Write vector X as a linear combinatio of vectors A and B in R2. (1) X = (0, 1), A = (1, 1), B = (0, 1);
(2) X = (2, 1), A = (1, −1), B = (1, 1);
(3) X = (1, 1), A = (2, 1), B = (−1, 0).
Exercise 1, answer
(1) X = B
(2) (2, 1) = a(1, −1) + b(1, 1) = (a + b, −a + b), whence
a + b = 2
−a + b = 1
It follows that 2b = 3, that is b = 32 and consequently a = 12:
X = 1 2A +3
2B
(3) (1, 1) = a(2, 1) + b(−1, 0) = (2a − b, a), whence a = 1 and consequently b = 1:
X = A + B
Find the coordinates of vector X with respect to vectors A, B, C in R3. (1) X = (1, 0, 0), A = (1, 1, 1), B = (−1, 1, 0), C = (1, 0, −1);
(2) X = (1, 1, 1), A = (0, 1, −1), B = (1, 1, 0), C = (1, 0, 2);
(3) X = (0, 0, 1), A = (1, 1, 1), B = (−1, 1, 0), C = (1, 0, −1).
Exercise 2, answer
(1) (1, 0, 0) = a(1, 1, 1)+b(−1, 1, 0)+c(1, 0, −1) = (a−b+c, a+b, a−c), whence
a − b + c = 1 a + b = 0 a − c = 0
From the second and third equations, it follows that c = −b = a, so that from the first equation one has a + a + a = 1, that is a = 13. This implies b = −13, c = 13.
The coordinates of X with respect to A, B, C are then 13, −13,13.
(2) (1, 1, 1) = a(0, 1, −1)+b(1, 1, 0)+c(1, 0, 2) = (b +c, a +b, −a +2c), whence
b + c = 1 a + b = 1
−a + 2c = 1
From the second equation, b = 1 − a, so that the first equation yields c = 1 − b = 1 − 1 + a = a. Substituting this value in the last equation, one has −a + 2a = a = 1, hence b = 0, c = 1.
The coordinates of X with respect to A, B, C are then (1, 0, 1).
Exercise 2, answer (cont.)
(3) (0, 0, 1) = a(1, 1, 1)+b(−1, 1, 0)+c(1, 0, −1) = (a−b+c, a+b, a−c), whence
a − b + c = 0 a + b = 0 a − c = 1
From the second equation b = −a, while the third yields c = a − 1.
Substituting this values in the first equation, a + a + a − 1 = 0, so that a = 13, and consequently b = −13, c = −23.
The coordinates of X with respect to A, B, C are then 13, −13, −23.
Exercise 3
Let V be a K-vector space and let u, v , w ∈ V be vectors. Prove or disprove the following assertion: Let u, v , w a be pairwise linearly independent (that is, u, v are linearly independent, u, w are linearly independent, v , w are linearly independent), then u, v , w are linearly independent as well.
u = (1K, 0K), v = (0K, 1K), w = (1K, 1K).
Then u, v , w are pairwise linearly independent, but the triple u, v , w is linearly dependent, as u + v − w = (0K, 0K).
Consequently, the assertion is false.
Exercise 3
Let V be a K-vector space and let u, v , w ∈ V be vectors. Prove or disprove the following assertion: Let u, v , w a be pairwise linearly independent (that is, u, v are linearly independent, u, w are linearly independent, v , w are linearly independent), then u, v , w are linearly independent as well.
Answer. The fact that u, v , w are pairwise linearly independent means that they are pairwise non-parallel; this does not rule out the possibility that u, v , w are coplanar, that is, linearly dependent.
For example, let V = K2, and set
u = (1K, 0K), v = (0K, 1K), w = (1K, 1K).
Then u, v , w are pairwise linearly independent, but the triple u, v , w is linearly dependent, as u + v − w = (0K, 0K).
Consequently, the assertion is false.
Given the matrices A and B, compute A + B, AB, and BA.
A =
1 2 1
2 0 0
1 0 3
, B =
2 2 1
2 5 2
1 2 4
.
Exercise 4, answer
A + B =
1 + 2 2 + 2 1 + 1 2 + 2 0 + 5 0 + 2 1 + 1 0 + 2 3 + 4
=
3 4 2
4 5 2
2 2 7
AB =
1 · 2 + 2 · 2 + 1 · 1 1 · 2 + 2 · 5 + 1 · 2 1 · 1 + 2 · 2 + 1 · 4 2 · 2 + 0 · 2 + 0 · 1 2 · 2 + 0 · 5 + 0 · 2 2 · 1 + 0 · 2 + 0 · 4 1 · 2 + 0 · 2 + 3 · 1 1 · 2 + 0 · 5 + 3 · 2 1 · 1 + 0 · 2 + 3 · 4
=
=
7 14 9
4 4 2
5 8 13
BA =
2 · 1 + 2 · 2 + 1 · 1 2 · 2 + 2 · 0 + 1 · 0 2 · 1 + 2 · 0 + 1 · 3 2 · 1 + 5 · 2 + 2 · 1 2 · 2 + 5 · 0 + 2 · 0 2 · 1 + 5 · 0 + 2 · 3 1 · 1 + 2 · 2 + 4 · 1 1 · 2 + 2 · 0 + 4 · 0 1 · 1 + 2 · 0 + 4 · 3
=
=
7 4 5
14 4 8
9 2 13
