Any basis can be straightened into an orthogonal one. Keep the first vector; from the second, remove its projection onto the first; from the third, remove its projections onto the two already made; and so on. What is left each time is at right angles to everything before it.

Building a staircase against a sloping wall, each new step is trimmed where it would lean into the steps already laid, so that every step sits square to the ones below.

One step of Gram–Schmidt. The projection of $\mathbf{x}_2=(2,3)$ onto $\mathbf{x}_1=(2,1)$ is $\mathbf{p}=\frac75\mathbf{x}_1=(2.8,1.4)$. What is left, $\mathbf{v}_2=\mathbf{x}_2-\mathbf{p}=(-0.8,1.6)$, is perpendicular to $\mathbf{x}_1$, and with $\mathbf{x}_1$ it spans the same plane as $\mathbf{x}_1,\mathbf{x}_2$.

Start from independent vectors $\mathbf{x}_1,\ldots,\mathbf{x}_k$. Put $\mathbf{v}_1=\mathbf{x}_1$, and for each later subtract from its projections onto the vectors already made. The are orthogonal, and dividing each by its length makes them orthonormal. It works with any inner product in place of the dot product.

• At every stage $\operatorname{span}\{\mathbf{v}_1,\ldots,\mathbf{v}_j\}=\operatorname{span}\{\mathbf{x}_1,\ldots,\mathbf{x}_j\}$: the theorem below.

Go one step at a time. The new vector $\mathbf{v}_j$ is orthogonal to the earlier ones, because subtracting the projection onto $\mathbf{v}_k$ removes exactly the part along $\mathbf{v}_k$, and the earlier are orthogonal to each other. It lies in the span of , and it is not zero, since otherwise would be a combination of the earlier . So there are orthogonal non-zero vectors, which are independent, inside a space of dimension : they span it.

The first vector is kept.

For $k<j$, the earlier $\mathbf{v}$ are orthogonal to one another, so only one term of the sum survives the dot product, and it cancels.

$\mathbf{v}_j$ is $\mathbf{x}_j$ minus a combination of earlier $\mathbf{v}$, which are combinations of the earlier .

If it were $\mathbf{0}$, $\mathbf{x}_j$ would be a combination of $\mathbf{x}_1,\ldots,\mathbf{x}_{j-1}$, against independence.

$j$ orthogonal non-zero vectors are independent, and inside a space of dimension $j$ they span it.

To make $\mathbf{x}_2$ orthogonal to $\mathbf{v}_1$, subtract its projection onto $\mathbf{v}_1$.

Divide each vector of the orthogonal basis by its own length. Directions and orthogonality are unchanged; every length becomes 1.

• $\mathbf{v}_j$ is $\mathbf{x}_j$ minus its projections onto the earlier $\mathbf{v}_i$.

Great work! You now know how to:

• run the Gram–Schmidt process on a list of vectors;
• normalise to an orthonormal basis;
• recognise dependence when a vector comes out zero;
• connect the process with $A=QR$.