Let A be a linear map from $R^n$ to $R^m$, Then
Prove that $null(A)^{\bot} = range(A^T)$.
So far, I can easily show that $range(A^T) \subset null(A)^{\bot}$. See below.
if $x \in range(A^T)$, then $(x,n)=(A^Tu,n)=(u^T,An)=(u^T,0)=0$ for all $n \in null(A)^{\bot}$.
But I cannot proof the opposite direction.
How can I prove that if $x \in null(A)^{\bot},$ then $ x \in range(A^T)$?
Suppose $x^T \in null(A)^\perp$, we want to show that $x^T \in range(A^T)$, in other words, $x^T$ is in the row space of $A$.
Let $d$ be the dimension of the row space of $A$ and $\{ x_1^T, \ldots, x_d^T\}$ be a basis of the row space of $A$.
Then the dimension of $null(A)$ is $n-d$. Construct the matrix $B=\begin{bmatrix} y_1 \ldots y_{n-d}\end{bmatrix}$ where each columns are independent and they are elements of nullspace of $A$. By definition, we have $AB=0$which is just $B^TA^T=0$.
Rank of $B^T$ is $n-d$, hence $null(B^T)=n-(n-d)=d$, hence $\{ x_1, \ldots, x_d\}$ is a basis of $null(B^T)$.
Since $x^T \in null(A)^\perp$, we must have $B^Tx=0$, that is $x \in null(B^T)$, which means $x$ can be expressed as $$x=\sum_{i=1}^d c_i x_i$$
$$x^T=\sum_{i=1}^d c_i x_i^T$$
That is $x^T$ is in the row space of $A$.
