# Linear Algebra HELP

1. Nov 17, 2004

### rayveldkamp

Ok need help with a question that could be in tomorrow's exam:
In Euclidean space R^4 equipped with Euclidean Inner Product, let W be subspace which has basis {(1,0,-1,0), (0,-1,0,1), (2,1,-3,0)}.
Determine the vector in W which is closest to v=(1,0,0,-1)

2. Nov 17, 2004

### arildno

That vector must be the orthogonal projection of v into W.
Proof:
Consider the squared distance vector between an arbitrary element in W and v:
$$(v-\alpha{W}_{1}-\beta{W}_{2}-\gamma{W}_{3})^{2}$$
This is minimized by differentiating with respect to $$\alpha,\beta,\gamma$$ and setting the three results equal to zero.
For example, we gain by differentiating with respect to $$\alpha$$
$$-2(v-\alpha{W}_{1}-\beta{W}_{2}-\gamma{W}_{3})\cdot{W}_{1}=0$$
That is, the distance vector is orthogonal to $$W_{1}$$

By solving the system of equations, you gain the minimizing values of $$\alpha,\beta,\gamma$$ and hence, the vector in W with shortest distance to v.
$$W_{1},W_{2},W_{3}$$ is of course the basis of W.

Last edited: Nov 17, 2004