I understand the concept of using HouseHolder transformations during QR factorization, but I'm not quite sure how to actually apply them to an example. If we had some matrix, for example
$$ \begin{matrix} 3 & 5 \\ 4 & 2 \\ \end{matrix} $$
I see that there is a considerable difference when doing this method versus the Gram-Schmidt process.
$qr$ decomposition using householder transformation is very similar to $lu$ decomposition. in $lu$ you use elementary lower triangular matrices to bring the matrix to an upper triangular matrix; in $qr$ you use reflection matrices of the form $Q = I- 2uu^t/u^tu$ to $Q =2vv^t/v^tv- I$ where $u = a-b$ and $v = a+b$ to transform $a$ to $b$ provided $a^ta = b^tb.$ that is $a = Qb, b = Qa.$ you could pick one, in fact convenient scalar multiple of one, that makes the computation easier. note that $Qx = x - 2\frac{u^tx}{u^tu}u$ or $Qx = 2\frac{v^tx}{v^tv} v - x$
enough of theory. let us see how to apply on your example. we take the first column $a = \pmatrix{3\\4}$ which is of length $5.$ we will pick $b = \pmatrix{5\\0}.$ i will use $u = (a-b)/2 = \pmatrix{-1\\2}, u^tu = 5$ so that $Q_1 = I - 2uu^t/u^tu.$ we need to compute the transform of the second column $x = \pmatrix{5\\2}, u^tx = -5+4 = -1$.
now let us compute $$Q_1x = x - 2(-1/5)u = x +\frac25 u = \pmatrix{5\\2} + \frac25\pmatrix{-1\\2} = \pmatrix{\frac{23}5\\\frac{14}5}.$$
we have now is $$Q_1\pmatrix{3&5\\4&2} = \pmatrix{5&\frac{23}5\\0&\frac{14}5} $$ the $qr$ decomposition of your matrix is $$ \pmatrix{3&5\\4&2} = Q_1 \pmatrix{5&\frac{23}5\\0&\frac{14}5}$$