Linear algebra, inner product and matrix Let $A\in M_{m \times n}(\mathbb{R})$, $x\in \mathbb{R}^n$ and $b,y\in \mathbb{R}^m$. Show that if $Ax=b$ and $A^ty=0_{\mathbb{R}^m}$, then $\langle b,y\rangle=0$. Also make a geometric interpretation.
I think I may have something to do with overdetermined / underdetermined system, but do not know how to prove it.
Take this opportunity to ask for the appointment of a linear algebra and matrix reference, I'm able to do the numerical exercises mostly, but those involving statements are a big problem.
 A: So, we use elementary properties of the transpose.
Let $Ax=b$. Then, take the transpose of both sides, we have:
$$(Ax)^T=b^T \implies x^TA^T=b^T$$
Now, multiply on the right by $y$.
$$x^T(A^Ty)=b^Ty \implies b^Ty=0$$
This is implied by the given, and we now note that $b^Ty=0$ is the definition of the inner product for matrices, and we see that it is zero. Thus, they are orthogonal.
A: We have
\begin{align*}
\langle{b,y}\rangle &= \langle{Ax,y}\rangle \\
&= \langle{x,A^\intercal y}\rangle \\
&= \langle{x,0}\rangle \\
&= 0
\end{align*}
where ${}^{\intercal}$ is the transpose.
Geometrically, you can think of the columns of $A$ being basis vectors for some subspace, let's call it $W$. We have $Ax=b$, so $b$ is in the image of $A$. That is, $b\in W$. We also have that $A^\intercal y=0$, so $y$ is in the kernel of $A^\intercal$. What does this mean? It means $y$ is orthogonal to the columns of $A$. Since the columns of $A$ are a basis for $W$, we must have then that $y\in W^\perp$. Since $b\in W$ and $y\in W^\perp$, it follows that $y$ and $b$ are orthogonal.
A: I am really bad at geometric representations but I think I can help you with the first part of your question (I will omit the index of the 0 matrix):
If $A^ty=0$ it follows by transposing and using $(A^t)^t$=A
$(A^ty)^t=0^t \Leftrightarrow y^tA=0$. Now $b^ty=\langle b,y \rangle=\langle y,b \rangle =y^tb$.
Left-multiply $Ax=b$ by $y^t$. This yields $y^tAx=y^tb=\langle b,y\rangle$.
Why does this already solve the question?
Sincerely 
slinshady
