"Shortcut" to find the projection of a vector onto a subspace

Question

I understand that to project a vector $v$ onto a subspace, I must find an orthogonal basis for the subspace before projecting $v$ onto each of the orthogonal vectors and add them all up. However, for higher dimension subspaces, I find it rather cumbersome to apply the Gran Schmidt process to find the orthonormal basis.

I was wondering if there is a method to find the projection of $v$ onto a subspace?

For example:

Find the projection of $v_{1} = (-5, 3, 18)$ onto the subspace W spanned by $v_{2} = (-7, 6, 2)$ and $v_{3} = (-1, -2, 1)$

Instead of finding the orthonormal basis using the Gran Schmidt process, I simply found the cross product $v_{4} = v_{2} \times v_{3}$ and subtracted the projection of $v_{1}$ onto $v_{4}$ from $v_{1}$ to get the answer.

Are there similar shortcuts when projecting onto higher dimensions subspaces with more than two basis vectors?

Answer 1

What you did is actually to project $v_1$ onto the null-space of $v_2, v_3$ and deduct the projection .
You can do the same for higher dimensions and more vectors.

Answer 2

If the subspace is exactly $n-1$ dimensional, you can use the same trick. What you need is a generalization of the cross product. An operation that takes $n-1$ vectors of dimension $n$, and returns a single vector orthogonal to their subspace.

You write the $n-1$ vectors as row vectors of a matrix, to which you add an additional row of "free terms". Eg. in our case we have: $$ M = \begin{pmatrix} -7 & 6 & 2\\ -1 & -2 & 1\\ \vec{t}_{1} & \vec{t}_{2} & \vec{t}_{3} \end{pmatrix}$$

We have: $$\det(M) = 10 \vec{t}_{1} + 5 \vec{t}_{2} + 20 \vec{t}_{3}\\ v_{2} \times v_{3} = (10, 5,20)$$

This generalizes to any number of dimensions. Ie. with $3$ vectors of dimension $4$ you can make a $M$ as a $4\times 4$ matrix.