Math 2270
Quiz 8
This is a take home quiz. You are allowed to use any book, computer program, or other reference material that
you would like, but you are not allowed to consult any person but the instructor about these problems. This
includes your classmates. You are on your honor as far as this is concerned, but if I become aware of anyone
violating these rules, that person will recieve a zero for the quiz.
Since you have significantly more time to work on this quiz than an in-class quiz, I expect a little more out of
your presentation. When writing up your solutions to these problems, please use complete sentences and correct
mathematical statements. For examples of what I’d like, see my solutions to the in-class quizzes. Write your
solutions on separate paper that has no notebook edges. If you require more than one page, use real (not oragami)
staples to attach them together.
To receive credit, this quiz must be in my hands before I leave class on Friday, March 7, 2008.
1. Let V be an m-dimensional subspace of Rn and let T : V → V be the linear transformation defined by
T (~x) = c~x for a fixed scalar c. Let B be any basis for V . Find the matrix of T with respect to B.
Let us assume that B = {~v1 , ~v2 , . . . , ~vm }. Recall that the formula for the matrix of a linear transformation with
respect to the basis B is
B = [[T (~v1 )]B · · · [T (~vm )]B ] .
This means that in order to calculate B, we just find T (~vi ) and then find its coordinates with respect to B to get
the ith column of B. We have that T (~vi ) = c~vi , so [T (~vi )]B = c~ei . This implies that


B =  c~e1
c~e2
· · · c~em  ,
so B = cIm .
2. Suppose that {~v1 , ~v2 , . . . , ~vn } is an orthonormal set in Rn . Let
A = [~v1~v2 · · · ~vn ].
Show that A is invertible.
Showing that A is invertible is equivalent to showing that its columns are linearly independent. This is easily
done (it was done in class) as follows: Suppose that c1 , . . . , cn are such that
c1~v1 + · · · + cn~vn = ~0.
If we take the dot product of both sides of this equation with ~vi for i = 1, . . . , n, then we get
ci~vi · ~vi = 0.
Since ~vi ·~vi = 1, it must be that ci = 0 for i = 1, . . . , n. This shows that the columns of A are linearly independent,
so A is invertible.