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So I am trying to solve this practice problem, and so far I managed to get that the kernel of transformation is zero.

Show that $L$($\vec{x}$)=$A\vec{x}$ from $Im(A^{T})$ to $Im(A)$ is an isomorphism.

Let L(x)=0. Then L(x)=$A$$A^{T}y$ for some y. We know that Ker(A)=$Im(A^{T})^{\perp}$. So if $A$$A^{T}y$=0, $A^{T}y$ is in $Im(A^{T})^{\perp}$, which is possible only if $A^{T}y$=0, since it is both in the $Im(A^{T})^{\perp}$ and $Im(A^{T})$

I don't want use the fact that dim of these two subspaces (Im($A$) and Im($A^{T}$)) are equal, as I am trying to use this problem as another proof of that. So I need to show that Im($L$)=Im($A$) to complete the proof of the isomorphism. What would be a way of showing that everything in Im($A$) gets hit given this transformation (seems intuitive, but can't show it)? Thanks

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What you have shown is that $L$ is an injective linear map from $Im(A^T)$ to $Im(A)$. In particular, $\dim Im(A^T)\leq\dim Im(A)$. If this were an equality, then $L$ would be necessarily be an isomorphism since it's injective. But now you can use the same argument with $A$ and $A^T$ interchanged to show that $\dim Im(A)\leq\dim Im(A^T)$ (since $(A^T)^T=A$).

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    Could you elaborate? I understand what you are saying, but I do not know how to apply the double transpose.2011-03-12
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    @LinAlgStudent: The transpose of the transpose is equal to the matrx: the $(i,j)$ entry of $(A^T)^T$ is the $(j,i)$ entry of $A^T$, which is the $(i,j)$ entry of $A$. That means that the entries of $(A^T)^T$ are identical to the entries of $A$, so $(A^T)^T = A$. Now, you have shown that for *any* matrix $B$ you have $\dim(\mathrm{Im}(B)) \leq \dim(\mathrm{Im}(B^T))$. Setting $B=A$ shows that $\dim(\mathrm{Im}(A))\leq\dim(\mathrm{Im}(A^T))$. Setting $B=A^T$ shows that $\dim(\mathrm{Im}(A^T)) \leq \dim(\mathrm{Im}((A^T)^T)$. Since $(A^T)^T = A$, the two inequalities give the equality you want.2011-03-12
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    @Arturo: I see, thanks.2011-03-12