Knowing the answer to this question would help me answer the following question:

$A$ is an $m\times n$ matrix with $m>n$, and let $A=\hat{Q}\hat{R}$ be a reduced QR factorization. Suppose $\hat{R}$ has $k$ nonzero diagonal entries for some $k$ with $0\leq k<n$, what does this imply about the rank of $A$? Exactly $k$? At least $k$? At most $k$?

  • $\begingroup$ Yes, the rank of $A$ and the rank of $\hat{R}$ are the same. Have a look at the relevant Wikipedia page. $\endgroup$ – A.P. Jul 18 '15 at 22:42

Answering the title question: No.

The rank of a matrix equals its column rank, which by definition is the number of linearly independent columns. Adding one column will increase the column rank by one if it is independent from the others and will leave the column rank unchanged if it is dependent on the others.

  • $\begingroup$ Then the answer to the second question is that the rank of $A$ is at exactly $k$? $\endgroup$ – Tian He Jul 18 '15 at 22:18
  • $\begingroup$ @TianHe: I don't know QR decomposition so I can't say. $\endgroup$ – Rory Daulton Jul 18 '15 at 22:20
  • $\begingroup$ Yes, it is exactly $k$. QR decomposition are used to determine the rank numerically. $\endgroup$ – user251257 Jul 18 '15 at 22:28

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