Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

  1. \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} ,{\rm{ }}{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} ,\,\,{\mathop{\rm and}\nolimits} \,\,\frac{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}\)

Short Answer

Expert verified

The values are \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} = 5\), \({\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} = 8\) and \(\frac{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }} = \frac{8}{5}\).

Step by step solution

01

Inner product

Consider \({\mathop{\rm u}\nolimits} ,v,\) and \({\mathop{\rm w}\nolimits} \) as the vectors in \({\mathbb{R}^n}\) and consider \(c\) as the scalar. Then;

  1. \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} = {\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} \)
  2. \(\left( {{\mathop{\rm u}\nolimits} + {\mathop{\rm v}\nolimits} } \right) \cdot {\mathop{\rm w}\nolimits} = {\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} + {\mathop{\rm v}\nolimits} \cdot {\mathop{\rm w}\nolimits} \)
  3. \(\left( {c{\mathop{\rm u}\nolimits} } \right) \cdot {\mathop{\rm v}\nolimits} = c\left( {{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} } \right) = {\mathop{\rm u}\nolimits} \cdot \left( {c{\mathop{\rm v}\nolimits} } \right)\)
  4. \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} \ge 0\)and \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} = 0\) if and only if \({\mathop{\rm u}\nolimits} = 0\).
02

Compute

\({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} ,{\rm{ }}{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} ,\,\,{\mathop{\rm and}\nolimits} \,\,\frac{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}\)

It is given that, \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right)\).

Compute \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} \) as shown below:

\(\begin{aligned}{c}{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} &= \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right)\\ &= {\left( { - 1} \right)^2} + {2^2}\\ &= 1 + 4\\ &= 5\end{aligned}\)

Compute \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} \) as shown below:

\(\begin{aligned}{c}{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} &= \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right)\\ &4\left( { - 1} \right) + 6\left( 2 \right)\\ &= - 4 + 12\\ &= 8\end{aligned}\)

Compute \(\frac{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}\) as shown below:

\(\frac{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }} = \frac{8}{5}\)

Thus, the values are \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} = 5,{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} = 8\) and \(\frac{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }} = \frac{8}{5}\).

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Most popular questions from this chapter

Suppose radioactive substance A and B have decay constants of \(.02\) and \(.07\), respectively. If a mixture of these two substances at a time \(t = 0\) contains \({M_A}\) grams of \(A\) and \({M_B}\) grams of \(B\), then a model for the total amount of mixture present at time \(t\) is

\(y = {M_A}{e^{ - .02t}} + {M_B}{e^{ - .07t}}\) (6)

Suppose the initial amounts \({M_A}\) and are unknown, but a scientist is able to measure the total amounts present at several times and records the following points \(\left( {{t_i},{y_i}} \right):\left( {10,21.34} \right),\left( {11,20.68} \right),\left( {12,20.05} \right),\left( {14,18.87} \right)\) and \(\left( {15,18.30} \right)\).

a.Describe a linear model that can be used to estimate \({M_A}\) and \({M_B}\).

b. Find the least-squares curved based on (6).

In Exercises 17 and 18, all vectors and subspaces are in \({\mathbb{R}^n}\). Mark each statement True or False. Justify each answer.

17. a.If \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\) is an orthogonal basis for\(W\), then multiplying

\({v_3}\)by a scalar \(c\) gives a new orthogonal basis \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},c{{\bf{v}}_3}} \right\}\).

b. The Gram–Schmidt process produces from a linearly independent

set \(\left\{ {{{\bf{x}}_1}, \ldots ,{{\bf{x}}_p}} \right\}\)an orthogonal set \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) with the property that for each \(k\), the vectors \({{\bf{v}}_1}, \ldots ,{{\bf{v}}_k}\) span the same subspace as that spanned by \({{\bf{x}}_1}, \ldots ,{{\bf{x}}_k}\).

c. If \(A = QR\), where \(Q\) has orthonormal columns, then \(R = {Q^T}A\).

In exercises 1-6, determine which sets of vectors are orthogonal.

\(\left[ {\begin{align} 2\\{-5}\\{-3}\end{align}} \right]\), \(\left[ {\begin{align}0\\0\\0\end{align}} \right]\), \(\left[ {\begin{align} 4\\{ - 2}\\6\end{align}} \right]\)

In Exercises 9-12, find a unit vector in the direction of the given vector.

9. \(\left( {\begin{aligned}{*{20}{c}}{ - 30}\\{40}\end{aligned}} \right)\)

Let \(\overline x = \frac{1}{n}\left( {{x_1} + \cdots + {x_n}} \right)\), and \(\overline y = \frac{1}{n}\left( {{y_1} + \cdots + {y_n}} \right)\). Show that the least-squares line for the data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\) must pass through \(\left( {\overline x ,\overline y } \right)\). That is, show that \(\overline x \) and \(\overline y \) satisfies the linear equation \(\overline y = {\hat \beta _0} + {\hat \beta _1}\overline x \). (Hint: Derive this equation from the vector equation \({\bf{y}} = X{\bf{\hat \beta }} + \in \). Denote the first column of \(X\) by 1. Use the fact that the residual vector \( \in \) is orthogonal to the column space of \(X\) and hence is orthogonal to 1.)

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