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.

  • $\begingroup$ 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? $\endgroup$ – Ludwig Nov 25 '17 at 8:38
  • $\begingroup$ @Ludwig Why would we not be able to apply the GS orthogonalization? $\endgroup$ – eepperly16 Nov 25 '17 at 8:39
  • $\begingroup$ My question asks if $Q$ is orthogonal "as it is", not after applying another transformation. $\endgroup$ – Ludwig Nov 25 '17 at 8:40
  • $\begingroup$ Or perhaps you mean that we can apply a GS orthogonalization only on $Q_2$ in order to make $Q$ orthogonal? $\endgroup$ – Ludwig Nov 25 '17 at 8:43
  • $\begingroup$ We can just choose an orthogonal basis of $Null(A)$, that would do $\endgroup$ – Abishanka Saha Nov 25 '17 at 9:44

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