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Chapter 4: Q16Q (page 191)

Determine the dimensions of \({\mathop{\rm Nul}\nolimits} A\) and \({\mathop{\rm Col}\nolimits} A\) for the matrices shown in Exercise 13-18

16. \(A = \left( {\begin{array}{*{20}{c}}3&4\\{ - 6}&{10}\end{array}} \right)\)

Short Answer

Expert verified

The dimension of \({\mathop{\rm Col}\nolimits} A\) is 2, and that of \({\mathop{\rm Nul}\nolimits} A\) is 0.

Step by step solution

01

State the condition for the dimensions of \({\mathop{\rm Nul}\nolimits} A\) and \({\mathop{\rm Col}\nolimits} A\)

Thedimension of \({\mathop{\rm Nul}\nolimits} A\) is thenumber of free variables in the equation \(A{\mathop{\rm x}\nolimits} = 0\), and the dimension of \({\mathop{\rm Col}\nolimits} A\)is thenumber of pivot columnsin \(A\).

02

Apply the row operation

Perform an elementary row operation to produce the row-reduced echelon form of the matrix.

At row 1, multiply row 1 by \(\frac{1}{3}\).

\(\left( {\begin{array}{*{20}{c}}1&{1.333}\\{ - 6}&{10}\end{array}} \right)\)

At row 2, multiply row 1 by 6 and add it to row 2.

\(\left( {\begin{array}{*{20}{c}}1&{1.333}\\0&{18}\end{array}} \right)\)

At row 2, multiply row 2 by \(\frac{1}{{18}}\).

\(\left( {\begin{array}{*{20}{c}}1&{1.333}\\0&1\end{array}} \right)\)

At row 1, multiply row 2 by \(1.3333\) and subtract it from row 1.

\(\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\)

03

Determine the dimensions of \({\mathop{\rm Nul}\nolimits} A\) and \({\mathop{\rm Col}\nolimits} A\)

\({\mathop{\rm Col}\nolimits} A\)

Matrix A has two pivot columns, so the dimension of \({\mathop{\rm Col}\nolimits} A\) is 2. The equation \(A{\mathop{\rm x}\nolimits} = 0\) has only a trivial solution since there are no columns without pivots.

Therefore, \({\mathop{\rm Nul}\nolimits} A = \left\{ 0 \right\}\) and the dimension of \({\mathop{\rm Nul}\nolimits} A\) is 0.

Thus, the dimension of \({\mathop{\rm Col}\nolimits} A\) is 2, and that of \({\mathop{\rm Nul}\nolimits} A\) is 0.

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Most popular questions from this chapter

Let \(B = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{4}}}\end{array}} \right),\,\left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{9}}\end{array}} \right)\,} \right\}\). Since the coordinate mapping determined by B is a linear transformation from \({\mathbb{R}^{\bf{2}}}\) into \({\mathbb{R}^{\bf{2}}}\), this mapping must be implemented by some \({\bf{2}} \times {\bf{2}}\) matrix A. Find it. (Hint: Multiplication by A should transform a vector x into its coordinate vector \({\left( {\bf{x}} \right)_B}\)).

Let \({M_{2 \times 2}}\) be the vector space of all \(2 \times 2\) matrices, and define \(T:{M_{2 \times 2}} \to {M_{2 \times 2}}\) by \(T\left( A \right) = A + {A^T}\), where \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\).

  1. Show that \(T\)is a linear transformation.
  2. Let \(B\) be any element of \({M_{2 \times 2}}\) such that \({B^T} = B\). Find an \(A\) in \({M_{2 \times 2}}\) such that \(T\left( A \right) = B\).
  3. Show that the range of \(T\) is the set of \(B\) in \({M_{2 \times 2}}\) with the property that \({B^T} = B\).
  4. Describe the kernel of \(T\).

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

15. Let \(A\) be an \(m \times n\) matrix, and let \(B\) be a \(n \times p\) matrix such that \(AB = 0\). Show that \({\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B \le n\). (Hint: One of the four subspaces \({\mathop{\rm Nul}\nolimits} A\), \({\mathop{\rm Col}\nolimits} A,\,{\mathop{\rm Nul}\nolimits} B\), and \({\mathop{\rm Col}\nolimits} B\) is contained in one of the other three subspaces.)

Use coordinate vector to test whether the following sets of poynomial span \({{\bf{P}}_{\bf{2}}}\). Justify your conclusions.

a. \({\bf{1}} - {\bf{3}}t + {\bf{5}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{5}}t - {\bf{7}}{t^{\bf{2}}}\), \( - {\bf{4}} + {\bf{5}}t - {\bf{6}}{t^{\bf{2}}}\), \({\bf{1}} - {t^{\bf{2}}}\)

b. \({\bf{5}}t + {t^{\bf{2}}}\), \({\bf{1}} - {\bf{8}}t - {\bf{2}}{t^{\bf{2}}}\), \( - {\bf{3}} + {\bf{4}}t + {\bf{2}}{t^{\bf{2}}}\), \({\bf{2}} - {\bf{3}}t\)

Consider the following two systems of equations:

\(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 1\\4{x_1} + {x_2} - 6{x_3} = 9\end{array}\) \(\begin{array}{c}5{x_1} + {x_2} - 3{x_3} = 0\\ - 9{x_1} + 2{x_2} + 5{x_3} = 5\\4{x_1} + {x_2} - 6{x_3} = 45\end{array}\)

It can be shown that the first system of a solution. Use this fact and the theory from this section to explain why the second system must also have a solution. (Make no row operations.)
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