Worked examples · Sheet 2

Studies of linear dependence and independence

Theory used: Linearly independent vectors Dependence and linear combinations Independence and rank Computations in coordinates

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Determine whether the vectors $v_{1} = (1, 2, 1)$ , $v_{2} = (0, 1, 3)$ , $v_{3} = (2, 3, - 1)$ of $R^{3}$ are linearly independent; if not, write a dependence relation.

a (1, 2, 1) + b (0, 1, 3) + c (2, 3, - 1) = (0, 0, 0) ⟺ ⎩ ⎨ ⎧ a + 2 c = 0 2 a + b + 3 c = 0 a + 3 b - c = 0

a = - 2 c ⟹ - 4 c + b + 3 c = 0, b = c

a + 3 b - c = - 2 c + 3 c - c = 0 for every c

(a, b, c) = c (- 2, 1, 1) ⟹ - 2 v_{1} + v_{2} + v_{3} = 0, v_{3} = 2 v_{1} - v_{2}

By the definition of linear independence we write the zero combination and solve the homogeneous system in the unknowns $a$ , $b$ , $c$ . The first equation gives $a = - 2 c$ ; substituting into the second we obtain $b = c$ . With these two expressions the third equation becomes $0 = 0$ , so it is a consequence of the others and $c$ remains free. Since the system has nonzero solutions, the vectors are linearly dependent. With $c = 1$ we obtain the relation $- 2 v_{1} + v_{2} + v_{3} = 0$ , from which $v_{3} = 2 v_{1} - v_{2}$ ; the direct check gives $2 (1, 2, 1) - (0, 1, 3) = (2, 3, - 1)$ .

Exercise 2

Determine whether the vectors $v_{1} = (1, 0, 1, 2)$ , $v_{2} = (0, 1, 1, - 1)$ , $v_{3} = (1, 1, 0, 1)$ of $R^{4}$ are linearly independent.

A = 1012 011 - 1 1101 R_{3} \to R_{3} - R_{1} R_{4} \to R_{4} - 2 R_{1} 1000 011 - 1 11 - 1 - 1

R_{3} \to R_{3} - R_{2} R_{4} \to R_{4} + R_{2} 10000100 11 - 2 0 rk A = 3

There are three vectors in $R^{4}$ , so the matrix $A$ that has them as columns is $4 \times 3$ and the determinant is not defined. We use the theorem on independence and rank: the vectors are independent if and only if $rk A = 3$ . After row reduction three nonzero rows remain, with pivots in columns $1$ , $2$ and $3$ , so $rk A = 3$ and the vectors are linearly independent.

Exercise 3

Determine whether the polynomials $p_{1} = 1 + x^{2}$ , $p_{2} = x - x^{2}$ , $p_{3} = 1 + x + 2 x^{2}$ of $R_{2} [x]$ are linearly independent.

p_{1} \mapsto (1, 0, 1), p_{2} \mapsto (0, 1, - 1), p_{3} \mapsto (1, 1, 2)

101 01 - 1 112 = 1 \cdot (1 \cdot 2 - 1 \cdot (- 1)) - 0 + 1 \cdot (0 \cdot (- 1) - 1 \cdot 1)

= 3 - 1 = 2 \neq = 0

We pass to coordinates with respect to the standard basis $(1, x, x^{2})$ : the polynomial $a + b x + c x^{2}$ corresponds to the vector $(a, b, c)$ . By the remark on computations in coordinates, the polynomials are independent if and only if their coordinate vectors are. There are three vectors in $R^{3}$ , so we compute the determinant of the matrix that has them as columns, expanded along the first row. Since the determinant equals $2 \neq = 0$ , the coordinate vectors are independent, therefore $p_{1}$ , $p_{2}$ , $p_{3}$ are linearly independent and form a basis of $R_{2} [x]$ .

Exercise 4

As $k \in R$ varies, determine for which values the vectors

v_{1} = (1, 1, k), v_{2} = (1, k, 1), v_{3} = (k, 1, 1)

are linearly dependent, and for those values write a dependence relation.

11 k 1 k 1 k 11 R_{2} \to R_{2} - R_{1} R_{3} \to R_{3} - k R_{1} 100 1 k - 1 1 - k k 1 - k 1 - k^{2}

det A = (k - 1) (1 - k^{2}) - (1 - k)^{2} = (1 - k) [(k - 1) (1 + k) - (1 - k)] = (1 - k) (k^{2} + k - 2) = - (k - 1)^{2} (k + 2)

det A = 0 ⟺ k = 1 or k = - 2

k = 1 : v_{1} = v_{2} = v_{3} = (1, 1, 1), v_{1} - v_{2} = 0

k = - 2 : (1, 1, - 2) + (1, - 2, 1) + (- 2, 1, 1) = (0, 0, 0), v_{1} + v_{2} + v_{3} = 0

There are three vectors in $R^{3}$ , so by the theorem on independence and rank they are dependent if and only if the determinant of the matrix $A$ that has them as columns is zero. The operations $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - k R_{1}$ do not change the determinant, and the resulting matrix is expanded along the first column. Factoring out $1 - k$ and factoring $k^{2} + k - 2 = (k - 1) (k + 2)$ we obtain $det A = - (k - 1)^{2} (k + 2)$ , which vanishes for $k = 1$ and for $k = - 2$ .

For $k \neq = 1$ and $k \neq = - 2$ the vectors are independent. For $k = 1$ the three vectors coincide, so they are dependent and span only the line $⟨(1, 1, 1)⟩$ . For $k = - 2$ the sum of the three vectors is zero, so they are dependent, with relation $v_{1} + v_{2} + v_{3} = 0$ ; since $v_{1}$ and $v_{2}$ are not proportional, in this case they span a plane.