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$Q$ is an orthogonal matrix. $R$ is an upper triangular matrix. $A \in \mathbb{R}^{m\times n}$ with $m > n$ and its QR-Factorizations is $A = QR$. Show that if $A$ has full rank, then the diagonal elements of $R$ are non-zero. Show also that the first $n$ columns of $Q$ are an orthonormal basis of the column space of $A$.

I tried to prove that, but with no success. Can someone help me?

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Hint: if $R$ is upper triangular and $R_{kk} = 0$, then its first $k$ columns must be linearly dependent, which makes $R$ have rank $< n$.

By "the spanning of $A$" you mean "the column space of $A$", i.e. the span of the columns of $A$.

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  • $\begingroup$ Yes, I meant column space. I edited that, thanks. I don't quite understand why when the diagonal of $R$ has zeros then the first $k$ columns must be linearly dependent? Also even if this is true and $R$ has rank $< n$, how does this prove the rank of $A$? $\endgroup$ – ladyeli555 May 23 at 0:09
  • $\begingroup$ 1) If $R_{kk} = 0$, the first $k$ columns of $R$ can have nonzero entries only in the first $k-1$ positions, and so are in the span of the vectors $e_1, \ldots e_{k-1}$ (where $e_j$ is the unit vector with $1$ in position $j$ and $0$ everywhere else). Thus their span has dimension $\le k-1$. 2) If $Rv = 0$ then $QRv = 0$. $\endgroup$ – Robert Israel May 23 at 2:00
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The column space of the matrix $A$ is the span of the columns. I.e

$$ \textrm{Span}(A) = \bigg\{ \sum_{i=1}^{n} c_{i}a_{i} | c_{i} \in \mathbf{K} , a_{i} \in A \bigg\} $$

or any linear combination of the columns $a_{i}$. Now the columns $q_{i}$ are formed by iteratively subtracting of the previous projections.

$$ q_{n} = \frac{a_{n} - \sum_{i=1}^{n-1} r_{in}q_{i}}{r_{nn}}$$

so $q_{n}$ is a linear combination of the columns of $A$. Note that

$$ v_{n} = a_{n} - \sum_{i=1}^{n-1} r_{in}q_{i} $$

is orthogonal and the coefficient $r_{nn}$ is actually $\| v_{n}\|$ which makes $q_{n}$ orthonormal.

Note: I think $\textrm{Span}(A)$ is a slight abuse of notation.

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