$Q$ 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?
On
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.
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 $A$.