1 Q1
Q1a
The first line of the maple output computes the QR factorisation. The columns
of the matrix Q make the orthonormal basis for the subspace W:
u1 =
(
1√
7
,
2√
7
,
1√
7
,− 1√
7
)
and u2 =
(
3√
14
, 0,−
√
14
7
,
1√
14
)
.
Q1b
The matrix of the projection is QQT . The second line of the maple output com-
putes this matrix. The product QQTe1 computes the first column of this matrix
and the product QQTe2 computes the second column of this matrix. Hence,
the answer is the sum of the first and the second column of the matrix QQT :
PW
(
e1 + e2
)
=
(
15
14
,
6
7
, 0,
−3
14
)
Q1c
We know that PW + PW⊥ = I, where I is the identity map. Hence, we find
PW⊥
(
e1 + e2
)
=
(
e1 + e2
)− PW (e1 + e2)
⇒ PW⊥
(
e1 + e2
)
=
(
− 1
14
,
1
7
, 0,
3
14
)
.
1
2 Q2
Q2a
The orthogonal complement S⊥ is defined as follows
S⊥ =
{
x ∈ Rn : x · y = 0, ∀y ∈ S
}
.
Q2b
Since x ∈ S⊥, the vector x is perpendicular to every element of S. Since x ∈ S,
the vector x is perpendicular to itself:
x · x = 0.
Assume that x =
(
x1, x2, . . . , xn
)
. We know that x · x = x21 + x22 + . . . + x2n.
Hence, since xj ∈ R, j = 1, . . . , n,
x · x = 0 ⇒ x21 + x22 + . . .+ x2n = 0
⇒ x1 = x2 = . . . = xn = 0
⇒ x = 0.
QED
2
3 Q3
Q3a
Using co-factor expansion by the second column, we compute
T
(
2e1
)
= det
a1 2 b1a2 0 b2
a3 0 b3
 = −2 det(a2 b2
a3 b3
)
= −2a2b3 + 2b2a3.
Q3b
It means that
T
(
λx+ µy
)
= λT
(
x
)
+ µT
(
y
)
, ∀λ, µ ∈ C and ∀x,y ∈ C3.
3
4 Q4
Q4a
Once possible choice is as follows
A =
1 0 00 1 0
0 0 0
 , B =
0 0 00 0 0
0 0 1
 and A+B = I3
⇒ detA = 0, detB = 0 and det
(
A+B
)
= 1 6= 0 = detA+ detB
Q4b
The above examples shows that this map is NOT linear.
4