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

17. \(A = \left( {\begin{array}{*{20}{c}}1&{ - 1}&0\\0&4&7\\0&0&5\end{array}} \right)\)

Short Answer

Expert verified

The dimension of \({\mathop{\rm Col}\nolimits} A\) is 3, 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 \({\mathop{\rm Col}\nolimits} A\)

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

02

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

The given matrix A is in the echelon form. As it has three pivot columns, the dimension of \({\mathop{\rm Col}\nolimits} A\) is 3. 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 3 and that of \({\mathop{\rm Nul}\nolimits} A\) is 0.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(A\) be an \(m \times n\) matrix of rank \(r > 0\) and let \(U\) be an echelon form of \(A\). Explain why there exists an invertible matrix \(E\) such that \(A = EU\), and use this factorization to write \(A\) as the sum of \(r\) rank 1 matrices. [Hint: See Theorem 10 in Section 2.4.]

Let S be a maximal linearly independent subset of a vector space V. In other words, S has the property that if a vector not in S is adjoined to S, the new set will no longer be linearly independent. Prove that S must be a basis of V. [Hint: What if S were linearly independent but not a basis of V?]

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

\({\bf{1}} - {\bf{2}}{t^{\bf{2}}} - {t^{\bf{3}}}\), \(t + {\bf{2}}{t^{\bf{3}}}\), \({\bf{1}} + t - {\bf{2}}{t^{\bf{2}}}\)

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 \(T\).
See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free