方程$Ax=b$有解的充要条件是$A^Ty=0$,$y^Tb=0$

其妙

$m\times n$的矩阵$A_{m\times n}$，则方程$Ax=b$有解的充要条件是方程$A^Ty=0$的任何解$\beta$，都有$\beta^Tb=0$.

hbghlyj

Fredholm alternative Linear algebra
If $V$ is an $n$-dimensional vector space and $T:V\to V$ is a linear transformation, then exactly one of the following holds:

• For each vector $v$ in $V$ there is a vector $u$ in $V$ so that $T(u) = v$. In other words: $T$ is surjective (and so also bijective, since $V$ is finite-dimensional).
• $\dim(\ker(T)) > 0$.

A more elementary formulation, in terms of matrices, is as follows. Given an $m\times n$ matrix $A$ and a $m\times 1$ column vector $\mathbf{b}$, exactly one of the following must hold:

• Either: $A\mathbf{x} = \mathbf{b}$ has a solution $\mathbf{x}$.
• Or: $A^T\mathbf{y} = 0$ has a solution $\mathbf{y}$ with $\mathbf{y}^T\mathbf{b}\neq 0$.
  

In other words, $A\mathbf{x} = \mathbf{b}$ has a solution $(\mathbf{b}\in\def\im{\operatorname{Im}}\im(A))$ if and only if for any $\mathbf{y}$ s.t. $A^T\mathbf{y} = 0$, $\mathbf{y}^T\mathbf{b} = 0$ $(\mathbf{b}\in(\ker(A^T))^{\bot})$.

hbghlyj

Image of adjoint equals orthogonal complement of kernel
Let ${\bf v}\in{\rm im}(T^\ast)$, then ${\bf v}=T^\ast({\bf w})$ for some ${\bf w}\in W$. Now, given ${\bf u}\in\ker T$, we see that $T({\bf u})={\bf 0}$ and therefore
$$\langle {\bf u}\mid{\bf v}\rangle
=\langle {\bf u}\mid T^\ast({\bf w})\rangle
=\langle T({\bf u})\mid {\bf w}\rangle
=\langle {\bf 0}\mid{\bf w}\rangle
=0.$$
That is, ${\bf v}\in(\ker T)^\perp$. Conversely, if ${\bf v}\notin{\rm im}(T^\ast)$, then there exists an ${\bf v}'\in{\rm im}(T^\ast)^\perp$ such that $\langle {\bf v}\mid{\bf v}'\rangle\ne 0$.
In fact, we have ${\bf v}'\in\ker T$ because $T^\ast T({\bf v}')\in{\rm im}(T^\ast)$, which implies
$$\langle T({\bf v}')\mid T({\bf v}')\rangle
=\langle {\bf v}'\mid T^\ast T({\bf v}')\rangle=0\quad\Longrightarrow\quad
T({\bf v}')={\bf 0}.$$
Therefore ${\bf v}\notin(\ker T)^\perp$, which completes the proof.

hbghlyj

As a side remark: If you know exact sequences and that taking duals is exact (actually left-exactness suffices, which is a general property of the $\hom$-functor), then you can show $\im(T^*)=(\ker T)^\perp$ as follows: The exact sequence
$$
0\longrightarrow \ker T\longrightarrow V\xrightarrow T\im T\longrightarrow 0
$$
gives an exact sequence
$$
0\longrightarrow (\im T)^*\longrightarrow V^*\xrightarrow f (\ker T)^*\longrightarrow 0.
$$
By definition, we have $(\ker T)^\perp = \ker f$ and exactness gives $\ker f = (\im T)^*$ (where we view $(\im T)^*\subseteq V^*$ via the injective map given by precomposing $T$). So it suffices to show $(\im T)^* = \im T^*$. But this follows directly from the definitions.  
Here, all the work done above is implicit in the exactness of taking duals.