A is any real matrix. Is true that $\operatorname{Im}{(A^tA)}=\operatorname{Im}{(A^t)}$?
I only proved that $\operatorname{Im}{(A^tA)}\subset\operatorname{Im}{(A^t)}$.
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A is any real matrix. Is true that $\operatorname{Im}{(A^tA)}=\operatorname{Im}{(A^t)}$? I only proved that $\operatorname{Im}{(A^tA)}\subset\operatorname{Im}{(A^t)}$. |
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You've shown that $\rm{Im}(A^\mathrm{T}A)$ is a subspace of $\rm{Im}(A^\mathrm{T})$. The general result follows by showing that $\rm{rank}(A) = \rm{rank}(A^\mathrm{T}A)$. This answer may be helpful to you. What can you conclude about a subspace in a vector space of equal dimension? |
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