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Thread: MTH501 Linear Algebra Assignments No. 1 Solutions Fall 2014

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    MTH501 Linear Algebra Assignments No. 1 Solutions Fall 2014

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    MTH501 Linear Algebra Assignments No. 1 Discussions and Solutions Fall 2014 Close Date Dec 11, 2014

    Assignment 1
    Linear Algebra (MTH501)
    Lecture covered: 08-15
    Total Marks: 30
    Question 1:
    Determine the linear dependence and independence of the vectors:

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    3 + p
    2
    1 + p
    2

    and
    7
    1 + 2p
    2

    in R
    2 over the Öelds of :
    i) Real Numbers R
    ii) Rational Numbers Q:
    Question 2:
    Determine the Linearity of the transformation:T : C ! C deÖned by,
    T(x + iy) = iy
    ,where i =
    p
    1:
    Also Önd the Kernel of this transformation.
    Question:3
    For the system of equations:A
    !
    X =
    !
    b ;determine t such that the system has:
    i) no solution.
    ii) inÖnite many solutions
    iii) unique solution,
    where A =

    1 2
    t 2

    ;
    !
    b =

    1
    5

    MTH501LinearAlgebraAssignmentsNo.1.jpg

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    1 Linear combinations
    Definition 1.1. Let V be a vector space. A vector v ∈ V is a linear combination of vectors
    u1, u2, . . . , un if there exist a1, a2, . . . , an ∈ k such that
    v = a1u1 + a2u2 + · · · + anun. (1)
    Sometimes it is possible to express a vector as a linear combination of other vectors. It
    can be done by solving a corresponding linear system. We’ll demonstrate it in the following
    example.
    Example 1.2. Consider the space R
    2 — the space of all pairs of numbers. Let v = (8, 13),
    u1 = (1, 2), and u2 = (2, 3). Let’s express v as a linear combination of u1 and u2. To do this we
    have to find a and b such that v = au1+bu2, i.e. (8, 13) = a(1, 2)+b(2, 3) = (a·1+b·2, a·2+b·3).
    So, we get the following system:
    (
    1a + 2b = 8
    2a + 3b = 13
    We can simply solve this system: subtracting the first equation multiplied by 2 from the second
    one we get −b = −3, so b = 3, and so a = 8−2b = 2. So we see that (8, 13) = 2·(1, 2)+3·(2, 3).
    Example 1.3. Consider the space P(t) — space of all polynomials. Let v = 5t
    2 + 2t + 1,
    u1 = t
    2 + t, u2 = t + 1, u3 = t
    2 + 1. Let’s express v as a linear combination of u1, u2
    and u3. We should find a, b and c such that v = au1 + bu2 + cu3, i.e. 5t
    2 + 2t + 1 =
    a(t
    2 + t) + b(t + 1) + c(t
    2 + 1) = t
    2
    (a + c) + t(a + b) + (b + c). So, we get the following system:

    
    
    a + c = 5
    a + b = 2
    b + c = 1
    2 Linear dependence and independence
    Now we’ll study one of the most important concepts of linear algebra and the theory of vector
    spaces. This is a concept of linear dependence and independence.
    Definition 2.1. Let u1, u2, . . . , un be a system of vectors. A linear combination of them is
    called nontrivial if there exists a nonzero coefficient. If all coefficients are equal to 0, the
    linear combination is called trivial.
    Example 2.2. u1 + 0u2 + 0u3 −3u4 is nontrivial linear combination, and 0u1 + 0u2 + 0u3 + 0u4
    is a trivial linear combination.
    Definition 2.3. A system of vectors u1, u2, . . . , un is called linearly dependent if there exists
    a nontrivial linear combination of these vectors which is equal to zero. Otherwise the system is
    called linearly independent.
    Example 2.4. Consider a vector space R
    3
    . Let u1 = (3, −5, 0), u2 = (5, 0, 1), and u3 =
    (8, −5, 1). Then linear combination with coefficients 1,1, and -1 is nontrivial and equals to
    zero:
    1 · (3, 5, 0) + 1 · (5, 0, 1) + (−1) · (8, −5, 1) = (0, 0, 0).
    Example 2.5. Consider a vector space R
    2
    . Let u1 = (1, 1), and u2 = (0, 0). The linear
    combination with coefficients 0 and 1 is nontrivial, and equals to zero:
    0 · (1, 1) + 1 · (0, 0) = (0, 0)
    Moreover, if one of the vectors in the system equals to 0, then this system is linearly dependent,
    since we can make a coefficient before it equal to some nonzero number, and all other coefficients
    we can make equal to zero.

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