Anyone know linear algebra?
Consider subspace W = { [x1, x2, x3] such that x1 = x3} of R3. And let x = (-2, 1, 3)
Find projW (x).
I'll solve it by creating a matrix equation, because I think it's clearer. You could do a Gram-Schmidt process, which to me is more mechanical.
Here's my picture.
Let's decompose the vector x into two vectors p and q, where the projection p lives in the subspace W, and q lives in the subspace that is orthogonal to W.
We have the equation,
(1)
x = p + q
Now, let me define the vectors (with respect to the standard basis),
e1 = [1; 0; 1]
e2 = [0; 1; 0]
e3 = [1; 0; -1]
These three vectors are mutually orthogonal, so they form a basis for R^3. In particular, the 1st and 3rd components of e1 are the same. Hence, span{e1, e2} is exactly the subspace W, and span{e3} is the subspace orthogonal to W.
This means, for some real numbers p1, p2, and q3, we can write,
(2)
p = p1 * e1 + p2 * e2
q = q3 * e3
Combine equations (1) and (2), and use what we know about x, e1, e2, and e3 (we know their components with respect to the standard basis).
We get,
[-2; 1; 3] = p1 * [1; 0; 1] + p2 * [0; 1; 0] + q3 * [1; 0; -1]
The RHS of this equation is a linear combination of vectors. We write the RHS as a matrix-vector multiplication:
[-2; 1; 3] = A * [p1; p2; q3]
where the matrix A is defined to be,
A = [e1, e2, e3] = [1, 0, 1; 0, 1, 0; 1, 0, -1]
Because the vectors e1, e2, and e3 are mutually orthogonal, the matrix A is invertible. Hence, we can solve for the vector [p1; p2; q3].
We find that,
p1 = 1/2, p2 = 1, p3 = -5/2
which means,
p = [1/2; 1; 1/2]
q = [-5/2; 0; 5/2]
We see that x = p + q, and the 1st and 3rd components of p are the same (i.e. p lives in W).
