page no. 138 Suppose we want to determine whether or not the follinsang matricies in V = M_2.3 are

Example 4.18. Suppose we want to determine whether or not the follinsang matricies in M are linearly dependent A = [[1, 2, - 3], [4, 0, 1]] 8 = [[1, 3, - 4], [6, 5, 4]] C = [[3, 8, - 11], [16, 10, g]]

The coordinate vectors of the matrices in the usual basis of M 2, j at as follows:

[A] = \{1.2, - 3.4, 0.1\} [B] = [1.3, - 4.6 * 0.5 * 0.4] \ C\ =\ 3.8, - 11.16 10.9

Form the matrix M whose rows are the above coordinate vectors and reduce M to an echelon form

Since the echelon matrix has only two nonzero rows, the coordinate vectors [A], [B] ,[C span a subspace of dimension 2 and so are linearly dependent. Accordingly, the original matrices A, B, C are linearly dependent.

Solved Problems

VECTOR SPACES, LINEAR COMBINATIONS

4.1. Suppose iu and ir belong to a vector space V. Simplify each of the following expressions

(a) E_{1} = 3(2u - 4v) + 5u + 7v_{v}

(c) E₁ = 2+3/2+4)

(b) E_{2} = 3u - 6(3u - 5v) + 7u

(d) E_{4} = 5u - 3/4 + 5u

Multiply out and collect terms:

(a) E.= 6u - 12v + 5u + 7v = 11u - 5v

(b) E3u-18+300+ 7-8 +30

(c) E is not defined since the product we of vectors is not defined.

(d) E_{4} is not defined since division by a vector is not defined.

4.2. Prove Theorem 4.1 Let V be a vector space over a field K (i ) k * 0 = 0 (ii) 0u = 0 (iii) If ku = 0 nk = 0oru = 0 (iv ) (- k) * u = k(- u) = - k * u

(i) By Axiom [A] with u = 0 we have y + 0 = 0 Hence, by Axiom [M₁), we have k * 0 = k(0 + 0) =k0+k0 Adding -40 to both sides gives the desired result.

(ii) For scalars 0 + 0 = 0 Hence, by Axiom [M2], we have 0u = (0 + 0) * u = 0u + 0u Adding -Ou to both sides gives the desired result

(iii) Suppose ku r = 0 and k ne0 Then there exists a scalar k ^ - 1 such that k ^ - 1 * k = 1 Thus u = 1u = (k ^ - 1 * k) * u = k ^ - 1 * (ku) = k ^ - 1 * 0 = 0

(iv) Using u (- u) = 0 and k + (- k) =0 yields 0 = k*beta = k[u + (- u)] = ku + k(- u) and 0 = 0u = [k + (- k)] * u = ku + (- k) * u

Adding -ku to both sides of the first equation gives-ku= k(-u), and adding ku to both sides second equation gives - k * u = (- k) * u Thus (- k) * u =k( u)= - k

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