Solved Problems

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Solved Problems for Chapter 9

Find an orthonormal basis for the span of the following set of vectors.

$\left [ \begin {array}{r} 3 \\ -4 \\ 0 \end {array} \right ] ,\left [ \begin {array}{r} 7 \\ -1 \\ 0 \end {array} \right ] ,\left [ \begin {array}{r} 1 \\ 7 \\ 1 \end {array} \right ]$

Click the arrow to see answer.

We will use Gram-Schmidt orthogonalization algorithm. $\vec {f}_1=\begin {bmatrix}3\\-4\\0\end {bmatrix}$ $\vec {f}_2=\begin {bmatrix}7\\-1\\0\end {bmatrix}-\frac {25}{25}\begin {bmatrix}3\\-4\\0\end {bmatrix}=\begin {bmatrix}4\\3\\0\end {bmatrix}$ $\vec {f}_3=\begin {bmatrix}1\\7\\1\end {bmatrix}-\frac {-25}{25}\begin {bmatrix}3\\-4\\0\end {bmatrix}-\frac {25}{25}\begin {bmatrix}4\\3\\0\end {bmatrix}=\begin {bmatrix}0\\0\\1\end {bmatrix}$

Normalizing each vector we get: $\left [ \begin {array}{c} \frac {3}{5} \\ -\frac {4}{5} \\ 0 \end {array} \right ] ,\left [ \begin {array}{c} \frac {4}{5} \\ \frac {3}{5} \\ 0 \end {array} \right ] ,\left [ \begin {array}{c} 0 \\ 0 \\ 1 \end {array} \right ]$

Using the Gram Schmidt process find an orthonormal basis for the following span: $\mbox {span} \left \{ \left [ \begin {array}{r} 1 \\ 2 \\ 1 \end {array} \right ] ,\left [ \begin {array}{r} 2 \\ -1 \\ 3 \end {array} \right ] , \left [ \begin {array}{r} 1 \\ 0 \\ 0 \end {array} \right ] \right \}$

Click the arrow to see the answer.

$\left [ \begin {array}{c} \frac {1}{6}\sqrt {6} \\ \frac {1}{3}\sqrt {6} \\ \frac {1}{6}\sqrt {6} \end {array} \right ] ,\left [ \begin {array}{c} \frac {3}{10}\sqrt {2} \\ -\frac {2}{5}\sqrt {2} \\ \frac {1}{2}\sqrt {2} \end {array} \right ] ,\left [ \begin {array}{c} \frac {7}{15}\sqrt {3} \\ -\frac {1}{15}\sqrt {3} \\ -\frac {1}{3}\sqrt {3} \end {array} \right ]$

Write $\vec {x}$ as the sum of a vector in $W$ and a vector in $W^\perp$ if $\vec {x} = \begin {bmatrix}1\\ 5\\ 7\end {bmatrix},\quad W = \mbox {span}\left (\begin {bmatrix}1\\ -2\\ 3\end {bmatrix}, \begin {bmatrix}-1\\ 1\\ 1\end {bmatrix}\right )$

Click the arrow to see the answer.

Observe that the two vectors that span $W$ are orthogonal. We will refer to these vectors as $\vec {f}_1$ and $\vec {f}_2$ . $\vec {w}=\text {proj}_W(\vec {x})=\text {proj}_{\vec {f}_1}(\vec {x})+\text {proj}_{\vec {f}_2}(\vec {x})=\frac {12}{14}\begin {bmatrix}1\\-2\\3\end {bmatrix}+\frac {11}{3}\begin {bmatrix}-1\\1\\1\end {bmatrix}=\frac {1}{21}\begin {bmatrix}-59\\41\\131\end {bmatrix}$ $\vec {w}^{\perp }=\vec {x}-\vec {w}=\frac {1}{21}\begin {bmatrix}80\\64\\16\end {bmatrix}$ $\vec {x}=\vec {w}+\vec {w}^{\perp }$

Write $\vec {x}$ as the sum of a vector in $W$ and a vector in $W^\perp$ if $\vec {x} = \begin {bmatrix}3\\ 1\\ 5\\ 9\end {bmatrix}, \quad W = \mbox {span}\left (\begin {bmatrix}1\\ 0\\ 1\\ 1\end {bmatrix}, \begin {bmatrix}0\\ 1\\ -1\\ 1\end {bmatrix}, \begin {bmatrix}-2\\ 0\\ 1\\ 1\end {bmatrix}\right )$

Click the arrow to see the answer.

Observe that the three vectors that span $W$ form an orthogonal set. We will refer to these vectors as $\vec {f}_1$ , $\vec {f}_2$ and $\vec {f}_3$ . $\vec {w}=\text {proj}_W(\vec {x})=\text {proj}_{\vec {f}_1}(\vec {x})+\text {proj}_{\vec {f}_2}(\vec {x})+\text {proj}_{\vec {f}_3}(\vec {x})=$ $=\frac {17}{3}\begin {bmatrix}1\\0\\1\\1\end {bmatrix}+\frac {5}{3}\begin {bmatrix}0\\1\\-1\\1\end {bmatrix}+\frac {8}{6}\begin {bmatrix}-2\\0\\1\\1\end {bmatrix}=\frac {1}{3}\begin {bmatrix}9\\5\\16\\26\end {bmatrix}$ $\vec {w}^{\perp }=\vec {x}-\vec {w}=\frac {1}{3}\begin {bmatrix}0\\-2\\-1\\1\end {bmatrix}$ $\vec {x}=\vec {w}+\vec {w}^{\perp }$

Bibliography

Some of the problems come from the end of Chapter 7 of Ken Kuttler’s A First Course in Linear Algebra. (CC-BY)

Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, pp. 433–438.

2024-09-06 23:26:22