Distance between vector and subspace

Question

I'm currently practicing for my Linear Algebra exam and I am trying to solve the following question:

I solved a and I'm pretty sure that's correct:

The orthonormal basis for V is:

• x1 = (1/3)[1 2 -2 0]
• x2 = 1/(sqrt(3)[0 1 1 1]
• x3 = 1/sqrt(6)[2 -1 0 1]

But I'm not sure if I have the correct answer for b. I first projected y on V which results in: [5 3 -8 -1]. Then i did ||y - projected_y|| which results in sqrt(18). Which would mean the distance between y and V is sqrt(18).

Could anyone tell me if this is correct? :)

Answer 1

The logic of projecting $y$ on $V$ and then analyzing the norm of $\lVert y - Proj_y \rVert$ is correct. Also, notice that another way that you could solve this problem is to project $y$ about any vector perpendicular to $V$ and take the norm of this projection itself! :)

Answer 2

Let $V=[v_1,v_2,v_3]$, then the subspace is exactly the set $\{Vx\mid x\in\Bbb R^3\}$. Now, consider the function $$f(x)=\|y-Vx\|_2^2.$$ Can you: 1) show that $f$ is differentiable and convex; 2) work out a local minimum (which by convexité is also the global minimum)?