I have the following task :
Find the orthogonal projection and orthogonal component of vector $ \overline x $ relatively to linear subspace generated by vectors $ \overline a_{1}, \overline a_{2}, \overline a_{3} ...\overline a_{n} $
$ \overline x = (5, -1, -4, 0)^T$
$ \overline a_{1} = (23, 14, -13, 0)^T $
$ \overline a_{2} = (-11, -11, 6, -2)^T $
$ \overline a_{3} = (-32, -12, 18, 1)^T $
I can find the orthogonal projection of a vector onto subspace, with the method of least squares
$ \overline x_{a} = A (A^T A)^{-1} A^T \overline x$
but I don't really understand how I can find the orthogonal component, nor what that means. I know about a projection matrix, but what am I supposed to do with it?
$\endgroup$ 11 Answer
$\begingroup$Let $M$ be the subspace of $\mathbb{R}^{4}$ spanned by your vector $\{ \overline{a}_{1},\overline{a}_{2},\overline{a}_{3}\}$. Then the orthogonal complement $M^{\perp}$ is the subspace of all vectors $\vec{x}$ which are orthogonal to everything in $M$. Equivalently, $\overline{x}\in M^{\perp}$ iff $\overline{x}\perp \overline{a}_{j}$ for $j=1,2,3$. Equivalently, you want to find all solutions $\overline{v}=(v_{1},v_{2},v_{3},v_{4})^{T}$ of $$ \left[\begin{matrix}23 & 14 & -13 & 0\\ -11 & -11 & 6 & -2 \\ -32 & -12 & 18 & 1\end{matrix}\right]\left[\begin{matrix}v_{1} \\ v_{2} \\ v_{3} \\ v_{4}\end{matrix}\right] = \left[\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right]. $$
$\endgroup$
