Notice: This final homework assigment is due after our final midterm, but the questions on it cover material that will be on the test. I encourage you to work these problems before the test, although you don’t need to have a set of solutions written up until after.
Let be the vector space of polynomials in of degree at most , and let be the derivative (a linear transformation on ).
If you’ve already answered part (b), the answer is .
If you haven’t: Derivatives of polynomials of degree at most have to be polynomials of degree at most . So, .
If you’ve already answered part (a), the answer is .
If you haven’t: The only polynomials which differentiate to 0 are constant polynomials (and all constant polynomials differentiate to 0). So, .
Let be the vector space of polynomials in of degree at most . Then is an inner product on . If is an orthonormal basis for under this inner product, compute:
There’s not enough information to solve this question using the formula for the inner product (what on earth is ?). However, since is orthonormal, the formula for the inner product of two -vectors is just the usual dot product formula, .
Find an orthonormal basis for the kernel of the matrix :
To find a basis for the kernel, we solve by Gaussian elimination to find,
It turns out that is already orthogonal (why?) so we just have to normalize it:
For the following parts, a matrix or linear transformation is described. Explain whether or not it is invertible.
The linear transformation diagonalizes as
It is not, since 0 is an eigenvalue of (why?) based on its diagonalization.
The linear transformation scales volumes of hypercubes by a factor of .
The determinant of a transformation is the quantity which expresses how it scales the volumes of hypercubes. So, , which is not zero, so is invertible.
The rank of the linear transformation is 5.
For a matrix to be invertible, it must be full rank. However, , so is not full rank, and so not invertible.
The linear transformation is surjective.
A surjective, square, linear transformation is invertible. So is.
Find the component of the velocity vector,
in the direction as the vector,
The component of a vector in the direction of is the scaling multiple of in the orthogonal decomposition of :
Find the least squares solution to the equation,
The least squares solution to an equation is the solution to . We have that is,
and is,
Solving the equation
yields the least squares solution,