# QR decomposition and the fundamental subspaces

I'm trying to wrap my head around how QR decomposition can be directly used to find orthogonal bases for the fundamental subspaces.

I know that the bases can be obtained directly from the full QR factorization of a matrix, but I am unsure how to "extract" that information.

What is the best approach for doing so? To make things easier, assume that the starting matrix is $A$ and that the QR decomposition of $A^T=[Q_1 Q_2]R$ and, if needed, $A=[Q_3 Q_4]R$.

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The four fundamental spaces are laid out completely with $A^T=[Q_1 Q_2]R$ and $A=[Q_3 Q_4]R$. I am supposing that you know this yourself since you actually split the $Q = [Q_1 Q_2]$ in the QR factorization.
The $R$ is the column mix of $Q$. $R$ shows exactly what columns from $Q$ form $A$ (and $A^T$ for the other factorization), and which columns don't, ie the null space of $A$. If $R$ is full rank, then $A$ is full rank.
$R$ will be triangular of course, and if $A$ has null space, $R$ will have rows that are zero. These rows if non-zero would select the columns from $Q$ in the formation of $A$, but since they are zero, then those respective columns from $Q$ are the null space of $A$. Row space if looking at $A^T$
I'm still a bit confused. I believe that the row space would be the span of the columns of $Q_1$, but what would the null space be? I believe it is related to $Q_2$, but I don't know how to write that basis given Q and R. – TehTechGuy Jan 30 '13 at 4:31
Say the last row of $R$ is zero. This means that the null space is the last column of $Q$, because with $A = QR$, $R$ is a right multiplication of $Q$, and thus is a column mix of $Q$. The case of a zero row in $R$ then gives the last column of $Q$ is not used in the composition of $A$, so it is the column null space of $A$. – adam W Jan 30 '13 at 4:42