# “Completing” an LQ decomposition

Let $A$ be a real $m\times n$ matrix with $n>m$. Let $Q_1$ be a $n\times m$ with orthonormal columns such that $$AQ_1 = L$$ where $L$ is of dimension $m\times m$ and lower triangular.

Question: Is it always possible to find a $n\times (n-m)$ matrix $Q_2$ such that $Q:=[Q_1\, |\, Q_2]$ is orthogonal and $$AQ = [L\, |\, \mathbf{0}_{m\times (n-m)}],$$ where $\mathbf{0}_{m\times (n-m)}$ denotes a $m\times (n-m)$ zero matrix?

The matrix $A$ has maximum rank $m$. So rank of its null space $\ge n-m$. Hence we can get $n-m$ independent vectors from $Null(A)$ which constitutes $Q_2$ such that $AQ_2=0$. The columns of $Q_1$ and $Q_2$ will be independent. We can carry out a Gram Schmidt orthogonalization to make the matrix $Q$ orthogonal.
• So, in general (i.e., without applying a Gram-Schmidt orthogonalization step), it is not possible to find $Q_2$ such that $Q$ is orthogonal. Right? – Ludwig Nov 25 '17 at 8:38
• My question asks if $Q$ is orthogonal "as it is", not after applying another transformation. – Ludwig Nov 25 '17 at 8:40
• Or perhaps you mean that we can apply a GS orthogonalization only on $Q_2$ in order to make $Q$ orthogonal? – Ludwig Nov 25 '17 at 8:43
• We can just choose an orthogonal basis of $Null(A)$, that would do – Abishanka Saha Nov 25 '17 at 9:44