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Chapter 2: Q2.8-3Q (page 93)

Exercises 1-4 display sets in \({\mathbb{R}^2}\). Assume the sets include the bounding lines. In each case, give a specific reason why the set H is not a subspace of \({\mathbb{R}^2}\). (For instance, find two vectors in H whose sum is not in H, or find a vector in H with a scalar multiple that is not in H. Draw a picture.)

3.

Short Answer

Expert verified

Set H is closed neither under addition nor under multiplication. Thus, itis not a subspace of \({\mathbb{R}^2}\).

Step by step solution

01

State the condition for a subspace

A subspaceof \({\mathbb{R}^n}\) is any set H in \({\mathbb{R}^n}\) that has three properties:

  1. The zero vector is in H.
  2. For each u and v in H, the sum \[{\mathop{\rm u}\nolimits} + v\] is in H.
  3. For each u in H and each scalar c, the vector \[c{\mathop{\rm u}\nolimits} \] is in H.

Asubspace is closed under addition and scalar multiplication.

02

Draw the diagram with vectors in H

The scalar multiple of the vector and the sum of vectors u and v are not in H, as shown below.

03

Explain that set H  is not a subspace of \({\mathbb{R}^2}\)

Set H is closed under neither addition nor multiplication. The subset of points on the line \({x_2} = {x_1}\) is a subspace. Each counterexample includes at least one point that is not on the line.

Thus, set His not a subspace of \({\mathbb{R}^2}\).

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Most popular questions from this chapter

(M) Read the documentation for your matrix program, and write the commands that will produce the following matrices (without keying in each entry of the matrix).

  1. A \({\bf{5}} \times {\bf{6}}\) matrix of zeros
  2. A \({\bf{3}} \times {\bf{5}}\) matrix of ones
  3. The \({\bf{6}} \times {\bf{6}}\) identity matrix
  4. A \({\bf{5}} \times {\bf{5}}\) diagonal matrix, with diagonal entries 3, 5, 7, 2, 4

In Exercises 27 and 28, view vectors in \({\mathbb{R}^n}\) as \(n \times 1\) matrices. For \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^n}\), the matrix product \({{\mathop{\rm u}\nolimits} ^T}v\) is a \(1 \times 1\) matrix, called the scalar product, or inner product, of u and v. It is usually written as a single real number without brackets. The matrix product \({{\mathop{\rm uv}\nolimits} ^T}\) is an \(n \times n\) matrix, called the outer product of u and v. The products \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm uv}\nolimits} ^T}\) will appear later in the text.

27. Let \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\3\\{ - 4}\end{aligned}} \right)\) and \({\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}a\\b\\c\end{aligned}} \right)\). Compute \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \), \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \),\({{\mathop{\rm uv}\nolimits} ^T}\), and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\).

Let Abe an invertible \(n \times n\) matrix, and let \(B\) be an \(n \times p\) matrix. Explain why \({A^{ - 1}}B\) can be computed by row reduction: If\(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right) \sim ... \sim \left( {\begin{aligned}{*{20}{c}}I&X\end{aligned}} \right)\), then \(X = {A^{ - 1}}B\).

If Ais larger than \(2 \times 2\), then row reduction of \(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right)\) is much faster than computing both \({A^{ - 1}}\) and \({A^{ - 1}}B\).

Use partitioned matrices to prove by induction that the product of two lower triangular matrices is also lower triangular. [Hint: \(A\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrix \({A_1}\) can be written in the form below, where \[a\] is a scalar, v is in \({\mathbb{R}^k}\), and Ais a \(k \times k\) lower triangular matrix. See the study guide for help with induction.]

\({A_1} = \left[ {\begin{array}{*{20}{c}}a&{{0^T}}\\0&A\end{array}} \right]\).

Prove Theorem 2(b) and 2(c). Use the row-column rule. The \(\left( {i,j} \right)\)- entry in \(A\left( {B + C} \right)\) can be written as \({a_{i1}}\left( {{b_{1j}} + {c_{1j}}} \right) + ... + {a_{in}}\left( {{b_{nj}} + {c_{nj}}} \right)\) or \(\sum\limits_{k = 1}^n {{a_{ik}}\left( {{b_{kj}} + {c_{kj}}} \right)} \).
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