Existence of solution for Ax=b under some conditions. The following is a problem on page 389 in Elementary Differential Equations and Boundary Value Problems by William E.Boyce and Richard C.Diprima
Problem. Suppose that det$\mathbf{A=0}$ and that $\mathbf{y}$ is a solution of $\mathbf{A^*y=0}$.  Show that if ($\mathbf{b,y}$)=$\mathbf{0}$ for every such $\mathbf{y}$, then $\mathbf{Ax=b}$ has solutions.
On the question, $\mathbf{A,x,y,b}$ has complex numbers as their components and ($\mathbf{b,y}$):=$\mathbf{b^T\bar y}$.
 A: Well, the condition $(\mathbf{b},\mathbf{y})=0$ for all $\mathbf{y}$ such that $\mathbf{A}^*\mathbf{y}=0$ means that $\mathbf{b}$ is orthogonal to the kernel of $\mathbf{A}^*$. Since the orthogonal complement of the kernel of $\mathbf{A}^*$ is equal to the image of $\mathbf{A}$, the vector $\mathbf{b}$ is in the image of $\mathbf{A}$ and thus there exists an $\mathbf{x}$ such that $\mathbf{b}=\mathbf{A}\mathbf{x}$, in other words, the system has a solution.
To show that $\mathrm{Im}(\mathbf{A})=\mathrm{Ker}(\mathbf{A}^*)^{\perp}$, suppose that $\mathbf{u}\in\mathrm{Im}(\mathbf{A})$ so that there exists a $\mathbf{v}$ such that $\mathbf{u}=\mathbf{A}\mathbf{v}$ and let $\mathbf{w}\in\mathrm{Ker}(\mathbf{A}^*)$. Then $(\mathbf{u},\mathbf{w})=(\mathbf{A}\mathbf{v},\mathbf{w})=(\mathbf{v},\mathbf{A}^*\mathbf{w})=0$ since $\mathbf{A}^*\mathbf{w}=0$ and hence $\mathbf{u}$ is orthogonal to $\mathrm{Ker}(\mathbf{A}^*)$ and consequently $\mathrm{Im}(\mathbf{A})\subset\mathrm{Ker}(\mathbf{A}^*)^{\perp}$. The other inclusion can be shown similarly.
A: I got an answer from someone when components are real, so I post here.
Suppose $\mathbf{A}$ is an $\mathbf{n}$ x $\mathbf{n}$ matrix and has rank $\mathbf{m}$. Then, if we collect $\mathbf{y}$s satisfying $\mathbf{y^T Ax=0}$ for all $\mathbf{x}$, these $\mathbf{y}$s satisfy $\mathbf{A^Ty=0}$, and each $\mathbf{y}$ is orthogonal to the elements of range$\mathbf{A}$, and the dimension of the collection of $\mathbf{y}$s is $\mathbf{n-m}$. Since $\mathbf{b}$ is orthogonal to each member of the collection of $\mathbf{y}$s, it belongs to the range of $\mathbf{A}$. Hence there exists $\mathbf{x}$ s.t. $\mathbf{Ax=b}$.
