Please show all your work! Total points: (longer than an actual exam)
Consider the space P2 with inner product defined as
<p(x),q(x)> = p(0)q(0).
Is this an inner product space?
No, does not satisfy positivity: p(x) = x has zero length
according to this inner product, but p(x) ≠ 0.
Consider the space F[0,∞) with inner product defined as
<f,g> = ∫0∞ f(x)g(x)dx.
Prove that this is an inner product space.
It is okay to assume that F[0,1] is a vector space,
and just prove the four axioms on the inner product.
Let fn(x) = e-nx.
Find the norm of fn for any n>0.
∫ e-2nxdx = (-1/2n) e-nx.
Eval at [0,∞): ||fn||2 = 1/2n
So ||fn|| = 1/√(2n).
What about fn for n≤0? Describe the graph of
the functions, and find their norms.
At n=0, f0 = 1 (constant function), which has
infinite norm in this space.
For n<0, fn increases without bound (exponential
growth), so the norm is also infinite.
Apply the Gram-Schmidt process to obtain an orthonormal
set from the set {f1, f2}.
Let v1 = f1; according to a previous part,
its norm is 1/√(2n) = 1/√2, so the normalized version is
q1 = √2 e-x.
Is f2 orthogonal to q1?
Their inner product is √2 ∫ e-3xdx = √2 /3 ≠ 0, so no.
The projection of f2 onto q1 is
(2/3)e-x, so the orthogonal complement is
v2 = f2 - proj(f2 onto q1)
= e-2x - (2/3)e-x.
To normalize this, we need to find its norm:
||v2||2
= ∫ ( e-2x - (2/3)e-x )2 dx
= ∫ ( e-4x - (4/3)e-3x + (4/9)e-2x ) dx
= (1/4) - (4/9) + (2/9) = 1/36.
So ||v2|| = 1/6,
so q2 = 6v2 = 6 e-2x - 4 e-x.
The orthonormal set is {q1, q2}
= { √2 e-x, 6 e-2x - 4 e-x }.
Is this set an orthonormal basis?
Not for all of F[0,∞), which is an infinite-dimensional space.
Find a QR-decomposition for the matrix, if possible:
1
0
1
-1
1
1
1
0
1
-1
1
1
.
Consider the set of points (0,1), (1,3), (2,2), (3,3).
Find the polynomial of lowest degree which interpolates these points.
Find the line which best fits these points using least-squares
approximation.
What is the error in this approximation? (i.e., ||Ax-b||).
Find the quadratic (parabola) which best fits these points using
least-squares approximation.
What is the error in this approximation? (i.e., ||Ax-b||).
Consider the matrix A:
1
0
0
0
-1
2
0
2
2
Does there exist an orthogonal matrix P for which P-1AP is
a diagonal matrix?
Yes, A is orthogonally diagonalizable, because it is symmetric.
Find all eigenvalues and eigenvectors of A.
@λ=1: x = (1,0,0)t. @λ=3: x = (0,1/2,1)t.
@λ=-2: x = (0,-1/2,1)t.
Find an orthogonal matrix P which diagonalizes A.
Just normalize the eigenvectors: