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Can I calculate the QR-decomposition of the matrix below, even if there are 2 linearly dependent column vectors? Or should I form the QR-decomposition of those 2 vectors, which are linearly independent to each other.

$\begin{bmatrix}0 & 0 & 4\\6 & 3 & 1\\-2 & -1 & -1\\2 & 1 & 5\\2 & 1 & 3\end{bmatrix}$

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  • $\begingroup$ Welcome to math SE. Have a look at mathjax to help you with mathematical expression. With mathjax, you will be able to include the matrix directly instead of a picture. $\endgroup$ – Alain Remillard Jan 16 at 14:26
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Your matrix is of the form of $\begin{bmatrix} v_1, \frac12v_1, v_2\end{bmatrix}$

We can find an orthogonal basis for $\operatorname{Span}\{v_1, v_2\}$, let it be $w_1, w_2$ where $v_1=\|v_1\|w_1$ and $v_2=r_{13}w_1+r_{23}w_2$.

We let $Q=\begin{bmatrix} w_1 & w_2, & \ldots, &w_5\end{bmatrix}$ be an orthogonal matrix and let $\hat{Q}$ be the matrix that only consists of the first two columns of $Q$.

Then we have

$$\begin{bmatrix}v_1 & v_2 & v_3 \end{bmatrix}=Q\begin{bmatrix} \|v_1\| & \frac12\|v_1\| & r_{13}\\ 0 & 0 & r_{23} \\ 0 & 0 & 0 \\ 0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix}$$

The thin QR decomposition can be written as

$$\begin{bmatrix}v_1 & v_2 & v_3 \end{bmatrix}=\hat{Q}\begin{bmatrix} \|v_1\| & \frac12\|v_1\| & r_{13}\\ 0 & 0 & r_{23} \end{bmatrix}$$

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