anonymous
  • anonymous
Suppose that V is a vector space over R (not necessarily finite dimensional), and that T1 : V −→ V and T2 : V −→ V are linear transformations from V to V with the property that T3 = T2 ◦ T1 is the identity transformation, i.e. that T3(v) = v for all vectors v in V. T1(x1, x2, x3, . . .) = (0, x1, x2, x3, . . .) T2(x1, x2, x3, x4, . . .) = (x2, x3, x4, . . .) Check that T2 ◦ T1 is the identity transformation from V to V.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
i have a question... how do you ask a questiom?
anonymous
  • anonymous
if you've asked a question already you have to close it so that you can ask another.
anonymous
  • anonymous
reallly... aw man i miss the old open study, but i guess its for the better because people over post questions:)

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TuringTest
  • TuringTest
yep, that's why we changed it
amistre64
  • amistre64
it looks like most people are just closing their questions to post new ones; hoping that people browse thru the closed questions ....
anonymous
  • anonymous
yea me tooo
amistre64
  • amistre64
what does t2.t1 mean in this context?
amistre64
  • amistre64
isnt t3(v) = v just a eugene type thing?
anonymous
  • anonymous
i'm not sure, T1 and T2 are 2 transformations.
amistre64
  • amistre64
the open dot is whats a little confusing; does ot mean a dot product? ie, multiplication
anonymous
  • anonymous
i think its T2 of T1 Like .. T2(T1(x)))
amistre64
  • amistre64
hmm, never even heard of that in matrix stuff before
anonymous
  • anonymous
maybe i'm wrong then
amistre64
  • amistre64
\[\begin{array}\ a_1x^1+a_2x^2+a_3x^3+a_4x^4+...+a_nx^n=0\\ b_1x^1+b_2x^2+b_3x^3+b_4x^4+...+b_nx^n=x\\ c_1x^1+c_2x^2+c_3x^3+c_4x^4+...+c_nx^n=x^2\\ .\\.\\.\\ n_1x^1+n_2x^2+n_3x^3+n_4x^4+...+n_nx^n=x^{n+1}\\ \end{array}\] im wondering if we would have to do an induction proof
amistre64
  • amistre64
might be easier with a different set of notations; using arxn and brxn for the coeefs
amistre64
  • amistre64
\[\begin{array}\ a_{11}b_{11}+a_{12}b_{21}+...+a_{1n}b_{n1}=1& a_{11}b_{12}+a_{12}b_{22}+...+a_{1n}b_{n2}=0\\ a_{21}b_{11}+a_{22}b_{21}+...+a_{2n}b_{n1}=0& a_{21}b_{12}+a_{22}b_{22}+...+a_{2n}b_{n2}=1\\ a_{31}b_{11}+a_{32}b_{21}+...+a_{3n}b_{n1}=0& a_{31}b_{12}+a_{32}b_{22}+...+a_{3n}b_{n2}=0\\ ... &...\\ a_{n1}b_{11}+a_{n2}b_{21}+...+a_{nn}b_{n1}=0& a_{n1}b_{12}+a_{n2}b_{22}+...+a_{nn}b_{n2}=0\\ \end{array}\] then again, might be just as much of a pain
anonymous
  • anonymous
maybe. ahh thanks anyways!

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