c01549…: “Beside a summer tree”

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Let $\{\vec {x}_1, \vec {x}_2\}$ be a basis for a subspace $W$ . Use the Gram-Schmidt process to find and orthogonal basis $\{\vec {v}_1,\vec {v}_2\}$ for $W$ . $\vec {x}_1 = \begin {bmatrix} 2\\1\\4\end {bmatrix}, \hspace {10pt} \vec {x}_2 = \begin {bmatrix} -1\\5\\0\end {bmatrix}$ $\vec {v}_1 = \begin {bmatrix} \answer {2}\\\answer {1}\\\answer {4}\end {bmatrix}$
$\vec {v}_2 = \frac {1}{7}\begin {bmatrix} \answer {-9}\\\answer {34}\\\answer {-4}\end {bmatrix}$

Suppose using the Gram-Schmidt process, you found that the set $\{\vec {v}_1,\vec {v}_2\}$ is an orthogonal basis for some subspace $W$ . Find an orthonormal basis $\{\vec {w}_1,\vec {w}_2\}$ i for $W$ , $\vec {v}_1 = \begin {bmatrix} 1\\1\\2\end {bmatrix}, \hspace {10pt} \vec {v}_2 = \begin {bmatrix} 10\\0\\-5\end {bmatrix}$

$\vec {w}_1 = \frac {1}{\sqrt {\answer {6}}}\begin {bmatrix} \answer {1}\\\answer {1}\\\answer {2}\end {bmatrix}$
$\vec {w}_2 = \frac {1}{\sqrt {\answer {5}}}\begin {bmatrix} \answer {2}\\\answer {0}\\\answer {-1}\end {bmatrix}$

The matrix $A$ below has three pivot columns.

$A = \begin {bmatrix}0&-2&3\\1&2&1\\1&2&4\\0&1&-1 \end {bmatrix}$

First, find a basis for $\Col A$ .

$\left \{ \begin {bmatrix} \answer {0}\\\answer {1}\\\answer {1}\\\answer {0}\end {bmatrix}, \begin {bmatrix} \answer {-2}\\\answer {2}\\\answer {2}\\\answer {1}\end {bmatrix} , \begin {bmatrix}\answer {3}\\\answer {1}\\\answer {4}\\\answer {-1}\end {bmatrix} \right \}$

Now, use the Gram-Schmidt process to find an orthogonal basis for $\Col A$ .

$\left \{ \begin {bmatrix} \answer {0}\\\answer {1}\\\answer {1}\\\answer {0}\end {bmatrix}, \begin {bmatrix} \answer {-2}\\\answer {0}\\\answer {0}\\\answer {1}\end {bmatrix}, \frac {1}{10}\begin {bmatrix} \answer {2}\\\answer {-15}\\\answer {15}\\\answer {4}\end {bmatrix}, \right \}$

Let $A = \begin {bmatrix} 1&2&4\\ 3&4&10\\ 0&1&1\end {bmatrix}$ . First, find a basis for $\Col A$ .

$\left \{ \begin {bmatrix} \answer {1}\\\answer {3}\\\answer {0}\end {bmatrix}, \begin {bmatrix} \answer {2}\\\answer {4}\\\answer {1}\end {bmatrix} \right \}$

Now, use the Gram-Schmidt process to find an orthogonal basis for $\Col A$ .

$\left \{ \begin {bmatrix} \answer {1}\\\answer {3}\\\answer {0}\end {bmatrix}, \begin {bmatrix} \answer {1}/2\\\answer {-1}/2\\\answer {1}\end {bmatrix}\right \}$

The column space of $A$ is only two dimensional in this problem. How do you find a basis for $\Col A$ ?