Let $T_1:V\to V$ and $T_2:W\to W$ are linear maps where $V$ and $W$ are isomorphic. Let $\{e_1, e_2,\dots, e_k\}$ be a basis for $V$. Then $\{te_1,te_2,\dots,te_k\}$ is a basis for $W$ where $t:V\to W$ linear and one-one onto. Then the matrix generated by $T_1$ under $\{e_1, e_2,\dots, e_k\}$ is same as the matrix generated by $T_2$ under $\{te_1,te_2,\dots,te_k\}$ if $t(T_1(e_j))=T_2(t(e_j))$. Now I want to find $t$ for the following problem.

Actually, in one book I have found the following argument to prove that the matrix generated by a linear map $(T^t)^t:V'' \to V''$ (where $(T^t)^t (f)$ is defined by $(T^t)^t (f)(g)=f(T^t(g))$ and $T^t:V' \to V'$ is defined by $T^t(f)(v)=f(T(v))$) with respect to a basis $\{e_1'', e_2'',\dots, e_k''\}$ is same as the matrix generated by a linear map $T:V \to V$ under the basis $\{e_1, e_2, \dots, e_k\}$ : It says that $V''$ is isomorphic to $V$. So the basis $\{e_1'', e_2'',\dots, e_k''\}$ of $V''$ can be identified with $\{e_1, e_2, \dots, e_k\}$ and hence the matrix generated by $(T^t)^t$ is same as the matrix generated by $T$. Now if I can find such a $t$ (as mentioned in the above paragraph) which relates $(T^t)^t$ and $T$ in that way then the proof is done. How to find such a $t$ ?

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    $\begingroup$ The hypotheses give no relation between $T_1$ and $T_2$, so there is no reason for the conclusion to be true. After all, it won't be the case that every $T_2:W\to W$ has the same matrix. $\endgroup$ – Gerry Myerson Sep 1 '13 at 7:18
  • $\begingroup$ @GerryMyerson : I have added something. Plz comment $\endgroup$ – aaaaaa Sep 1 '13 at 11:25
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    $\begingroup$ I think you've missed to mention that $T_2=g\circ T_1$, otherwise the statement is false, as @GerryMyerson already mentioned. $\endgroup$ – TZakrevskiy Sep 1 '13 at 11:29
  • $\begingroup$ @TZakrevskiy: still the doubt remains. elaborated the problem definition. $\endgroup$ – aaaaaa Sep 1 '13 at 13:01
  • $\begingroup$ @Prasenjit You know, the title of a question doesn't have to start with "A basic question on". $\endgroup$ – Marc van Leeuwen Sep 1 '13 at 15:25

[This answer addresses the initial version of the question]

I suppose that in your second paragraph $V'$ denotes $\mathcal L(V,k)$ the dual vector space of the $k$-vector space$~V$, that $(e'_1,\ldots,e'_n)$ denotes the dual basis of $(e_1,\ldots,e_n)$, and that similar things hold for double primes. In that case the proof suggested just amounts to showing that the transpose of the transpose of a (square) matrix gives back the matrix itself (you can check that the matrix of $T^t$ with respect to $(e'_1,\ldots,e'_n)$ is the transpose of the matrix of$~T$ with respect to $(e_1,\ldots,e_n)$). But then everything is defined in terms of a single operator$~T$, but that is not the case in your first paragraph, where $T_1$ and $T_2$ are completely unrelated (there is also a third linear map$~t$, but it links the choices of basis in $V$ and $W$, so at least that gives some relation for$~t$, which is absent for $T_1,T_2$). So what you want to prove in the first paragraph simply doesn't hold. Try taking for $T_1$ the identity and for $T_2$ the zero operator, and you get a contradiction from what you want to prove.

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  • $\begingroup$ thanks for your reply. But I want to show that the matrix generated by $(T^t)^t$ is same as $T$ by finding the map $t$ and using the argument in the first paragraph(I corrected my mistakes). I am stuck there. $\endgroup$ – aaaaaa Sep 1 '13 at 15:47
  • $\begingroup$ The statement of the second paragraph (if I interpreted it correctly) is true, but the one in the first paragraph is false. So while I can sympathise with your desire to prove the second one using the first, it won't work that way. You might be able to prove the second under the assumption that the first holds, but that is useless since you will never be able to prove the first. So you are better off trying to prove the second without assuming that the first holds; in fact this is not hard to do. $\endgroup$ – Marc van Leeuwen Sep 1 '13 at 15:52
  • $\begingroup$ if $t(T_1(e_j)=T_2(t(e_j))$ then both the transforms give the same matrix. I have a proof of that. Am I still wrong? $\endgroup$ – aaaaaa Sep 1 '13 at 15:59
  • $\begingroup$ I see that the first paragraph has been changed, and that $t(T_1(e_j)=T_2(t(e_j))$ has changed status from conclusion to hypothesis. Indeed with that condition the matrices if $T_1$ and $T_2$ will be the same. You can apply that in the second paragraph with $t:V\to V''$ the natural map from $V$ to the double dual $\def\L{\mathcal L}\L(\L(V,k),k)$. It is given by $v\in V\mapsto\Bigl(\phi\in\L(V,k)\mapsto\phi(v)\Bigr)$. $\endgroup$ – Marc van Leeuwen Sep 1 '13 at 18:38

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