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Let $A$ be an $n$ by $n$ real matrix, and $b$ an $n$ by $1$ column vector with real entries. Prove that there exists an $n$ by $1$ column vector solution $x$ to the equation $Ax=b$, if and only if $b$ is in the orthocomplement of the kernel of the transpose of $A$.

So my idea was to first assume that there is such an $x$, so I want to show that $\langle b,y\rangle=0$ for all $y\in \mbox{ker}A^t$, but I dont know how to tackle it.

This is not homework nor anything, so I dont mind if you spoil it.

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There exists $x$ such that $Ax = b$ $\Longleftrightarrow$

$b$ is in the column space of $A$ $\Longleftrightarrow$

$b$ is in the row space of $A^{T}$ $\Longleftrightarrow$

Every element in the kernel of $A^{T}$ is orthogonal to $b$ $\Longleftrightarrow$

$b$ is in the orthogonal complement of the kernel of $A^{T}$

To see that the orthogonal complement of the kernel of $A^{T}$ is its row space, let $x \in \ker(A^{T})$. Then the column vector $A^{T}x$ has entries $r_{i} \cdot x$, the dot product of $x$ with the row vectors of $A^{T}$. If these are all zero, then $x$ is orthogonal to the rowspace of $A^{T}$. Conversely, if $x$ is orthogonal to the rowspace, all these $r_{i} \cdot x$ will be zero, so $x$ will be in the kernel of $A^{T}$.

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