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Elementary Linear Algebra: An Active Learning Approach

Worksheet 6.5.1 Worksheet
A4US

  1. Suppose we wish to construct a matrix \(A\) whose column space and nullspace both contain \(\left(\begin{array}{r}-3\\4\\-1 \end{array} \right)\text{.}\)
    1. For \(\left(\begin{array}{r}-3\\4\\-1 \end{array} \right)\in N(A)\text{,}\) how many columns must \(A\) have?
    2. For \(\left(\begin{array}{r}-3\\4\\-1 \end{array} \right)\in\) col\((A)\text{,}\) how many rows must \(A\) have?
    3. Thus: \(A\in\underline{\hspace{1in}}\)
    4. Dimension check: Determine the possibilities for dim\((N(A))\) and dim(col\((A)\)) and make sure they do not violate Theorem 6.1.40.
    5. Install \(\left(\begin{array}{r}-3\\4\\-1 \end{array} \right)\) as the first column of \(A\) and write the other two columns with \(a,b,c\text{,}\) etc. At this point
      \begin{align} A=\qquad\amp \nonumber\tag{6.5.1}\\ \amp { }^{ } { }^{ }\tag{6.5.2}\\ \amp { }^{ }\nonumber\tag{6.5.3}\\ \amp \underline{\hspace{3in}}\nonumber\tag{6.5.4} \end{align}
    6. You may have some freedom; use it. Remember a linear equation in two unknowns has a lot of solutions. Now use the fact that \(\left(\begin{array}{r}-3\\4\\-1 \end{array} \right)\in N(A)\) to solve for the remaining unknowns in (6.5.2).
      \begin{align*} A=\qquad\amp\\ \amp\\ \amp\\ \amp \underline{\hspace{3in}}\nonumber \end{align*}
  2. Suppose we want to construct a matrix \(A\) whose column space and nullspace are both spanned by \(\left(\begin{array}{r}-3\\4\\-1 \end{array} \right)\text{.}\)
    1. For \(\left(\begin{array}{r}-3\\4\\-1 \end{array} \right)\in N(A)\text{,}\) how many columns must \(A\) have?
    2. For \(\left(\begin{array}{r}-3\\4\\-1 \end{array} \right)\in\) col\((A)\text{,}\) how many rows must \(A\) have?
    3. What do you conclude? \(A\in\underline{\hspace{1in}}\)
    4. Since col \((A)\) is spanned by \(\left(\begin{array}{r}-3\\4\\-1 \end{array} \right)\text{,}\) what is dim(col\((A))\text{?}\)
    5. Since \(N(A)\) is spanned by \(\left(\begin{array}{r}-3\\4\\-1 \end{array} \right)\text{,}\) what is dim\((N(A))\text{?}\)
    6. What does Theorem 6.1.40 (dim\((N(A))+\) dim(col\((A))=n\)) have to say about this?
    7. What do you conclude?
  3. Suppose we want to construct a matrix \(A\) whose column space and nullspace both contain \(\left(\begin{array}{r}2\\1\\-3 \end{array} \right)\) and \(\left(\begin{array}{r}4\\-4\\8 \end{array} \right)\text{.}\)
    1. For \(\left(\begin{array}{r}2\\1\\-3 \end{array} \right)\in N(A)\text{,}\) how many columns must \(A\) have?
    2. For \(\left(\begin{array}{r}2\\1\\-3 \end{array} \right)\in\) col\((A)\text{,}\) how many rows must \(A\) have?
    3. What do you conclude? \(A\in\underline{\hspace{1in}}\)
    4. Since col \((A)\) contains \(\left(\begin{array}{r}2\\1\\-3 \end{array} \right)\) and \(\left(\begin{array}{r}4\\-4\\8 \end{array} \right)\text{,}\) what is dim(col\((A))\text{?}\)
    5. Since \(N(A)\) contains \(\left(\begin{array}{r}2\\1\\-3 \end{array} \right)\) and \(\left(\begin{array}{r}4\\-4\\8 \end{array} \right)\text{,}\) what is dim\((N(A))\text{?}\)
    6. What does Theorem 6.1.40 have to say about this?
    7. What do you conclude?
    8. Would the outcome be different if instead we want to construct a matrix \(A\) whose column space and nullspace both contain \(\left(\begin{array}{r}2\\1\\-3 \end{array} \right)\) and \(\left(\begin{array}{r}-4\\-2\\6 \end{array} \right)\text{.}\) Why or why not?
  4. Suppose we want to construct a matrix \(A\) whose column space contains \(\left(\begin{array}{r}-2\\3\\-7 \end{array} \right)\) and whose nullspace consists of all multiples of (i.e. is spanned by) \(\left(\begin{array}{r}1\\-3\\3\\6 \end{array} \right)\text{.}\)
    1. How many columns must \(A\) have?
    2. How many rows must \(A\) have?
    3. Thus: \(A\in\underline{\hspace{1in}}\)
    4. Dimension check:
    5. Installing \(\left(\begin{array}{r}-2\\3\\-7 \end{array} \right)\) as the first column of \(A\) and writing the other columns with \(a,b,c\text{,}\) etc., leads to solving a \(3\times 3\) system. We can be more efficient: we are free to choose two of the columns at will, as long as the three columns we specify are linearly independent, and let the nullspace relation find us the fourth column. So, install \(\left(\begin{array}{r}-2\\3\\-7 \end{array} \right)\) as the first column of \(A\) and choose any two linearly independent columns you like for \(A_{*2}\) and \(A_{*3}\text{.}\) At this point
      \begin{align} A=\qquad\amp \nonumber\tag{6.5.5}\\ \amp { }^{ } { }^{ }\tag{6.5.6}\\ \amp { }^{ }\nonumber\tag{6.5.7}\\ \amp \underline{\hspace{3in}}\nonumber\tag{6.5.8} \end{align}
    6. Now use the fact that \(\left(\begin{array}{r}1\\-3\\3\\6 \end{array} \right)\in N(A)\) to solve for the unknowns in (6.5.6).
      \begin{align*} A=\qquad\amp\\ \amp\\ \amp\\ \amp \underline{\hspace{3in}}\nonumber \end{align*}
  5. Suppose we want to construct a matrix \(A\) whose column space contains \(\left(\begin{array}{r}-2\\3\\-7 \end{array} \right)\) and \(\left(\begin{array}{r}-4\\1\\6 \end{array} \right)\) and whose nullspace is spanned by \(\left(\begin{array}{r}1\\-2\\6 \end{array} \right)\text{.}\)
    1. What size matrix must \(A\) be, i.e., \(A\in\underline{\hspace{1in}}\)
    2. Do a dimension check. Is this possible? If so, find such an \(A\text{.}\)
  6. If possible, construct a matrix \(A\) whose nullspace is spanned by \(\left(\begin{array}{r} 1\\0\\2\\-3 \end{array} \right)\text{.}\) If this is not possible, clearly explain why. (Hint: See Remark 6.4.12 and Example 6.4.13.)
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