Find an orthogonal basis for the space spanned by the columns of the given matrix. Let $$X = \begin{pmatrix} 
1 & 1 & 4 \\ 
1 & 2 & 1 \\ 
1 & 3 & 0 \\ 
1 & 4 & 0 \\ 
1 & 5 & 1 \\ 
1 & 6 & 4 \end{pmatrix}$$
It is immediately clear to me that the columns of $X$ form a basis for the space spanned by the columns of $X$.
How does one generate an orthogonal basis for the space spanned by the columns of $X$? I know that for each element $x \neq y$ in this basis that $x\cdot y = 0$, but this doesn't help me actually generate the basis.

To apply Gran-Schmidt, I used
$$u_k = v_k - \sum\limits_{j=1}^{k-1}\operatorname{proj}_{u_j}(v_k)$$
where $$\operatorname{proj}_{u_j}(v_k) = \dfrac{v_k \cdot u_j}{u_j \cdot u_j}u_j\text{.}$$
We have $$u_1 = v_1 = \begin{pmatrix} 
1 \\
1 \\
1 \\
1 \\
1 \\
1\end{pmatrix}\text{.}$$
Now $$\operatorname{proj}_{u_1}(v_2) = \dfrac{v_2 \cdot u_1}{u_1 \cdot u_1}u_1 = \dfrac{21}{6}u_1\text{.}$$
Thus, $$u_2 = v_2 - \dfrac{21}{6}u_1 = \begin{pmatrix}
-5/2 \\
-3/2 \\
-1/2 \\
1/2 \\
3/2 \\
5/2\end{pmatrix}\text{.}$$
Also,
\begin{align*}
\operatorname{proj}_{u_1}(v_3) &= \dfrac{v_3 \cdot u_1}{u_1 \cdot u_1}u_1 = \dfrac{10}{6}u_1 \\
\operatorname{proj}_{u_2}(v_3) &= \dfrac{v_3 \cdot u_2}{u_2 \cdot u_2}u_2 = \dfrac{0}{17.5}u_2 = 0\text{.}
\end{align*}
Thus, $$u_3 = v_3 - \dfrac{10}{6}u_1 = \begin{pmatrix}
7/3 \\
-2/3 \\
-5/3 \\
-5/3 \\
-2/3 \\
7/3\end{pmatrix}\text{.}$$
Hence $\{u_1, u_2, u_3\}$ is an orthogonal basis for the space spanned by the columns of $X$.
