB(x,ϵ) such that f ( z ) = y f ( z ) = y f(z)=yf(z) = yf(z)=y.
    • Since z B ( x , ϵ ) z B ( x , ϵ ) z in B(x,epsilon)z \in B(x, \epsilon)zB(x,ϵ), we have z x < ϵ z x < ϵ ||z-x|| < epsilon\|z – x\| < \epsilonzx<ϵ.
    • Using linearity and boundedness of f f fff, we get | f ( z ) f ( x ) | C z x < C ϵ | f ( z ) f ( x ) | C z x < C ϵ |f(z)-f(x)| <= C||z-x|| < C epsilon|f(z) – f(x)| \leq C \|z – x\| < C \epsilon|f(z)f(x)|Czx<Cϵ.
    • This means y = f ( z ) y = f ( z ) y=f(z)y = f(z)y=f(z) lies in an open interval ( f ( x ) C ϵ , f ( x ) + C ϵ ) ( f ( x ) C ϵ , f ( x ) + C ϵ ) (f(x)-C epsilon,f(x)+C epsilon)(f(x) – C \epsilon, f(x) + C \epsilon)(f(x)Cϵ,f(x)+Cϵ) which is contained in f ( B ( x , ϵ ) ) f ( B ( x , ϵ ) ) f(B(x,epsilon))f(B(x, \epsilon))f(B(x,ϵ)).
Thus, f ( B ( x , ϵ ) ) f ( B ( x , ϵ ) ) f(B(x,epsilon))f(B(x, \epsilon))f(B(x,ϵ)) contains an open interval around every point y y yyy in it, making f ( B ( x , ϵ ) ) f ( B ( x , ϵ ) ) f(B(x,epsilon))f(B(x, \epsilon))f(B(x,ϵ)) an open set. Therefore, f f fff is an open map.
In conclusion, the statement “Any non-zero bounded linear functional on a Banach space is an open map” is true, and we have justified it with a short proof.

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c) Every bounded linear map on a complex Banach space has an eigen value.
Answer:

Statement: Every bounded linear map on a complex Banach space has an eigenvalue.

Justification:

The statement is false.
An eigenvalue λ λ lambda\lambdaλ for a bounded linear map T : X X T : X X T:X rarr XT: X \to XT:XX on a complex Banach space X X XXX is a complex number such that there exists a non-zero vector x X x X x in Xx \in XxX satisfying T ( x ) = λ x T ( x ) = λ x T(x)=lambda xT(x) = \lambda xT(x)=λx.

Counterexample:

Consider the right shift operator T : 2 2 T : 2 2 T:ℓ^(2)rarrℓ^(2)T: \ell^2 \to \ell^2T:22 on the complex Banach space 2 2 ℓ^(2)\ell^22 of square-summable sequences. The operator T T TTT is defined as follows:
T ( ( x 1 , x 2 , x 3 , ) ) = ( 0 , x 1 , x 2 , x 3 , ) T ( ( x 1 , x 2 , x 3 , ) ) = ( 0 , x 1 , x 2 , x 3 , ) T((x_(1),x_(2),x_(3),dots))=(0,x_(1),x_(2),x_(3),dots)T((x_1, x_2, x_3, \ldots)) = (0, x_1, x_2, x_3, \ldots)T((x1,x2,x3,))=(0,x1,x2,x3,)
Let’s assume, for the sake of contradiction, that T T TTT has an eigenvalue λ λ lambda\lambdaλ and corresponding eigenvector x = ( x 1 , x 2 , x 3 , ) x = ( x 1 , x 2 , x 3 , ) x=(x_(1),x_(2),x_(3),dots)x = (x_1, x_2, x_3, \ldots)