Exam 6: Orthogonality and Least Squares

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Given A and b, determine the least-squares error in the least-squares solution of Ax = b. - A=[432132],b=[301]A = \left[ \begin{array} { l l } 4 & 3 \\2 & 1 \\3 & 2\end{array} \right] , b = \left[ \begin{array} { l } 3 \\0 \\1\end{array} \right]

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Write the word or phrase that best completes each statement or answers the question. -As a function, q3(t)=12t212t+2\mathrm { q } _ { 3 } ( \mathrm { t } ) = 12 \mathrm { t } ^ { 2 } - 12 \mathrm { t } + 2 . The orthogonal basis for the subspace W\mathrm { W } is {q1,q2,q3}\left\{ q _ { 1 } , q _ { 2 } , q _ { 3 } \right\} .

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Find a unit vector in the direction of the given vector. - [442]\left[ \begin{array} { r } - 4 \\ 4 \\ - 2 \end{array} \right]

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Write the word or phrase that best completes each statement or answers the question. -For f,gf , g in C[a, b], set f,g=abf(t)g(t)dt\langle f , g \rangle = \int _ { a } ^ { b } f ( t ) g ( t ) d t . Show that f,g\langle f , g \rangle defines an inner product of C[a,b]C [ a , b ] .

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 Find the least-squares line y=β0+βzx that best fits the given data. \text { Find the least-squares line } y = \beta _ { 0 } + \beta _ { z } x \text { that best fits the given data. } -Given: The data points (-3, 2), (-2, 5), (0, 5), (2, 3), (3, 3). Suppose the errors in measuring the y-values of the last two data points are greater than for the Other points. Weight these data points half as much as the rest of the data. X=[1312101213],β=[β1β2],y=[25533]X = \left[ \begin{array} { r r } 1 & - 3 \\1 & - 2 \\1 & 0 \\1 & 2 \\1 & 3\end{array} \right] , \beta = \left[ \begin{array} { l } \beta _ { 1 } \\\beta _ { 2 }\end{array} \right] , y = \left[ \begin{array} { l } 2 \\5 \\5 \\3 \\3\end{array} \right]

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Let W be the subspace spanned by the uʹs. Write y as the sum of a vector in W and a vector orthogonal to W. - y=[19311],u1=[101],u2=[212]\mathbf { y } = \left[ \begin{array} { r } 19 \\3 \\11\end{array} \right] , \mathbf { u } _ { 1 } = \left[ \begin{array} { r } 1 \\0 \\- 1\end{array} \right] , \mathbf { u } _ { 2 } = \left[ \begin{array} { l } 2 \\1 \\2\end{array} \right]

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Find the distance between the two vectors. - u=(0,0,0),v=(5,3,4)\mathbf { u } = ( 0,0,0 ) , \mathbf { v } = ( - 5 , - 3 , - 4 )

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Determine whether the set of vectors is orthogonal. - [204020],[20020],[202020]\left[ \begin{array} { l } - 20 \\- 40 \\- 20\end{array} \right] , \left[ \begin{array} { r } 20 \\0 \\- 20\end{array} \right] , \left[ \begin{array} { l } - 20 \\- 20 \\- 20\end{array} \right]

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Compute the length of the given vector. - p(t)=13t2\mathrm { p } ( \mathrm { t } ) = 13 \mathrm { t } ^ { 2 } and q(t)=t1\mathrm { q } ( \mathrm { t } ) = \mathrm { t } - 1 , where t0=0,t1=12,t2=1\mathrm { t } _ { 0 } = 0 , \mathrm { t } _ { 1 } = \frac { 1 } { 2 } , \mathrm { t } _ { 2 } = 1

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Compute the dot product u · v. - u=[100],v=[143]\mathbf { u } = \left[ \begin{array} { r } 10 \\0\end{array} \right] , \mathbf { v } = \left[ \begin{array} { c } 14 \\- 3\end{array} \right]

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 Find the least-squares line y=β0+βzx that best fits the given data. \text { Find the least-squares line } y = \beta _ { 0 } + \beta _ { z } x \text { that best fits the given data. } -Given: The data points (-2, 2), (-1, 5), (0, 5), (1, 3), (2, 5). 44) Suppose the errors in measuring the y-values of the last two data points are greater than for the Other points. Weight these data points twice as much as the rest of the data. X=[1211101112],β=[β1β2],y=[25535]\mathrm { X } = \left[ \begin{array} { r r } 1 & - 2 \\1 & - 1 \\1 & 0 \\1 & 1 \\1 & 2\end{array} \right] , \beta = \left[ \begin{array} { l } \beta _ { 1 } \\\beta _ { 2 }\end{array} \right] , \mathrm { y } = \left[ \begin{array} { l } 2 \\5 \\5 \\3 \\5\end{array} \right]

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Find the distance between the two vectors. - u=(24,14,19),v=(4,4,7)\mathbf { u } = ( 24,14,19 ) , \mathbf { v } = ( - 4,4,7 )

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Let W be the subspace spanned by the uʹs. Write y as the sum of a vector in W and a vector orthogonal to W. - y=[17712],u1=[221],u2=[134]\mathbf { y } = \left[ \begin{array} { r } 17 \\ 7 \\ 12 \end{array} \right] , \mathbf { u } _ { 1 } = \left[ \begin{array} { r } 2 \\ 2 \\ - 1 \end{array} \right] , \mathbf { u } _ { 2 } = \left[ \begin{array} { r } - 1 \\ 3 \\ 4 \end{array} \right]

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Express the vector x as a linear combination of the uʹs. - u1=[24],u2=[126],x=[324]\mathbf { u } _ { 1 } = \left[ \begin{array} { r } 2 \\- 4\end{array} \right] , \mathbf { u } _ { 2 } = \left[ \begin{array} { r } 12 \\6\end{array} \right] , \mathbf { x } = \left[ \begin{array} { c } 32 \\- 4\end{array} \right]

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Solve the problem. -Let C[0,π]C [ 0 , \pi ] have the inner product f,g=0πf(t)g(t)dt\langle f , g \rangle = \int _ { 0 } ^ { \pi } f ( t ) g ( t ) d t , and let mm and nn be unequal positive integers. Prove that cos(mt)\cos ( \mathrm { mt } ) and cos(nt)\cos ( \mathrm { nt } ) are orthogonal.

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Compute the dot product u · v. - u=[117],v=[31]\mathbf { u } = \left[ \begin{array} { r } 1 \\17\end{array} \right] , \mathbf { v } = \left[ \begin{array} { l } 3 \\1\end{array} \right]

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Find the closest point to y in the subspace W spanned by u1 and u2. - y=[1602],u1=[101],u2=[212]\mathbf { y } = \left[ \begin{array} { r } 16 \\0 \\2\end{array} \right] , \mathbf { u } _ { 1 } = \left[ \begin{array} { r } 1 \\0 \\- 1\end{array} \right] , \mathbf { u } _ { 2 } = \left[ \begin{array} { l } 2 \\1 \\2\end{array} \right]

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Find the distance between the two vectors. - u=(6,12),v=(12,12)\mathbf { u } = ( 6 , - 12 ) , \mathbf { v } = ( - 12,12 )

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Find a unit vector in the direction of the given vector. - [1326]\left[ \begin{array} { r } 13 \\ - 26 \end{array} \right]

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Find the orthogonal projection of y onto u. - y=[1236],u=[84]\mathbf { y } = \left[ \begin{array} { r } - 12 \\36\end{array} \right] , \mathbf { u } = \left[ \begin{array} { r } - 8 \\4\end{array} \right]

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