Arman B. answered • 05/26/19

UC Berkeley Graduate tutoring Physics, Calculus, and Chemistry

Once you have a set of orthogonal basis representing a subspace, then you can project any vector with the same dimension as your subspace to you subspace. Projecting a vector into a subspace is the act adding up all the vectors that align with your basis subspace. Than means, you can write the projection of a vector **v** onto a subspace spanned by the orthogonal set of vectors **{a**_{n}**}** as:

{a_{n}}

proj_{{a_n}}(**v**) = (a_{1}·v/||a_{1}||^{2}) **a**_{1}** **+ (a_{2}·v/||a_{2}||^{2}) **a**_{2}+ ... + (a_{n}·v/||a_{n}||^{2}) **a**_{n}

As an example, think about a plane spanned by two orthogonal 3 dimensional vectors in real euclidean space of 3 dimensions (ℜ^{3} ), then an arbitrary vector in ℜ^{3} can be projected onto this plane (subspace) by the process described above.

For further utilities, you can compose a matrix A, whose column vectors are the **orthonormal **basis of the subspace **{a**_{n}**} **(that means if the basis were not normalized, you have to normalize before constructing the matrix), then the projector to the subspace P = AA^{T}. To convince yourself of this, you can first check that P has the property of a projector (idemptotence), i.e. P^{2}=P. You can work out the details of how the projection described in the equation above is captured in the matrix vector multiplication P**v**.