§1.8:
x1
−3
7
(20.) Let ~x =
, ~v1 =
, and ~v2 =
, and let T : R2 → R2 be a linear transformation
x2
5
−2
that maps ~x into x1~v1 + x2~v2 . Find a matrix A such that T (~x) = A~x for each ~x ∈ R2 .
SOL’N: Without using the theorem from the next section we can actually see exactly what the
matrix will be by what is given. We are given that T is defined by the vector equation
−3
7
T (~x) = x1~v1 + x2~v2 = x1
+ x2
.
5
−2
But we know that a vector equation like this can be rewritten as a matrix equation A~x where the
columns of A are just ~v1 and ~v2 , i.e.
−3 7
A = [~v1 ~v2 ] =
.
5 −2
(24.) An affine transformation T : Rn → Rm has the form T (~x) = A~x + ~b, with A an m × n matrix and
~b ∈ Rm . Show that T is not a linear transformation when ~b 6= ~0.
SOL’N: Remember that T being linear means that it satisfies two properties, namely
T (~u + ~v ) = T (~u) + T (~v ) and T (c~u) = cT (~u) for all ~u, ~v ∈ Rn and c ∈ R. So to show that an affine
transformation is not linear we just have to show that one of these properties does not hold.
So let T : Rn → Rm be defined by T (~x) = A~x + ~b with ~b 6= ~0. Then by the properties of the
product of a matrix and a vector we have
T (c~u) = A(c~u) + ~b = cA(~u) + ~b
whereas
cT (~u) = c(A(~u) + ~b) = cA(~u) + c~b.
So we see that T (c~u) 6= cT (~u) if ~b 6= ~0.
(30.) Suppose vectors ~v1 , . . . , ~vp span Rn , and let T : Rn → Rm be a linear transformation. Suppose
T (~vi ) = ~0 for i = 1, . . . , p. Show that T is the zero transformation. That is, show that if ~x is any
vector in Rn , then T (~x) = ~0.
SOL’N: Again we are going to use the properties of linearity for this problem. Let ~x ∈ Rn . Since
the vectors ~v1 , . . . , ~vp span Rn we know that there exist weights c1 , . . . , cp ∈ R such that
~x = c1~v1 + · · · + cp~vp .
Therefore, by the linearity of the transformation T we have
T (~x) = T (c1~v1 + · · · + cp~vp )
= T (c1~v1 ) + · · · + T (cp~vp )
= c1 T (~v1 ) + · · · + cp T (~vp )
= c1~0 + · · · + cp~0
= ~0
by assumption.
§1.9:
1
−2
3
, T (~e2 ) =
, and T (~e3 ) =
.
4
9
−8
Determine if the transformation is (a) one-to-one and (b) onto. Justify your answer.
(26.) Let T : R3 → R2 be defined so that T (~e1 ) =
SOL’N: By Thm 10 from §1.9 we know that this transformation is a matrix transformation with
standard matrix
1 −2 3
A=
.
4 9 −8
Straight away, we know that we should turn this matrix into reduced echelon form (you should
check my arithmetic here).
11
1 −2 3
1 0
17
∼ ··· ∼
4 9 −8
0 1 − 20
17
We also have from Thm 11 from this section that T is one − to − one if and only if the equation
T (~x) = A~x = ~0 has the trivial solution. We know now that this happens if and only if there are
no free variables. Well, we see from the reduced echelon form of A that there is a free variable so
T is NOT one-to-one.
For determining whether or not T maps onto R2 we know that this happens if and only if the
colums of A span R2 which in turn happens if and only if there is a pivot in every row. Oh, there
is a pivot in every row so we know that the linear transformation T must be onto.
(32.) Let T : Rn → Rm be a linear transformation, with A its standard matrix. Complete the following
statement to make it true: “T maps Rn onto Rm if and only if A has
pivot columns.”
Find some theorems that explain why the statement is true.
SOL’N: We actually talked about this one in class because I forgot that I assigned it for your
homework. We know the statement is true by filling in m into the blank. This is true because we
know, again, that T is onto if and only if the columns span Rm which happens if and only if there
is a pivot in every row. As there are m rows, there must be m pivot columns for the statement to
be true.
(34.) Let S : Rp → Rn and T : Rn → Rm be linear transformations. Show that the mapping
~x 7→ T (S(~x)) is a linear transformation (from Rp to Rm ).
SOL’N: From the yellow-ish box on pg. 66 in the book and the discussion thereafter, we know
that a transformation T is linear if and only if
T (c~u + d~v ) = cT (~u) + dT (~v ) ∀ vectors ~u, ~v and scalars c, d.
So let ~u, ~v ∈ Rp and c, d ∈ R. Then
T (S(c~u + d~v )) = T (cS(~u) + dS(~v ))
| {z }
|
{z
}
∈Rp
∈Rn
by the linearity of S. Note at this point that both cS(~u) and dS(~v ) are vectors living in Rn so
that it makes sense to evaluate T , whose domain is Rn , at their sum. Furthermore, because we
also know T is linear we get
T (cS(~u) + dS(~v )) = cT (S(~u)) + dT (S(~v ))
{z
}
|
∈Rm
which is exactly what we wanted to show to show that the composition T ◦ S is linear.