Math 2270, Fall 2015
Instructor: Thomas Goller
22 September 2015
Quiz #3 Solutions


1
0
2
(1) Suppose T : R ! R is linear and that T (~e1 ) =
, T (~e2 ) =
, and T (~e3 ) =
.
1
4
2
Compute the matrix A such that T (~x) = A~x. (2 points)
3

2
Solution:
A=

1 0 2
1 4 2
2
3
1 2
(2) Let A = 4 4 75 and let T : R2 ! R3 be the linear function defined by T (~x) = A~x.
1 2
Determine whether T is (a) one-to-one and (b) onto. Justify each answer. (3 points)
Possible solution: First, we row-reduce A to echelon form to find the pivot positions:
2
3
2
3
2
3
1 2
1 2
1 2
4 4 75 4R1 ! 40
15
15
! 40
1 2 +R1
0 4 +4R2
0 0
(a) T is one-to-one because every column of A contains a pivot position.
(b) T is not onto because the third row of the echelon form of A does not have a pivot
position.
Comments:
• Intuitively, it is easy to see that T is one-to-one because the columns of A are
independent and that T is not onto because the two columns of A have no chance
of spanning R3 (no two vectors can span a three-dimensional space!). But finding
the pivot positions is the best way (so far) of making this intuition rigorous.
• Don’t mix up one-to-one and onto by trying to memorize which one has a pivot
in each column and which has a pivot in each row! Know the definitions of oneto-one and onto in terms of T and try to develop an intuition as in the previous
comment. You should understand why looking for pivots works. (Recall that the
pivot positions pick out independent columns: if the set of all columns of A is
independent, then T is one-to-one; if every row of A has a pivot, then A has enough
independent columns to span the codomain, so T is onto.)
• Expect questions about the di↵erent ways of seeing one-to-one and onto on Test
#2!
Math 2270, Fall 2015
Instructor: Thomas Goller
2
3
0 
2 4 0
35
using any method. (2 points)
7 0 1
2
3
(3) Compute the matrix product 4 2
1
Solution:
22 September 2015
2
6 12
4 17
8
12
4
3
0
35
2
(4) Bonus problem: Let T : R3 ! R3 be a linear transformation defined by T (~x) = A~x for a
3 ⇥ 3 matrix A. One possible reduced echelon form of A is
2
3
1 ⇤ 0
40 0 15 T is not one-to-one; T is not onto.
0 0 0
Compute the other seven possible reduced echelon forms of A (using ⇤ to denote entries
that could be any real number). In each case, state whether T is one-to-one and whether
T is onto. (1 bonus point)
Solution:
2
0 0
40 0
0 0
The other seven possible reduced echelon forms of A are
3 2
3 2
3 2
3 2
3 2
3 2
3
0
0 0 1
0 1 ⇤
1 ⇤ ⇤
0 1 0
1 0 ⇤
1 0 0
05 , 40 0 05 , 40 0 05 , 40 0 05 , 40 0 15 , 40 1 ⇤5 , 40 1 05
0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 1
T is one-to-one and onto only in the last case and is not one-to-one and not onto in the
first six cases.
Comments: One-to-one and onto are equivalent in the special case when the domain
and codomain of T are the same (when the matrix for T is square). We will talk about
this in Sections 2.2 and 2.3.