Notes on Homework 7
1.
(a) Two things to show:



y1
x1
additivity Suppose ~x =  x2  , ~y =  y2  ∈ R3 . Then
y3
x3

f (~x + ~y) =
=
=
=
   
y1
x1
f ( x2  +  y2 )
y3
x3


x1 + y1
f ( x2 + y2 )
x3 + y3
2( x1 + y1 ) + ( x2 + y2 )
4( x3 + y3 )
2x1 + x2
2y1 + y2
+
4x3
4y3
= f (~x) + f (~y).
In fact, this isn’t the way one would probably derive this solution; in practice, the way
to solve this part is probably to compute f (~x + ~y) and f (~x) + f (~y) separately, and then
observe that they’re the same vector.
homogeneity Suppose ~x ∈ R3 , λ ∈ R. Then


λx1
f (λ~x) = f (λx2 )
λx3
2(λx1 + λx2 )
=
4λx3
2x1 + x2
=λ
4x3
= λ f (~x).
See remarks above.
(b) Fix a matrix B. Two things to show:
additivity Suppose A, C ∈ Matn,n (F). Note that B is a fixed matrix throughout this problem, so we’re not allowed to modify it. Then
g( A) + g(C ) = AB − BA + CB − BC
Professor Jeff Achter
Colorado State University
M369 Linear Algebra
Fall 2008
while
g( A + C ) = ( A + C ) B − B( A + C )
= AB + CB − BA − BC start using matrix properties
= AB − BA + CB − BC
= g( A) + g(C ) from the first line.
homogeneity Suppose A ∈ Matn,n (F) and λ ∈ F. Using properties of matrix multiplication from about a month ago, we have
g(λA) = λAB − BλA
= λAB − λBA
= λ ( AB − BA)
= λg( A)
(c) Same story:
additivity Suppose p( z) = az2 + bz + c, q( z) = dz2 + ez + f , p( z), q( z) ∈ P2 (R)[ z].
Then
h( p( z) + q( z)) = h( az2 + bz + c + dz2 + ez + f )
= h(( a + d) z2 + (b + e) z + (c + f ))


( a + d) + (b + e)
= (b + e) − (c + f )
2(c + f )
while

 

a+b
d+e
h( p( z)) + h(q( z)) =  b − c  + e − f 
2c
2f


( a + d) + (b + e)
= (b + e) − (c + f ) .
2(c + f )
Professor Jeff Achter
Colorado State University
M369 Linear Algebra
Fall 2008
homogeneity Suppose λ ∈ R, p( z) = az2 + bz + c ∈ P2 (R)[ z]. Then
h(λp( z)) = h(λ ( az2 + bz + c))
= h(λaz2 + λbz + λc)


λa + λb
=  λb − λc 
2λc


a+b
= λ b − c
2c
= λh( p( z)).
2.
(a) By definition, we have
x1
1 3
x
T(
)=
· 1
x2
2 4
x2
x1 + 3x2
=
.
2x1 + 4x2
(b) Our formula yields
3+3·1
2·3+4
=
6
,
10
while the definition yields
3
1 3
3
T(
)=
·
1
2 4
1
6
=
.
10
3.
(a) Note that
 


x1
x1 − x2
f ( x2 ) =  x1 + 4x3 
x3
− x1
 
 
 
1
−1
0
= x1  1  + x2  0  + x3 4 .
−1
0
0
So in fact, we can compute f (~x) via

 
1 −1 0
x1
0 4  x2  .
f (~x) =  1
−1 0 0
x3
Professor Jeff Achter
Colorado State University
M369 Linear Algebra
Fall 2008
(b) Compute in two different ways; using the given formula,
 


1
1−3
T ( 3 ) = 1 + 4(−1)
−1
−1
 
−2
= −3
−1
while multiplying by our matrix yields

   
1 −1 0
1
−2
 1
0 4  3  = −3 .
−1 0 0
−1
−1
4.
(a) The reduced row echelon form of A is
1 0 − 17
2
9
0 1
2
− 19
2
11
2
.
(Computation omitted here.) Therefore, the most general solution to the equation Ax =
0 is

 17
19
2 x3 + 2 x4
− 9 x3 − 11 x4 

 2
2


x3
x4
and the null space of A is
17   19 
2
2
− 9  − 11 



2
2 ).
span(
,

null( A) =
 1   0 
0
1
These two vectors are linearly independent check this, and dim ker( A) = 2. A twodimensional subspace maps to zero, and f is not injective.
Remark: Actually, you can figure out that f is not injective without computing anything! Since
dim R4 = 4 > dim R2 = 2, there is no injective linear transfromation R4 → R2 .


1 0 0
(b) Since the reduced row echelon form of B is 0 1 0 (calculation omitted), for any
0 0 1
3
3
b ∈ R there is a unique x ∈ R such that Ax = b. In particular, the equation Ax = 0
has a unique solution, and g is injective.
Professor Jeff Achter
Colorado State University
M369 Linear Algebra
Fall 2008
 
b1
5. The key observation for this problem is that for any vector b = b2 , the equation Ax = b
b3
is consistent. Moreover, each variable is a pivot variable, so there is exactly one solution to
Ax = b for any particular b. (Explicitly, the solution is x3 = b3 ; x2 = b2 − 2x3 = b2 − 2b3 ,
and x1 = b1 − 2x2 − 3x3 = b1 − 2b2 + b3 .)
(a) Since Ax = b always has a solution, each element b ∈ R3 is in the column space of A,
and the associated linear transformation T is surjective.
(b) Since T is surjective, we know that T is bijective if and only if it is injective, too. From
class, we know that T is injective if and only if the nullspace of A is trivial, that is, if
and only if the solution to T (~x) = ~0 is the zero vector.
This last condition is true; we can read this off from the echelon form, as discussed
above. Alternatively, since
dim im( T ) + dim ker( T ) = dim(source) = 3,
and since dim im( T ) = 3, the kernel has dimension zero, and thus T is injective.
Professor Jeff Achter
Colorado State University
M369 Linear Algebra
Fall 2008