Linear Algebra HELP: Closest Vector to v in R^4

[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)

 

[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.

 

[M98Ranger]

To find the vector in W that is closest to v, we can use the orthogonal projection formula:

proj_w(v) = ((v · u1)/(u1 · u1))u1 + ((v · u2)/(u2 · u2))u2 + ((v · u3)/(u3 · u3))u3

Where u1, u2, and u3 are the basis vectors for W.

First, we need to find the dot products of v with each of the basis vectors:

v · u1 = (1,0,0,-1) · (1,0,-1,0) = 1
v · u2 = (1,0,0,-1) · (0,-1,0,1) = -1
v · u3 = (1,0,0,-1) · (2,1,-3,0) = 2

Next, we need to find the dot products of each basis vector with itself:

u1 · u1 = (1,0,-1,0) · (1,0,-1,0) = 2
u2 · u2 = (0,-1,0,1) · (0,-1,0,1) = 2
u3 · u3 = (2,1,-3,0) · (2,1,-3,0) = 14

Now we can plug these values into the projection formula:

proj_w(v) = (1/2)(1,0,-1,0) + (-1/2)(0,-1,0,1) + (2/14)(2,1,-3,0)
= (1/2, 0, -1/2, 0) + (0, 1/2, 0, -1/2) + (4/14, 2/14, -6/14, 0)
= (7/14, 1/2, -7/14, -1/2)

Therefore, the vector in W that is closest to v is (7/14, 1/2, -7/14, -1/2).

This can also be verified by finding the distance between v and each vector in W and choosing the vector with the smallest distance. The distance formula is given by:

d(v, w) = ||v - w|| =