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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

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

17. A submatrix of a matrix A is any matrix that results from deleting some (or no) rows and/or columns of A. It can be shown that A has rank \(r\) if and only if A contains an invertible \(r \times r\) submatrix and no longer square submatrix is invertible. Demonstrate part of this statement by explaining (a) why an \(m \times n\) matrix A of rank \(r\) has an \(m \times r\) submatrix \({A_1}\) of rank \(r\), and (b) why \({A_1}\) has an invertible \(r \times r\) submatrix \({A_2}\).

The concept of rank plays an important role in the design of engineering control systems, such as the space shuttle system mentioned in this chapter’s introductory example. A state-space model of a control system includes a difference equation of the form

\({{\mathop{\rm x}\nolimits} _{k + 1}} = A{{\mathop{\rm x}\nolimits} _k} + B{{\mathop{\rm u}\nolimits} _k}\)for \(k = 0,1,....\) (1)

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(\left\{ {{{\mathop{\rm x}\nolimits} _k}} \right\}\) is a sequence of “state vectors” in \({\mathbb{R}^n}\) that describe the state of the system at discrete times, and \(\left\{ {{{\mathop{\rm u}\nolimits} _k}} \right\}\) is a control, or input, sequence. The pair \(\left( {A,B} \right)\) is said to be controllable if

\({\mathop{\rm rank}\nolimits} \left( {\begin{array}{*{20}{c}}B&{AB}&{{A^2}B}& \cdots &{{A^{n - 1}}B}\end{array}} \right) = n\) (2)

The matrix that appears in (2) is called the controllability matrix for the system. If \(\left( {A,B} \right)\) is controllable, then the system can be controlled, or driven from the state 0 to any specified state \({\mathop{\rm v}\nolimits} \) (in \({\mathbb{R}^n}\)) in at most \(n\) steps, simply by choosing an appropriate control sequence in \({\mathbb{R}^m}\). This fact is illustrated in Exercise 18 for \(n = 4\) and \(m = 2\). For a further discussion of controllability, see this text’s website (Case study for Chapter 4).

In Exercise 6, find the coordinate vector of x relative to the given basis \({\rm B} = \left\{ {{b_{\bf{1}}},...,{b_n}} \right\}\).

6. \({b_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right),{b_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{ - {\bf{6}}}\end{array}} \right),x = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{\bf{0}}\end{array}} \right)\)

In Exercises 27-30, use coordinate vectors to test the linear independence of the sets of polynomials. Explain your work.

Explain what is wrong with the following discussion: Let \({\bf{f}}\left( t \right) = {\bf{3}} + t\) and \({\bf{g}}\left( t \right) = {\bf{3}}t + {t^{\bf{2}}}\), and note that \({\bf{g}}\left( t \right) = t{\bf{f}}\left( t \right)\). Then, \(\left\{ {{\bf{f}},{\bf{g}}} \right\}\) is linearly dependent because g is a multiple of f.

What would you have to know about the solution set of a homogenous system of 18 linear equations 20 variables in order to understand that every associated nonhomogenous equation has a solution? Discuss.

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