# The rank of QR factorization

If A is a $m\times n$ matrix $m\geq n$,A=QR is a reduced QR factorization. If R has k nonzero diagonal entries ($0\leq k<n$). I want to know what is the rank of A.Is it at least k?

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There seems to be a typo -- you probably want to know the rank of something other than $k$? –  joriki Sep 19 '12 at 22:31
Thanks.you are right –  89085731 Sep 19 '12 at 22:56

The rank of $A$ will be equal to $k$. A constructive way to see this is to consider the Graham-Schmidt algorithm that produces the QR factorization. In the j'th step of GS, $R_{jj} = 0$ when $a_j \in span(a_1, \dots ,a_{j-1})$, where $a_j$ is the j'th column of A. So each linearly independant column of A corresponds to a non-zero diagonal entry of R.

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$A=QR=[Q_1,Q_2]\begin{bmatrix}R_1 \\ 0\end{bmatrix}=Q_1R_1$

Since $Q_1$ has full column rank, so $\operatorname{rank}(A)=\operatorname{rank}(Q_1R_1)=\operatorname{rank}(R_1)=k$.