B.xml

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    <article xml:id="B">
        <title>Sage and Linear Algebra Worksheet</title>
        <subtitle>FCLA Section B</subtitle>

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        <frontmatter>
            <xi:include href="../header.xml" />
        </frontmatter>

        <section>
            <title>Bases</title>

            <p>Five <q>random</q> vectors, each with 4 entries, collected into a set <c>S</c>.</p>

            <sage><input>
            v1 = vector(QQ, [-4, -2,  3, -11])
            v2 = vector(QQ, [-2,  7,  3,   9])
            v3 = vector(QQ, [ 6, -4, -7,   5])
            v4 = vector(QQ, [-1,  0,  3,  -4])
            v5 = vector(QQ, [-4,  5, -5,  11])
            S = [v1, v2, v3, v4, v5]
            </input></sage>

            <p>Consider the subspace spanned by these five vectors.  We will make these vectors the <em>rows</em> of a matrix and row-reduce to see a basis for the space (subspace, or row space, take your pick).  This is an application of Theorem BRS.</p>

            <sage><input>
            A = matrix(S)
            A
            </input></sage>

            <sage><input>A.rref()</input></sage>

            <p>Sage does this semi-automatically, tossing zero rows for us.</p>

            <sage><input>
            W = span(S)
            B = W.basis()
            B
            </input></sage>

            <exercise>
                <statement>
                    <p>Construct a <em>random</em> vector, <c>w</c>, in this subspace by choosing scalars for a linear combination of the vectors we used to build <c>W</c> as a span originally.</p>

                    <p>Then use the three <em>basis</em> vectors in <c>B</c> to recreate the vector <c>w</c>.  Question: how many ways can you do this?  By Theorem VRRB there should always be exactly one way to create <c>w</c> using a linear combination of a basis of <c>W</c>.</p>
                </statement>
            </exercise>

            <sage><input>
            w = *v1 + *v2 + *v3 + *v4 + *v5
            w
            </input></sage>

            <sage><input>w in W</input></sage>

            <sage><input>*B[0] + *B[1] + *B[2]</input></sage>
        </section>

        <section>
            <title>Nonsingular Matrices</title>

            <p>We will obtain a basis of <m>\mathbb{C}^{10}</m> from the columns of a <m>10\times 10</m> nonsingular matrix.</p>

            <sage><input>
            entries = [[ 1,  1,  1, -1, -2,  4,  2, -3,  1, -6],
                       [-2, -1, -2,  2,  4, -7, -4,  5, -1,  7],
                       [ 1, -1,  2, -2, -5,  8,  5, -3,  4, -4],
                       [-1, -2,  0,  1,  0, -5,  0, -3, -5,  6],
                       [ 0, -2,  1, -1, -2,  3,  2,  3,  3,  7],
                       [ 1,  0,  1, -1, -2,  4,  2,  0,  2,  0],
                       [-1,  0, -1,  1,  3, -1, -2,  7,  5,  1],
                       [ 1,  1,  1, -1, -2,  8,  3,  2,  8, -6],
                       [ 0,  2, -1,  1,  2, -1, -2,  2,  2, -6],
                       [ 1,  3,  0,  0,  1,  3,  0,  0,  3, -8]]
            M = matrix(QQ, entries)
            M
            </input></sage>

            <sage><input>
            not M.is_singular()
            </input></sage>

            <p>A totally random vector with 10 entries:</p>

            <sage><input>
            v = random_vector(ZZ, 10, x=-9, y=9)
            v
            </input></sage>

            <exercise>
                <statement>
                    <p>By Theorem CNMB, the columns of the matrix are a basis of <m>\mathbb{C}^{10}</m>.  So the vector <c>v</c> should be a linear combination of the columns of the matrix.  Verify this fact in three ways.<ol>
                        <li>First, the old-fashioned way, thus exposing Theorem NMUS.</li>

                        <li>Then, the modern way, with an inverse, since a nonsingular matrix is invertible, thus exposing Theorem SNCM.</li>

                        <li>Finally, the Sage way, as described below.</li>
                    </ol></p>
                </statement>
            </exercise>

            <sage><input>
            aug = M.augment(v)
            aug.rref()
            </input></sage>

            <sage><input>M.inverse()*v</input></sage>

            <p>The Sage way: first create a space with a <term>user basis</term>.</p>

            <sage><input>
            X = (QQ^10).subspace_with_basis(M.columns())
            X
            </input></sage>

            <p>Sage still carries an <term>echelonized basis</term>, in addition to the <term>user-installed</term> basis.</p>

            <sage><input>
            X.basis()
            </input></sage>

            <sage><input>
            X.echelonized_basis()
            </input></sage>

            <p>Now ask for a coordinatization, relative to the basis in <c>X</c>, thus exposing Theorem VRRB.</p>

            <sage><input>
            X.coordinates(v)
            </input></sage>
        </section>

        <section>
            <title>Orthonormal Bases</title>

            <p>A particularly simple orthonormal basis of <m>\mathbb{C}^3</m>, collected into the set <c>S</c>.</p>

            <sage><input>
            v1 = vector(QQ, [1/3, 2/3, 2/3])
            v2 = vector(QQ, [2/3, -2/3, 1/3])
            v3 = vector(QQ, [2/3, 1/3, -2/3])
            S = [v1, v2, v3]
            </input></sage>

            <exercise>
                <statement>
                    <p>If these vectors are an orthonormal basis, then as the columns of a matrix they should create an orthonormal basis.</p>
                </statement>
            </exercise>

            <sage><input>
            Q = column_matrix(S)
            Q
            </input></sage>

            <sage><input>Q.conjugate_transpose()*Q</input></sage>

            <sage><input>Q.is_unitary()</input></sage>

            <exercise>
                <statement>
                    <p>Build a random vector of size <m>3</m> and find our ways to express the vector as a (unique) linear combination of the basis vectors.  Which method is most efficient?</p>
                </statement>
            </exercise>

            <p>A totally random vector with 3 entries.</p>

            <sage><input>
            v = random_vector(ZZ, 3, x=-9, y=9)
            v
            </input></sage>

            <p>First, the old-fashioned way, thus exposing Theorem NMUS.</p>

            <sage><input>
            aug = Q.augment(v)
            aug.rref()
            </input></sage>

            <p>Now, the modern way, with an inverse, since a nonsingular matrix is invertible,  thus exposing Theorem SNCM.</p>

            <sage><input>Q.inverse()*v</input></sage>

            <p>The Sage way. Create a space with a <q>user basis</q> and ask for a coordinatization, thus exposing Theorem VRRB.</p>

            <sage><input>
            X = (QQ^3).subspace_with_basis(Q.columns())
            X.coordinates(v)
            </input></sage>

            <p>Finally, exploiting the orthonormal basis, and computing scalars for the linear combination with an inner product, thus exposing Theorem COB.  (Sage's <c>.inner_product()</c> does not conjugate the entries of either vector, so we use the more careful <c>.hermitian_inner_product()</c> vector method instead.)</p>

            <sage><input>
            a1 = v1.hermitian_inner_product(v)
            a2 = v2.hermitian_inner_product(v)
            a3 = v3.hermitian_inner_product(v)
            a1, a2, a3
            </input></sage>
        </section>

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    </article>
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