# A problem on linear maps and matrix

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 $t$ ?

• 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. – Gerry Myerson Sep 1 '13 at 7:18
• @GerryMyerson : I have added something. Plz comment – aaaaaa Sep 1 '13 at 11:25
• I think you've missed to mention that $T_2=g\circ T_1$, otherwise the statement is false, as @GerryMyerson already mentioned. – TZakrevskiy Sep 1 '13 at 11:29
• @TZakrevskiy: still the doubt remains. elaborated the problem definition. – aaaaaa Sep 1 '13 at 13:01
• @Prasenjit You know, the title of a question doesn't have to start with "A basic question on". – Marc van Leeuwen Sep 1 '13 at 15:25

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.
• 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. – aaaaaa Sep 1 '13 at 15:47
• 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? – aaaaaa Sep 1 '13 at 15:59
• 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 $v\in V\mapsto\Bigl(\phi\in\L(V,k)\mapsto\phi(v)\Bigr)$. – Marc van Leeuwen Sep 1 '13 at 18:38