# finding the distance between a vector and image

• Dec 9th 2010, 12:23 PM
finding the distance between a vector and image
I have a vector b
[1
1
1
0]
and a matrix A
[1 0 1
2 -1 1
-1 2 1
0 -1 -1]

it says
1. the sum of the first two columns is equal to the third and conclude that the rank(A) = 2
2. replace A by a matrix B with maximal rank and same image
3. compute the peojection P on Im(A) = Im(B) by u sing the method of least square solution
4. compute the projection b0 = Pb of b on Im(A)
5. compute the distance between b and b0

I get the first two steps but I don't understand how to do the 3rd. I'm pretty sure I can do the last two steps once i get the result from step 3. can anyone clarify on what I'm supposed to do in step 3?
• Dec 10th 2010, 03:43 AM
HallsofIvy
I presume you have recently been introduced to finding "least squares" approximations by using the "adjoint" matrix.

If A is a linear operator from vector space U to vector space V, and y is NOT in the image of A, then we cannot solve Ax= y. But we can find x such that Ax is "closest" to y. We argue that Ax will be closest to y when y- Ax is perpendicular to Ax. That means that \$\displaystyle <Ax, y- Ax>_V= 0\$ where \$\displaystyle < , >_V\$ is the inner product in V. Since the "adjoint" of A is the operator, \$\displaystyle A^T\$, from U back to V such that, for all u in U, v, in V, \$\displaystyle <Au, v>_V= <u, A^Tv>_U\$ where the second \$\displaystyle < , >_U\$ is the inner product in U.

Specifically, we have \$\displaystyle <x, A^T(y- Ax>_U= 0\$. But now, x can be any vector in U so we must have \$\displaystyle A^T(y- Ax)= 0\$. The "Ax" satisfying \$\displaystyle A^Ty= A^TAx\$ is the projection of y onto the image of A.