7.1.a Suppose that f and g are in the set. Then f (0) = f (1) and g(0) = g(1),
therefore f (0) + g(0) = f (1) + g(1), so f + g is also in the set.
If c is an arbitrary scalar, then cf (0) = cf (1), hence cf is also in the set.
The set is non-empty, since, for example, the constant zero function is in the
set.
Consequently, the set is a subspace
7.1.b No, since if we multiply an increasing function by −1, we get a decreasing function.
7.2. Reducing the matrix


1 −1 1
 2
1 7 
2
2 10
to the row echelon form, we get





1 −1 1
1 −1 1
1
 0
3 5 → 0
3 5 → 0
0
4 8
0
1 3
0


−1 1
1
1 3 → 0
3 5
0

−1
1
1
3 ,
0 −4
we see that the vectors are independent, hence the dimension is 3. (The row
operations also show that the determinant of the matrix is non-zero.)
7.3.




1 2 2
3 1
4 2
1 2 2 3 1 4
2
 1 2 3 2 3 5
1 4 →
6  →  0 0 1 −1 2
0 0 1 −1 2 −3 4
3 6 7 8 5 9 10




1 2 2
3 1 4 2
1 2 2
3 1
4 2
 0 0 1 −1 2
1 4  →  0 0 1 −1 2 1 4  →
0 0 0
0 0 1 0
0 0 0
0 0 −4 0




1 2 0
5 −3 0 −6
1 2 2
3 1 0 2
 0 0 1 −1 2 0 4  →  0 0 1 −1
2 0
4 .
0 0 0
0 0 1 0
0 0 0
0
0 1
0
We see that the first, third, and sixth columns form a basis, hence a basis
of the columns space is
     
2
4 
 1
 1 , 3 , 5  .


3
7
9
A basis of the row space is
1 2 0 5 −3 0 −6 ,
0
0
1
1
−1
2
0
4
,
0
0
0
0
0
1
0
.










The solution of the homogeneous system is









6
3
−5
−2
−2x2 − 5x4 + 3x5 + 6x7
 0 
 0 
 0 
 1 

x2









 −4 
 −2 
 1 
 0 

x4 − 2x5 − 4x7









 = x2  0 +x4  1 +x5  0 +x7  0  ,
x4









 0 
 1 
 0 
 0 

x5









 0 
 0 
 0 
 0 

0
1
0
0
0
x7
therefore a basis of the nullspaces is

 
−5
−2



 1   0

 



 0   1

 

 0 , 1
 


 0   0


 




 0   0



0
0
 
 
 
 
 
,
 
 
 
 
3
0
−2
0
1
0
0
 
 
 
 
 
,
 
 
 
 
6
0
−4
0
0
0
1




















7.4. a) For any two 4 × 4 matrices A, B, we have L(A + B) = (A + B) +
2(A + B)> = A + B + 2A> + 2B > = A + 2A> + B + 2B > = L(A) + L(B). For
every scalar c we have L(cA) = (cA) + 2(cA)> = cA + c2A> = cL(A), hence L
is a linear transformation (operator).
b) L(cA) = (cA)2 = c2 A2 6= cA2 = cL(A), hence L is not a linear operator.
7.5. It is


1 2 0
 1 1 1 .
2 2 2
It is just the coefficients in the formula.
7.6. Let us diagonalize the matrix. Its characteristic polynomial is
1−x
1
1 −1
−1 − x
−1 = (1−x)(−1−x)(1−x)−1−1+1+x+1−x+1−x =
1
1
1−x x2 − x3 = x2 (1 − x),
hence the eigenvalues are 0 (double) and 1.
For λ = 0 the eigenvectors are found solving the system


1
1
1
 −1 −1 −1  → 1 1 1 ,
1
1
1
solution:






−x2 − x3
−1
−1

 = x2  1  + x3  0  ,
x2
x3
0
1
2

 

−1 
 −1
hence a basis of the eigenspace is  1  ,  0  .


0
1
For λ = 1 we have




0
1
1
1
1
0
1 1 0
1 0 −1
 −1 −2 −1  →  0
1
1 →
→
,
0 1 1
0 1
1
1
1
0
0 −1 −1


1
and an eigenvector is  −1 .
1




−1 −1
1
0 0 0
0 −1  and D =  0 0 0 .
We get A = SDS −1 for S =  1
0
1
1
0 0 1
Let us find S −1 :

−1
 1
0

−1
1 1
0 −1 0
1
1 0
1 1
 0 1
0 1

−1
0
1

1
 0
0
−1
−1
0
1
1
0


0
−1 −1 1 1 0
0  →  0 −1 0 1 1
1
0
1 1 0 0


1 1 −1 −1
0 0
−1 0  →  0 1
0 −1
0 1
0 0
1
1


0
1 0 0
0
1 1
0 −1 −1 0  →  0 1 0
1
1
1 1
0 0 1

0
1
0
1
2 1
so S −1 =  −1 −1 0 .
1
1 1
It follows that


−1 −1
1
0 −1  
eA = SeD S −1 =  1
0
1
1


−1 −1
1
1
2
 1
0 −1   −1 −1
0
1
1
e
e
e0
0
0

0
0 →
1

0 0
−1 0  →
1 1
1
−1
1

2 1
−1 0  ,
1 1


1
2 1
0
0   −1 −1 0  =
1
1 1
e1
 

1
e
e−1 e−1
0  =  1 − e 2 − e 1 − e .
e
e−1 e−1
e
0 0
7.7. The eigenvalues of A are 0, 1, hence A = S
S −1 for some S.
0 1
Then
√
√
√
0 √0
0 0
−1
−1
A = S DS = S
S =S
S −1 = A.
0 1
0
1
3
0
e0
0