Proving the Orthogonal Projection Formula for Vector Subspaces

[member 428835]

Hi PF!

I've been reading and it appears that the orthogonal projection of a vector $v$ to the subspace spanned by $e_1,...,e_n$ is given by $$\sum_j\langle e_j,v \rangle e_j$$ ($e_j$ are unit vectors, so ignore the usual inner product denominator for simplicity) but there is never a proof in the texts. It's always given by definition or I see "trivial" next to the proof. Surely this is something we prove, right?

 

[StoneTemplePython]

it really depends on what you're trying to do here, as there are lots of flavors of this. Suppose we have vectors in $\mathbb C^m$ with $m \geq n$ and use the standard inner product.

you have some vector $\mathbf u$ and you project it to a subspace with these n mutually orthornormal vectors. You can do this explicitly with $\mathbf v = P\mathbf u$ where $P$ is a projector -- i.e. $P^2 = P$ and $P = P^*$.

so we have a generating set for this subspace and use it to write $\mathbf v$ as a linear combination of these. (Note: ignoring the zero vector: if you have an inner product, orthogonality implies linear independence, why?)

so
$\mathbf v = \sum_{j=1}^n \alpha_j \mathbf e_j$

multiply on the left by $\mathbf e_k^*$ (conjugate transpose) to see

$\mathbf e_k^*\mathbf v = \mathbf e_k^*\sum_{j=1}^n \alpha_j \mathbf e_j = \sum_{j=1}^n \alpha_j \mathbf e_k^*\mathbf e_j = \alpha_k \mathbf e_k^*\mathbf e_k = \alpha_k$

now re-write this with general inner product notation -- same result.

 

[member 428835]

(Note: ignoring the zero vector: if you have an inner product, orthogonality implies linear independence, why?)

This is simple to show: take $c_1v_1+...+c_nv_n = 0$ and dot this into some $v_j$. Then orthogonality implies for non-zero $v$, $c_j=0$, and thus linear independence.

And thanks! Good to see it.