به انگلیسی

Same map, new coordinates

Mathematics · Linear Algebra · ages 19-20

Name ______________________ Date ____________

In basis B = {(2, 0), (0, 3)}, what is the first B-coordinate of (8, 6)?
Answer: ______________
In basis B = {(1, 1), (1, minus 1)}, the vector (3, 1) has coordinates (a, b). What is a?
Answer: ______________
What are the columns of the transition matrix that takes C-coordinates to B-coordinates?
- The B basis vectors written in C-coordinates
- The C basis vectors written in B-coordinates
- The standard basis vectors in order
A vector has B-coordinates (4, 1) with B = {(1, 1), (1, minus 1)}. What is its first standard coordinate?
Answer: ______________
A map has matrix [[1, 4], [2, 3]] in the standard basis. What is its determinant in any other basis?
Answer: ______________
With A = [[1, 1], [0, 2]] and P = [[1, 2], [0, 1]], what is the (1, 2) entry of P inverse A P?
Answer: ______________
Changing the basis changes the determinant of the matrix of a fixed map. True or false?
Circle one: True False
B = {(1, 0), (1, 1)}. Which vector has B-coordinates (2, 3)?
- (2, 3)
- (3, 5)
- (5, 3)
- (5, 5)
A map has eigenvalues 2 and 5 in one basis. What are they in a new basis?
- still 2 and 5
- doubled to 4 and 10
- both become 7
With A = [[2, 0], [0, 1]] and P = [[1, 1], [0, 1]], what is the (1, 2) entry of P inverse A P?
Answer: ______________

Answer key

For grown-ups. Fold this page away before handing over the rest.

4 · (8, 6) = 4(2, 0) + 2(0, 3), so the B-coordinates are (4, 2).
2 · a + b = 3 and a minus b = 1 give a = 2, b = 1.
The C basis vectors written in B-coordinates · Each column says where one old basis vector lands in the new coordinate language.
5 · 4 times (1, 1) plus 1 times (1, minus 1) is (5, 3), so the first standard coordinate is 5.
-5 · The determinant is 1 times 3 minus 4 times 2, which is -5, and a basis change keeps it.
-1 · P inverse is [[1, minus 2], [0, 1]]. P inverse A P works out to [[1, minus 1], [0, 2]], so the (1, 2) entry is minus 1.
False · False: the determinant belongs to the map, so every matrix of the same map shares it.
(5, 3) · 2(1, 0) + 3(1, 1) = (5, 3).
still 2 and 5 · Eigenvalues belong to the map, so a new basis keeps them at 2 and 5.
1 · P inverse is [[1, -1], [0, 1]]. P inverse A = [[2, -1], [0, 1]], and multiplying by P gives [[2, 1], [0, 1]].