!If f(Σ=1a;x; ) =f(Σ+1 aix; ) = Σ+1 af(x;)TrueO False*Any finite dimension vector space with dim(X) is equal ton is isomorphism with Fdows bamiTrueFalseΣ=1a;f(x;) thenThere exist two method of directsum of any two sub space of vectorspaceTrueO False

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Answered: (2i=1 αix Zi=1 f(Σ+1 aix; ) = Σ+1…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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If A is an adjoint operator on H=K2 and for x=(x(1), x(2)), we define A(x)=(x(1)-x(2), x(1)+x(2)), how do we derive that A*(x) = (x(1)+x(2), -x(1)+x(2))?
In R3x1, let U = span ( 1 Find bases for u- and (U -) -.
b) For in R", we say that is in the kernel of Tif T(x) 0. If x₁ and 2 are both in the kernel of T, show that ax₁ + bx2 is also in the kernel of T for all scalars a and b. =
O Let (R, +) is aring. and f:R→R defind as XER Show that f is isomorPhi'sM and is said tobe identity isomorphism. OLet (e>+)is aring of complex numbers and Mz be aringof2 matrices aud g &–DM, défined as follows- ollarb}= ( ) isemesphim or not. ab show that whether gis -ba) isoMexphism or not.
Once you get away from vector spaces of finite dimension, it is no longer true that every linear map is automatically continuous. For example, let Coo {f E F(N, R) : f(n) = 0 for all sufficiently large n}, where F(N, R) is the vector space of all func- tions from N to R with the usual pointwise operations. Another way of expressing the condition for f to belong to Coo is that {n : f(n) # 0} is finite. Then define L: C00R by L(f) = Enf(n). Notice that the sum on the right hand side is in fact a finite sum. Now (a) Check that L is linear. (b) Show that {|L(f)| : ||F|| < 1} is not bounded above. Then deduce that L is not continuous, from the following theorem: For each continuous linear map L:V W the set {I|L(x)|lw: x|lv <1} is bounded above. 1 n=r (Hint: consider the functions e,, defined by e, (n) = and define a %3D 0 otherwise norm on e, by ||e,|| := sup{e,(n)|: n < 1} = le,(1)|. Show that for all r E N, |le,|| < 1. On the other hand {|L(e,)|:r€ N} is not bounded above.]
Let V be a finite dimensional inner product space and T a non negative operator on V. Then T has a unique non negative square root, that is, there is one and only one non negative operator N on V such that N^2=T.
im submitting same question again please answe #2
All the listed vector spaces are constructed over the field of real numbers R and equipped with the usual operations. In which case are the vector spaces isomorphic to each other? The correct answer is (b), or why?

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(2i=1 αix Zi=1 f(Σ+1 aix; ) = Σ+1 af(x;) * True False Any finite dimension vector space with dim(X) is equal to nis isomorphism with F O True False