1 1 0 1 1 2 3 and yo = 5 . We seek to nd the column vector x0 E R3 of smallest 1 0 1 0 norm such that Axe is of minimal distance from yo , Where the…

Let A:=[1 1 1,1 -2 0, 0 3 1] and y0 =[1 5 0]. We seek to find the column vector x0 ∈R3 of smallest

101 0

norm such that Ax0 is of minimal distance from y0 , where the norm on R3 is the usual Euclidean norm coming from the dot product.

To keep the ideas straight, you should keep in mind the facts that as a linear function R3 → R3 , the kernel of the linear function L(x) := Ax is the null space Nul(A) and the image of L is the column space Col(A) of A.

(a) Use row and column operations to put A in standard form, and hence show that the nullity of A is 1.

(b) We need to find the orthogonal projection of y0 onto the column space of A, but the latter space has dimension 2 so it is easier to find the orthogonal projection of y0 onto (Nul(At)) . Find a vector spanning Nul(At), and use this to find the orthogonal projection y1 of y0 on Col(A).

(c) Find x1 ∈ R3 satisfying Ax2 = y.

(d) Find a vector spanning Nul(A), and by calculating the orthogonal projection of x1 onto Nul(A) , find the vector x0 ∈ R3 of minimal norm for which Ax0 is of minimal possible distance from y0.

1 1 0 11 —2 3 and yo = 5 . We seek to find the column vector x0 E R3 of smallest1 0 1 0 norm such that Axe is of minimal distance from yo , Where the norm on R3 is the usual Euclidean Let A :2 norm coming from the dot product. To keep the ideas straight, you should keep in mind the factsthat as a linear function R3 —> R3 , the kernel of the linear function L(x) :2 Ax is the null spaceN ul(A) and the image of L is the column space Col (A) of A.