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Chapter 1: Q39E (page 1)

Let \(A\) be an \(m \times n\) matrix and let \({\mathop{\rm u}\nolimits} \) be a vector in \({\mathbb{R}^n}\) that satisfies the equation \(Ax = 0\). Show that for any scalar \(c\), the vector \(c{\mathop{\rm u}\nolimits} \) also satisfies \(Ax = 0\). (That is, show that \(A\left( {c{\mathop{\rm u}\nolimits} } \right) = 0\).)

Short Answer

Expert verified

The vector \(cu\) also satisfies that \(Ax = 0\), i.e., \(A\left( {cu} \right) = 0\) is

Step by step solution

01

Consider the given part

Let \(A\) be an \(m \times n\) matrix and let \(u\) be a vector in \({\mathbb{R}^n}\) that satisfies the equation \(Ax = 0\).

02

Show that vector \(cu\) satisfies \(Ax = 0\)

Theorem 5tells that \(A\) is an \(m \times n\) matrix; \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) are vectors in \({\mathbb{R}^n}\), and \(c\) is a scalar, then

a. \(A\left( {u + v} \right) = Au + Av\)

b. \(A\left( {cu} \right) = c\left( {Au} \right)\)

If \(u\) satisfies the equation \(Ax = 0\), then \(Au = 0\). For any scalar \(c\),

\(\begin{array}{c}A\left( {cu} \right) = c \cdot Au\\ = c \cdot 0\\ = 0.\end{array}\)

Thus, the vector \(cu\) also satisfies \(Ax = 0\), i.e., \(A\left( {cu} \right) = 0\) is proved.

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Most popular questions from this chapter

Let \(A = \left[ {\begin{array}{*{20}{c}}1&0&{ - 4}\\0&3&{ - 2}\\{ - 2}&6&3\end{array}} \right]\) and \(b = \left[ {\begin{array}{*{20}{c}}4\\1\\{ - 4}\end{array}} \right]\). Denote the columns of \(A\) by \({{\mathop{\rm a}\nolimits} _1},{a_2},{a_3}\) and let \(W = {\mathop{\rm Span}\nolimits} \left\{ {{a_1},{a_2},{a_3}} \right\}\).

  1. Is \(b\) in \(\left\{ {{a_1},{a_2},{a_3}} \right\}\)? How many vectors are in \(\left\{ {{a_1},{a_2},{a_3}} \right\}\)?
  2. Is \(b\) in \(W\)? How many vectors are in W.
  3. Show that \({a_1}\) is in W.[Hint: Row operations are unnecessary.]

In Exercises 9, write a vector equation that is equivalent to

the given system of equations.

9. \({x_2} + 5{x_3} = 0\)

\(\begin{array}{c}4{x_1} + 6{x_2} - {x_3} = 0\\ - {x_1} + 3{x_2} - 8{x_3} = 0\end{array}\)

In Exercises 6, write a system of equations that is equivalent to the given vector equation.

6. \({x_1}\left[ {\begin{array}{*{20}{c}}{ - 2}\\3\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}8\\5\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}1\\{ - 6}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\)

Question:Let A be the n x n matrix with 0's on the main diagonal, and 1's everywhere else. For an arbitrary vector b→in □n, solve the linear system Ax→=b⇀, expressing the components x1,.......,xnof x→in terms of the components of b⇀. See Exercise 69 for the case n=3 .

Determine the value(s) of \(a\) such that \(\left\{ {\left( {\begin{aligned}{*{20}{c}}1\\a\end{aligned}} \right),\left( {\begin{aligned}{*{20}{c}}a\\{a + 2}\end{aligned}} \right)} \right\}\) is linearly independent.
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