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Let $A$ be an $m\times n$ real matrix. Then show that, $N(A)\cap Im(A^T)=\{0\}$.

To prove that first I take $x\in N(A)\cap Im(A^T)$. Then, $Ax=0$ and $x=A^Ty$ for some $y\in \Bbb R^m$. Then, $AA^Ty=0$ , i.e. $y\in N(AA^T)=N(A)$.

From here how I can proceed further ?

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    And what is denoted $N(A)$, if you don't mind? – 2017-01-23
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    I think you meant the null space of $A$. “Nullity” is a number, the dimension of this space in fact. – 2017-01-23
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    Yeah..Null space. – 2017-01-23

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$$\|x\|^2 = \langle x,x\rangle = \langle x, A^Ty\rangle = \langle Ax, y\rangle = 0 \\ \therefore x = 0$$