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Chapter 2: Q2.8-32Q (page 93)

In Exercises 31-36, respond as comprehensively as possible, and justify your answer.

32. If R is a \(6 \times 6\) matrix and Nul R is not the zero subspace, what can you say about Col R?

Short Answer

Expert verified

Col \(R\) is a subspace of \({\mathbb{R}^6}\), whereas Col \(R \ne {\mathbb{R}^6}\).

Step by step solution

01

Condition for the column space

Thecolumn spaceof matrix A is the set Col A of all linear combinationsof the columns of A.

Col Aequals \({\mathbb{R}^m}\) only when the columns of A span \({\mathbb{R}^m}\). Otherwise, Col A is only a part of \({\mathbb{R}^m}\).

02

Explanation for the column space of matrix R

When Nul \(R\) has nonzero vectors, equation \(Rx = 0\) has nontrivial solutions. According to the Invertible Matrix theorem, \(R\) is not invertible and the columns of \(R\) do not span \({\mathbb{R}^6}\) because \(R\) is a square matrix. Therefore, Col \(R\) is a subspace of \({\mathbb{R}^6}\), but Col \(R \ne {\mathbb{R}^6}\).

Thus, Col \(R\) is a subspace of \({\mathbb{R}^6}\), whereas Col \(R \ne {\mathbb{R}^6}\).

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Most popular questions from this chapter

[M] For block operations, it may be necessary to access or enter submatrices of a large matrix. Describe the functions or commands of your matrix program that accomplish the following tasks. Suppose A is a \(20 \times 30\) matrix.

  1. Display the submatrix of Afrom rows 15 to 20 and columns 5 to 10.
  2. Insert a \(5 \times 10\) matrix B into A, beginning at row 10 and column 20.
  3. Create a \(50 \times 50\) matrix of the form \(B = \left[ {\begin{array}{*{20}{c}}A&0\\0&{{A^T}}\end{array}} \right]\).

[Note: It may not be necessary to specify the zero blocks in B.]

In exercise 11 and 12, mark each statement True or False.Justify each answer.

a. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{\bf{1}}}}&{{A_{\bf{2}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_{\bf{1}}}}&{{B_{\bf{2}}}}\end{array}} \right]\), with \({A_{\bf{1}}}\) and \({A_{\bf{2}}}\) the same sizes as \({B_{\bf{1}}}\) and \({B_{\bf{2}}}\), respectively then \(A + B = \left[ {\begin{array}{*{20}{c}}{{A_1} + {B_1}}&{{A_{\bf{2}}} + {B_{\bf{2}}}}\end{array}} \right]\).

b. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_1}}\\{{B_{\bf{2}}}}\end{array}} \right]\), then the partitions of \(A\) and \(B\) are comfortable for block multiplication.

Let \(A = \left( {\begin{aligned}{*{20}{c}}3&{ - 6}\\{ - 1}&2\end{aligned}} \right)\). Construct a \({\bf{2}} \times {\bf{2}}\) matrix Bsuch that ABis the zero matrix. Use two different nonzero columns for B.

1. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{6}}\\{\bf{5}}&{\bf{4}}\end{aligned}} \right)\).

Suppose the transfer function W(s) in Exercise 19 is invertible for some s. It can be showed that the inverse transfer function \(W{\left( s \right)^{ - {\bf{1}}}}\), which transforms outputs into inputs, is the Schur complement of \(A - BC - s{I_n}\) for the matrix below. Find the Sachur complement. See Exercise 15.

\(\left[ {\begin{array}{*{20}{c}}{A - BC - s{I_n}}&B\\{ - C}&{{I_m}}\end{array}} \right]\)
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