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Let $A\in M_n(\mathbb R)$ and $b\in M_{n1}(\mathbb R)$ column vector. Prove $\exists x\in M_{n1}$ column vector solution to the equation $Ax=b\iff b=(\ker(A^{\tau}))^{\bot}$.

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 $y\in \ker(A^{\tau})$, but I dont know how to tackle it.

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

2 Answers 2

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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 $A^{T}$.