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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

Compute the least-squares error associated with the least square solution found in Exercise 3.

(M) Use the method in this section to produce a \(QR\) factorization of the matrix in Exercise 24.

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

Exercises 19 and 20 involve a design matrix \(X\) with two or more columns and a least-squares solution \(\hat \beta \) of \({\bf{y}} = X\beta \). Consider the following numbers.

(i) \({\left\| {X\hat \beta } \right\|^2}\)โ€”the sum of the squares of the โ€œregression term.โ€ Denote this number by .

(ii) \({\left\| {{\bf{y}} - X\hat \beta } \right\|^2}\)โ€”the sum of the squares for error term. Denote this number by \(SS\left( E \right)\).

(iii) \({\left\| {\bf{y}} \right\|^2}\)โ€”the โ€œtotalโ€ sum of the squares of the \(y\)-values. Denote this number by \(SS\left( T \right)\).

Every statistics text that discusses regression and the linear model \(y = X\beta + \in \) introduces these numbers, though terminology and notation vary somewhat. To simplify matters, assume that the mean of the -values is zero. In this case, \(SS\left( T \right)\) is proportional to what is called the variance of the set of -values.

19. Justify the equation \(SS\left( T \right) = SS\left( R \right) + SS\left( E \right)\). (Hint: Use a theorem, and explain why the hypotheses of the theorem are satisfied.) This equation is extremely important in statistics, both in regression theory and in the analysis of variance.

In Exercises 1-6, the given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.

6. \(\left( {\begin{aligned}{{}}3\\{ - 1}\\2\\{ - 1}\end{aligned}} \right),\left( {\begin{aligned}{{}}{ - 5}\\9\\{ - 9}\\3\end{aligned}} \right)\)

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