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Let \(W\) be a subspace of \({\mathbb{R}^n}\), and let \({W^ \bot }\) be the set of all vectors orthogonal to \(W\). Show that \({W^ \bot }\) is a subspace of \({\mathbb{R}^n}\) using the following steps.

  1. Take \({\bf{z}}\) in \({W^ \bot }\), and let \({\bf{u}}\) represent any element of \(W\). Then \({\bf{z}} \cdot {\bf{u}} = 0\). Take any scalar \(c\) and show that \(c{\bf{z}}\) is orthogonal to \({\bf{u}}\). (Since \({\bf{u}}\) was an arbitrary element of \(W\), this will show that \(c{\bf{z}}\) is in \({W^ \bot }\).)
  2. Take \({{\bf{z}}_1}\) and \({{\bf{z}}_2}\) in \({W^ \bot }\), and let \({\bf{u}}\) be any element of \(W\). Show that \({{\bf{z}}_1} + {{\bf{z}}_2}\) is orthogonal to \({\bf{u}}\). What can you conclude about \({{\bf{z}}_1} + {{\bf{z}}_2}\)? Why?
  3. Finish the proof that \({W^ \bot }\) is a subspace of \({\mathbb{R}^n}\).

Short Answer

Expert verified

It is verified that \({W^ \bot }\) is a subspace of \({\mathbb{R}^n}\).

Step by step solution

01

Definition of Orthogonal sets

The two vectors \({\bf{u}}{\rm{ and }}{\bf{v}}\) are Orthogonal if:

\(\begin{aligned}{l}{\left\| {{\bf{u}} + {\bf{v}}} \right\|^2} &= {\left\| {\bf{u}} \right\|^2} + {\left\| {\bf{v}} \right\|^2}\\{\rm{and}}\\{\bf{u}} \cdot {\bf{v}} &= 0\end{aligned}\)

02

 Computing for required proof (a)

Since,in step (a), \({\bf{z}}{\rm{ and }}{\bf{u}}\)are in \({W^ \bot }\) for \(c\)as any scalar,

Then, we have:

\(\begin{aligned}{c}\left( {c{\bf{z}}} \right) \cdot {\bf{u}} &= c\left( {{\bf{z}} \cdot {\bf{u}}} \right)\\ &= c\left( 0 \right)\\ &= 0\end{aligned}\)

Thus, \(c{\bf{z}}\)is orthogonal to \({\bf{u}}\).

03

 Computing for required proof (b)

Similarly, for \({{\bf{z}}_1}{\rm{ and }}{{\bf{z}}_2}\)are in \({W^ \bot }\), which implies,

\(\begin{aligned}{l}{{\bf{z}}_1} \cdot {\bf{u}} = 0\\{{\bf{z}}_2} \cdot {\bf{u}} = 0\end{aligned}\)

Then, we have:

\(\begin{aligned}{c}\left( {{{\bf{z}}_1} + {{\bf{z}}_2}} \right) \cdot {\bf{u}} &= {{\bf{z}}_1} \cdot {\bf{u}} + {{\bf{z}}_2} \cdot {\bf{u}}\\ &= 0 + 0\\ &= 0\end{aligned}\)

Thus, \({{\bf{z}}_1} + {{\bf{z}}_2}\) is orthogonal to \({\bf{u}}\).

04

 Computing for required proof (c)

As 0 orthogonal to all the vectors. So, \({W^ \bot }\)is a subspace in \({\mathbb{R}^n}\).

Hence these are the required proofs.

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Most popular questions from this chapter

In Exercises 1-4, find a least-sqaures solution of \(A{\bf{x}} = {\bf{b}}\) by (a) constructing a normal equations for \({\bf{\hat x}}\) and (b) solving for \({\bf{\hat x}}\).

2. \(A = \left( {\begin{aligned}{{}{}}{\bf{2}}&{\bf{1}}\\{ - {\bf{2}}}&{\bf{0}}\\{\bf{2}} {\bf{3}}\end{aligned}} \right)\), \(b = \left( {\begin{aligned}{{}{}}{ - {\bf{5}}}\\{\bf{8}}\\{\bf{1}}\end{aligned}} \right)\)

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 1-4, find the equation \(y = {\beta _0} + {\beta _1}x\) of the least-square line that best fits the given data points.

  1. \(\left( { - 1,0} \right),\left( {0,1} \right),\left( {1,2} \right),\left( {2,4} \right)\)

In Exercises 13 and 14, the columns of Q were obtained by applying the Gram-Schmidt process to the columns of A. Find an upper triangular matrix R such that \(A = QR\). Check your work.

13. \(A = \left( {\begin{aligned}{{}{}}5&9\\1&7\\{ - 3}&{ - 5}\\1&5\end{aligned}} \right),{\rm{ }}Q = \left( {\begin{aligned}{{}{}}{\frac{5}{6}}&{ - \frac{1}{6}}\\{\frac{1}{6}}&{\frac{5}{6}}\\{ - \frac{3}{6}}&{\frac{1}{6}}\\{\frac{1}{6}}&{\frac{3}{6}}\end{aligned}} \right)\)

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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