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Let $E$ be a finite dimensional vector space over $F$. Let $A:E\to E$ be a linear transformation. Define $T_{A}:L(E)\to L(E)$ given by $T_A(X))=AX$ for all $X\in L(E)$. Prove that $T_A$ is invertible if and only if $A$ is invertible.

Note: $L(E)$ Stands for the Linear transformations from $E\to E$.

($\Leftarrow$): We assume that $A$ is invertible, so $T_A(X(v)))=A(X(v))$ then $A^{-1}(T_A(X(v)))=X(v)$ for $v\in E$, this shows that $T_{A}^{-1}=A^{-1}$.

($\Rightarrow$): I think I should asume that $T_A$ is invertible a somehow get that exists $B$ such that $A(B(v))=B(A(v))=v$ for this is the meaning that $A$ is invertible. Yet I could only get $T_A(X(v)))=A(X(v))\Rightarrow T_{A}^{-1}(A(X(v)))=X(v)$ but is not equal to $v$. Perhaps I am seeing something wrong?

Share your opinion, It would be most welcomed.

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Your notation is confusing. Here's what I think you're going for:

$\Longleftarrow:$

Suppose $A$ is invertible. We note that for all $X \in L(E)$, we have $$ (T_A \circ T_{A^{-1}})(X) = AA^{-1}X = X $$ which is to say that $T_A \circ T_{A^{-1}}$ is the identity map $I_{L(E)}$. Similarly, $T_{A^{-1}} \circ T_A = I_{L(E)}$. It follows that $T_{A^{-1}} = (T_{A})^{-1}$, so $T_A$ is invertible.

$\Longrightarrow:$

Suppose that $T_A$ is invertible. Then there exists an $X \in L(E)$ such that $T_A(X) = I_{E}$, which is to say that $AX = I_{E}$. Because $E$ is finite dimensional and $A \in L(E)$, $AX = I_{E}$ implies that $A$ is invertible with $A^{-1} = X$. So, $A$ is invertible.

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  • $\begingroup$ Yeah I agree, what you wrote is more understandable and clear. Thanks! $\endgroup$ – César Rosendo Nov 28 '16 at 22:12
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$T_A$ is a linear map over a finite dimensional space. Hence it is invertible if and only if its kernel is reduced to the $0$ matrix.

If $A$ is not invertible, there exists a vector $v \neq 0$ such that $Av=0$. The matrix $B$ having $n$ columns equal to $v$ is not $0$ but $T_A(B)=AB=0$.

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