Question: Show that in the full $QR$ decomposition of the full rank $m\times n$ matrix $A,m ≥ n$, the vectors $q\_{n+1}, \ldots, q\_m$ are an
orthonormal basis for the null space of $A^T$.

Full $QR$ :$Q\in \mathbb R^{m\times m} , R\in \mathbb R^{m\times n}$

Thanks!

Answer: 
If $A=QR$ is a full decomposition, then $R= \begin{bmatrix} R\_1 \\ 0 \end{bmatrix}$ where $R\_1$ is invertible.

Then $\ker R^T = \operatorname{sp} \{e\_{n+1},\cdots, e\_m \}$ and so
$\ker A^T = \ker R^T Q^T = Q \ker R^T $ and
hence
$\ker A^T = \operatorname{sp} \{Qe\_{n+1},\cdots, Qe\_m \} = \operatorname{sp} \{q\_{n+1},\cdots, q\_m \}$.
$rank(A^T) + nullity(A^T) = m$. But $rank(A) = rank(A^T) = n$ since $A$ is full rank. So $nullity(A^T) = m - n$. So basis for $A^T$ must contain $m - n$ vectors.

The set $\{q\_{n+1}, \dots, q\_m\}$ contains right number of vectors and they are already orthonormal. So all that remains to show is $A^T{q\_j} = \mathbf{0}$ for all $j \in \{n+1, \dots, m\}$.

$A^T = R^T Q^T$ and $Q^T q\_j = e\_j$ since $Q$ is orthogonal. Thus $A^T q\_j = R^T e\_j = \text{$j^{th}$ column in $R^T$} = (\text{$j^{th}$ row in $R$})^T$.

Consider that $A$ is full rank and $R$ is upper triangular $\implies$ all diagonal entries in $R$ are non-zero. Thus, $n+1$ to $m$ rows in $R$ are the only zero rows in $R$. Thus $A^T q\_j = \mathbb{0}$ for all $j \in \{n+1, \dots, m\}$.