MS-C1342 Linear algebra, V/2023 Noferini / Puska
Linear algebra
Exercise sheet 1 / Model solutions
1. Consider the vectors x = [2, 3, 4]T , y = [1, 0, 2]T and z = [0, 1, 0]T in R3 .
(a) Are the vectors x, y and z linearly dependent?
(b) Find a vector w ∈ R3 such that x, y and w are linearly independent.
(c) Find a vector v ∈ R3 such that x, z and v are linearly independent.
Solution.
(a) Yes, the three vectors are linearly dependent. To see this, let A = [y, z, x] and let us
find a nonzero coefficient vector α ∈ R3 such that Aα = 0. Gaussian elimination
yields that this linear system is equivalent to
(
α1 + 2α3 = 0
α2 + 3α3 = 0
which has solution α = [−2t, −3t, t]T where t ∈ R. So we can find a nonzero α and
the three vectors are linearly dependent.
2 1 w1
(b) One way to achieve this is to impose that the determinant of 3 0 w2 is nonzero.
4 2 w3
This yields 2w1 6= w3 , so for example we can pick w = [1, 0, 1]T .
(c) This part can be solved with the same method as part (b). We obtain the condition
v3 − 2v1 6= 0; one possible solution is v = [1, 0, 1]T .
2. Let x = [1, 2, 3]T . Represent x as a linear combination of the basis vectors
(a) e1 = [1, 0, 0]T , e2 = [0, 1, 0]T and e3 = [0, 0, 1]T (so-called cartesian basis vectors),
(b) q 1 = [1, 1, 0]T , q 2 = [1, 0, 1]T and q 3 = [1, 1, 1]T ,
(c) v 1 = [−1, 1, −1]T , v 2 = [1, 2, 2]T and v 3 = [1, −2, 1]T .
Solution.
(a) Clearly x = e1 + 2e2 + 3e3 .
(b) If Q = [q 1 q 2 q 3 ], we must solve the linear system Qα = x. Any method (for instance
Gaussian Elimination) yields as solution α = [−2, −1, 4]T , so x = −2q 1 − q 2 + 4q 3 .
(c) This question can be solved similarly to part (b). The solution is x = 4v 1 + 2v 2 + 3v 3 .
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MS-C1342 Linear algebra, V/2023 Noferini / Puska
3. Let
1 −1 0 1
A = 1 0 −1 and b = −3 .
0 −2 2 8
Compute a basis for the nullspace N (A) by hand and find all solutions to the equation
Ax = b.
Solution. The set N (A) is defined as
N (A) := {x ∈ R3 | Ax = 0}.
Hence, a basis for N (A) can be found by solving the linear system Ax = 0 by Gaussian
elimination:
1 −1 0 0 1 −1 0 0 1 −1 0 0
1 0 −1 0 → 0 1 −1 0 → 0 1 −1 0
0 −2 2 0 0 −2 2 0 0 0 0 0
Back substitution gives
1 1
x = t 1
and N (A) = span 1 .
1 1
Gaussian elimination is used to find the solutions to Ax = b:
1 −1 0 1 1 −1 0 1 1 −1 0 1
1 0 −1 −3 → 0 1 −1 −4 → 0 1 −1 −4 .
0 −2 2 8 0 −2 2 8 0 0 0 0
Back substitution gives
−3 1
x = −4 + t 1 .
0 1
4. Let 0 6= b, c ∈ Rn and define A = bcT ∈ Rn×n . Show that
(a) R(A) = span{b},
(b) N (A) = x ∈ Rn | cT x = 0 .
Solution. To prove that two sets V and W are equal, first pick v ∈ V and show that v ∈ W .
Then pick w ∈ W and show that w ∈ V .
(a) Let x ∈ R(A). Then there exists z ∈ Rn such that x = bcT z. Denote α = cT z so
that x = αb ∈ span{b}.
c T
Let x ∈ span{b}. Choose z = kck 2 , so that bc z = b. Hence x ∈ R(A).
2
(b) Let v ∈ N (A). This means that bcT v = 0. Because b 6= 0, we have that cT v = 0 and
v ∈ {x ∈ Rn | cT x = 0}.
Let v ∈ {x ∈ Rn | cT x = 0}. Then bcT v = 0. Hence, v ∈ N (A).
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