Orthogonality proof by row-column partioning I'm stuck at the following problem:

Let $A$ be a $n \times n$ matrix. Show that if $x \in Null(A)$ and $b \in col(A^{T})$, then $b$ and $x$ are orthogonal.

Here is my approach:

$x \in Null(A)$ means: $Ax=0$ and also $x^{T}A^{T} = 0$.

By $Ax=0$ we have $\displaystyle\sum_{k=1}^{n} col_{k}(A) row_{k}(x) = 0 ~~~~~~~~~~~~~~~~~(*)$.

By $x^{T}A^{T} = 0$ we have $\displaystyle\sum_{k=1}^{n} col_{k}(x^{T}) row_{k}(A^{T}) = 0 ~~~~~~(**)$.

Since $b \in col(A^{T})$, we can conclude $b^{T} \in row(A)$. 

This is the point that I cannot proceed anymore, since $b$ and $b^{T}$ correspond to $col(A^{T})$ and $row(A)$, but $(*)$ and $(**)$ don't include them, so I have no idea how to replace $b$ or $b^{T}$ into $(*)$ or $(**)$.

Since I am looking for orthogonality, I must end up with something like this in reverse:

$b.x=0 \Leftarrow b^{T}x=0 \Leftarrow \displaystyle\sum_{k=1}^{n} col_{k}(b^{T}) row_{k}(x) = 0 ~~~~~~(***)$, but I can't build $(***)$.

Any help?
Let me make it clear. Define:


$Col_k(A)$ is k-th column vector of matrix A, n-by-1 vector
$Row_k(A)$ is k-th row vector of matrix A, 1-by-n vector
$A_{ij}$ is i-th row, j-th column element of matrix A
$Col(A)$ is column space of matrix A
$Row(A)$ is row space of matrix A.


the basis of $Col(A^T)$ are n $Row_k(A)$ vectors. Since $b \in Col(A^T)$, so b is linear combination of these n $Row_k(A)$ vectors. That is:

$b=\sum_{k=1}^nc_kRow_k(A)$  (b is 1-by-n vector)

Since $x \in Null(A)$, so $Ax=0$, where $x$ is n-by-1 vector. Therefore

$\sum_{k=1}^nA_{kj}x_j=0$ for every k. That is $Row_k(A).x=0$ for every k.

So

$b.x=\sum_{k=1}^nc_kRow_k(A).x$=$\sum_{k=1}^nc_k*0=0$

b and x are orthogonal.
Let me make it clear. Define:


$Col_k(A)$ is k-th column vector of matrix A, n-by-1 vector
$Row_k(A)$ is k-th row vector of matrix A, 1-by-n vector
$A_{ij}$ is i-th row, j-th column element of matrix A
$Col(A)$ is column space of matrix A
$Row(A)$ is row space of matrix A.


the basis of $Col(A^T)$ are n $Row_k(A)$ vectors. Since $b \in Col(A^T)$, so b is linear combination of these n $Row_k(A)$ vectors. That is:

$b=\sum_{k=1}^nc_kRow_k(A)$  (b is 1-by-n vector)

Since $x \in Null(A)$, so $Ax=0$, where $x$ is n-by-1 vector. Therefore

$\sum_{k=1}^nA_{kj}x_j=0$ for every k. That is $Row_k(A).x=0$ for every k.

So

$b.x=\sum_{k=1}^nc_kRow_k(A).x$=$\sum_{k=1}^nc_k*0=0$

b and x are orthogonal.