Find the orthogonal projection of $(x,y,z)$ onto the subspace spanned by $(1,2,2)\ \text{and}\ (-2,2,1)$.
The answer is $\frac{(5x-2y+4z,-2x+8y+2z,4x+2y+5z)}{9}$. Why is it that I do not have to orthonormalize the basis of the subspace? If I apply Gram-Schmidt to the basis, then my answer would be $\frac{(5x-2y+4z,-2x+8y+2z,4x+2y+5z)}{3}$.
Is it only when the bases are not orthogonal I apply Gram-Schmidt to the basis?
Edit: my bad, turns out the answers to both are the same, I miscalculated.
The vectors which span the subspace are linearly independent, which I believe is all that is necessary to compute the orthogonal projection. Orthogonality is nice but not required. The "orthogonal" only refers to the projection operation, and not to the vectors which span the subspace.
Gram-Schmidt converts a set of linearly independent vectors to an orthogonal/orthonormal set, but this is not needed to determine the projection as mentioned.
Additionally, the general expression to compute the projection matrix is:
$$ P=A\left(A^{T}A\right)^{-1}A^{T} $$
where $A$ is the matrix made up of the vectors which span the subspace, which in this case is
$$ A= \begin{bmatrix} 1 & -2 \\ 2 & 2 \\ 2 & 1 \\ \end{bmatrix} $$.
The calculated $P$ would then be applied to the example vector:
$$ P \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} $$
yielding the orthogonal projection onto the subspace spanned by the vectors in the matrix $A$.
I hope this helps.