NM.xml

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

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


        <p>First, a guaranteed nonsingular <m>5\times 5</m> matrix, created at random.</p>

        <sage>
            <input>
            A = random_matrix(QQ, 5, algorithm="unimodular", upper_bound=20)
            A
            </input>
        </sage>

        <exercise>
            <statement>
                <p>Augment with the zero vector, using the matrix method <c>.augment()</c> and the vector constructor <c>zero_vector(QQ, 5)</c>.  Then row-reduce to use Definition NM.  Or instead, do not augment and apply Theorem NMRRI.</p>
            </statement>
        </exercise>

        <sage />

        <exercise>
            <statement>
                <p>Build some random vectors with <c>random_vector(QQ, 5)</c>, augment the matrix and row-reduce.  There will always be a unique solution to the linear system represented by the augmented matrix.  This is Theorem NMUS.</p>
            </statement>
        </exercise>

        <sage />

        <p>Instead<mdash />cheap, easy and powerful:</p>

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

        <p>Now, a carefully crafted singular matrix.</p>

        <sage>
            <input>
            B = matrix(QQ, [[ 7, -1, -12, 21, -8],
                            [-3,  3,   0, -9,  6],
                            [ 3,  3, -12,  9,  0],
                            [ 2,  3, -10,  6,  1],
                            [-2,  2,   0, -6,  4]])
            B
            </input>
        </sage>

        <exercise>
            <statement>
                <p>Augment with the zero vector and row-reduce (Definition NM), or don't augment and row-reduce (Theorem NMRRI).]</p>
            </statement>
        </exercise>

        <sage />

        <exercise>
            <statement>
                <p>A random vector of constants will only rarely build a consistent system when paired with <c>B</c>.  Try it.  But this is not a theorem, see the vector <c>c</c> below.</p>
            </statement>
        </exercise>

        <sage />

        <p>Instead<mdash />cheap, easy and powerful:</p>

        <sage>
            <input>B.is_singular()</input>
        </sage>

        <p>Two carefully crafted vectors for linear systems with <c>B</c> as coefficient matrix.</p>

        <sage>
            <input>
            c = vector(QQ, [17,-15,-3,-5,-10])
            d = vector(QQ, [-3,1,-2,1,2])
            </input>
        </sage>

        <exercise>
            <statement>
                <p>Which of these two column vectors will create a consistent system for this singular coefficient matrix? (Stay tuned.)</p>
            </statement>
        </exercise>

        <sage />

        <p>A null space is called a <term>right kernel</term> in Sage.  It's description contains a lot of things we do not understand yet.</p>

        <sage>
            <input>
            NS = B.right_kernel()
            NS
            </input>
        </sage>

        <exercise>
            <statement>
                <p>But we can test membership in the null space, which is the most basic property of a set.  Try <c>u in NS</c> and then repeat with <c>v</c>.</p>
            </statement>
        </exercise>

        <sage>
            <input>u = vector(QQ, [0,0,3,4,6])</input>
        </sage>

        <sage />

        <sage>
            <input>v = vector(QQ, [1,0,0,5,-2])</input>
        </sage>

        <sage />

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    </article>

</pretext>