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

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        <section>
            <title>Vector Spaces</title>

            <p>It is easy in Sage to make a reasonable facsimile of <m>\mathbf{C}^n</m>.  We just restrict our attention to rational entries rather than complex entries.  This vector space contains vectors with 4 slots, each filled with a rational number.</p>

            <sage>
                <input>
                V = QQ^4
                V
                </input>
            </sage>

            <exercise>
                <statement>
                    <p>We can test membership using the word/command <c>in</c>. Try vectors with different numers of slots, and perhaps include the complex number <c>2 + 3*I</c> as an entry.</p>
                </statement>
            </exercise>

            <sage />

        </section>

        <section>
            <title>Vector Form of Solutions to Homogeneous Systems</title>

            <p>These are the coefficient matrix and vector of constants from yesterday's big system that led to a colored matrix in reduced row-echelon form.</p>

            <p>The <c>.right_kernel()</c> method will give the vectors of the vector form of the solutions to a homogeneous system when used with the <c>basis='pivot'</c> option.</p>

            <sage>
                <input>
                A = matrix(QQ, [[  1,  2, 12,   1,  13,  5,  2],
                                [ -2, -3, -21,  0, -13,  2, -5],
                                [  1,  3,  15,  4,  28, 25,  0],
                                [ -2, -3, -21, -1, -15, -6, -3],
                                [  1,  1,   9,  1,   4,  9,  1]])
                b = vector(QQ, [8, -15, 7, -10, 3])
                </input>
            </sage>

            <sage>
                <input>A.right_kernel(basis='pivot')</input>
            </sage>

            <p>Rows of the <q>basis matrix</q> are vectors in yesterday's linear combination (with scalars <m>x_3</m>, <m>x_5</m>, <m>x_6</m>).  This is a spanning set for the null space of the matrix <m>A</m>.  See Theorem VFSLS and Theorem SSNS.</p>

            <p>Theorem PSPHS can explain how to use a single solution to the non-homogeneous system and the spanning set of the null space of the coefficient matrix to arrive at all solutions to the system.  Here is a single solution to the system.</p>

            <sage>
                <input>A.solve_right(b)</input>
            </sage>

            <p>Notice that this vector is the solution when we set each free variable to zero, which is the <q>other</q> vector from yesterday that is not part of the linear combination.</p>
        </section>

        <section>
            <title>Spanning Sets</title>

            <p>Example ABS from FCLA.</p>

            <sage>
                <input>
                x1 = vector(QQ,[1,1,3,1])
                x2 = vector(QQ,[2,1,2,-1])
                x3 = vector(QQ,[7,3,5,-5])
                x4 = vector(QQ,[1,1,-1,2])
                x5 = vector(QQ,[-1,0,9,0])
                W = span([x1, x2, x3, x4, x5])
                W
                </input>
            </sage>

            <exercise>
                <statement>
                    <p>Make a <q>random</q> linear combination of the five vectors and test for membership (which will be trivially true, repeatedly).  Remember to use the <c>*</c> operator for vector scalar multiplication.</p>
                </statement>
            </exercise>

            <sage />

            <p>But not any old vector is in <m>W</m>.</p>

            <sage>
                <input>
                v = vector(QQ, [1, 1, -3, 2])
                v in W
                </input>
            </sage>

            <p>It should make sense that arbitrary linear combinations are in the span.  How did we manufacture a vector <em>not</em> in the span?  Stay tuned.</p>
        </section>

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