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Suppose $A$ is a $m \times n$ matrix and $x \in K^n$. Does $A^T\cdot Ax=0 \implies Ax=0$? I am trying hard to prove this but it's really difficult, i want to show the intersection of $Ker(A^T)$ and $im(A)$ is $0$ to prove the statement but I just end up with a loop.

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  • $\begingroup$ That depends on $K$, and what you mean by $T$. Can we have $K = \Bbb C$? If so, does $T$ refer to the conjugate-transpose, or just the "entrywise transpose" of the complex matrix? $\endgroup$ – Ben Grossmann Nov 2 '17 at 22:12
  • $\begingroup$ If $K=\mathbb R$: notice that $A^T Ax = 0 \implies x^T A^T A x = \| Ax \|^2 = 0$. $\endgroup$ – user251257 Nov 2 '17 at 22:15
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Assuming $\Bbb K = \Bbb Q$, $K =\Bbb R$, or $K = \Bbb C$ and $A^T$ denotes the ordinary transpose (in the cases $\Bbb K = \Bbb Q, \Bbb R$) or the "conjugate transpose" or hermitian adjoint (also denoted $A^\dagger$, when $\Bbb K = \Bbb C$), we may say:

If

$A^T \cdot Ax = 0, \tag 1$

then

$x^T A^T \cdot Ax = 0, \tag 2$

whence

$\Vert Ax \Vert^2 = (Ax)^T \cdot (Ax) = x^TA^T \cdot Ax = 0, \tag 3$

which forces

$Ax = 0. \tag 4$

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Assume $Ax\neq0$. Then $Ax\in Col(A)$. The row space of $A^T$ is $Col(A)$. So there is at least one row $R_i$ of $A^T$, such that $<Ax,R_i>\neq 0$ and it results in $A^TAx\neq 0$. Therefore, $Ax$ should be equal to zero, if we want $A^TAx$ to be zero.

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  • $\begingroup$ That requires an inner product on $K^n$ which is compatible with transpose. So you basically silently assume $K$ is a subfield of $\mathbb R$. $\endgroup$ – user251257 Nov 2 '17 at 22:44

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