so for any Matrix above $R$ I'd like to prove this equation. So far I've been able to prove $\operatorname{Im}(A) \subset Ker(A^T)^\perp$ however I can't figure how I can prove the other way around. All help is highly appreciated.
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For $x\in\ker(A^T)$, by definition $A^Tx=0$. Hence, for any $y$, $\langle y,A^Tx\rangle=0$. Can you continue from here? – user427574 Apr 19 '19 at 12:36
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no, not entirely – crommy Apr 19 '19 at 12:46
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We have $0=\langle y,A^Tx\rangle=\langle Ay,x\rangle$. As $y$ was arbitrary, $x\in \im(A)^\perp$. Notice that $(\im(A)^\perp)^\perp=\im(A)$. – user427574 Apr 19 '19 at 12:50
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$\dim\operatorname{Ker}(A^T)^{\perp}=n-\dim\operatorname{Ker}(A^T)=n-\dim\operatorname{Ker}(A)=\dim\operatorname{Im}(A)$
Bernard
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Tsemo Aristide
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Note that the kernel of $A^T$ is exactly the perpendicular complement of the image, since
$$A^Tx = \begin{bmatrix} v_1^T \\ v_2^T \\ \vdots \\ v_n^T\end{bmatrix}x = \begin{bmatrix} v_1^Tx \\ v_2^Tx \\ \vdots \\ v_n^Tx\end{bmatrix}$$
must be the zero vector if and only if $x$ is orthogonal to the columns of $A$. So, we have
$$\operatorname{im}(A)^{\perp} = \ker(A^T).$$
Since we’re in a finite-dimensional vector space, this implies that
$$\operatorname{im}(A) = (\operatorname{im}(A)^{\perp})^{\perp} = \ker(A^T)^{\perp}.$$
Santana Afton
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