Gram-Schmidt process for $$$\left[\begin{array}{c}1\\0\\0\end{array}\right]$$$, $$$\left[\begin{array}{c}2\\1\\1\end{array}\right]$$$, $$$\left[\begin{array}{c}2\\1\\1\end{array}\right]$$$

The calculator will orthonormalize the set of the vectors $$$\left[\begin{array}{c}1\\0\\0\end{array}\right]$$$, $$$\left[\begin{array}{c}2\\1\\1\end{array}\right]$$$, $$$\left[\begin{array}{c}2\\1\\1\end{array}\right]$$$ using the Gram-Schmidt process, with steps shown.
A
$$$\mathbf{\vec{v_{1}}}$$$ $$$\mathbf{\vec{v_{2}}}$$$ $$$\mathbf{\vec{v_{3}}}$$$

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Your Input

Orthonormalize the set of the vectors $$$\mathbf{\vec{v_{1}}} = \left[\begin{array}{c}1\\0\\0\end{array}\right]$$$, $$$\mathbf{\vec{v_{2}}} = \left[\begin{array}{c}2\\1\\1\end{array}\right]$$$, $$$\mathbf{\vec{v_{3}}} = \left[\begin{array}{c}2\\1\\1\end{array}\right]$$$ using the Gram-Schmidt process.

Solution

According to the Gram-Schmidt process, $$$\mathbf{\vec{u_{k}}} = \mathbf{\vec{v_{k}}} - \sum_{j=1}^{k - 1} \operatorname{proj}_{\mathbf{\vec{u_{j}}}}\left(\mathbf{\vec{v_{k}}}\right)$$$, where $$$\operatorname{proj}_{\mathbf{\vec{u_{j}}}}\left(\mathbf{\vec{v_{k}}}\right) = \frac{\mathbf{\vec{u_{j}}}\cdot \mathbf{\vec{v_{k}}}}{\mathbf{\left\lvert\vec{u_{j}}\right\rvert}^{2}} \mathbf{\vec{u_{j}}}$$$ is a vector projection.

The normalized vector is $$$\mathbf{\vec{e_{k}}} = \frac{\mathbf{\vec{u_{k}}}}{\mathbf{\left\lvert\vec{u_{k}}\right\rvert}}$$$.

Step 1

$$$\mathbf{\vec{u_{1}}} = \mathbf{\vec{v_{1}}} = \left[\begin{array}{c}1\\0\\0\end{array}\right]$$$

$$$\mathbf{\vec{e_{1}}} = \frac{\mathbf{\vec{u_{1}}}}{\mathbf{\left\lvert\vec{u_{1}}\right\rvert}} = \left[\begin{array}{c}1\\0\\0\end{array}\right]$$$ (for steps, see unit vector calculator).

Step 2

$$$\mathbf{\vec{u_{2}}} = \mathbf{\vec{v_{2}}} - \operatorname{proj}_{\mathbf{\vec{u_{1}}}}\left(\mathbf{\vec{v_{2}}}\right) = \left[\begin{array}{c}0\\1\\1\end{array}\right]$$$ (for steps, see vector projection calculator and vector subtraction calculator).

$$$\mathbf{\vec{e_{2}}} = \frac{\mathbf{\vec{u_{2}}}}{\mathbf{\left\lvert\vec{u_{2}}\right\rvert}} = \left[\begin{array}{c}0\\\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\end{array}\right]$$$ (for steps, see unit vector calculator).

Step 3

$$$\mathbf{\vec{u_{3}}} = \mathbf{\vec{v_{3}}} - \operatorname{proj}_{\mathbf{\vec{u_{1}}}}\left(\mathbf{\vec{v_{3}}}\right) - \operatorname{proj}_{\mathbf{\vec{u_{2}}}}\left(\mathbf{\vec{v_{3}}}\right) = \left[\begin{array}{c}0\\0\\0\end{array}\right]$$$ (for steps, see vector projection calculator and vector subtraction calculator).

Since $$$\mathbf{\vec{u_{3}}}$$$ is a zero vector, we just skip it.

Answer

The set of the orthonormal vectors is $$$\left\{\left[\begin{array}{c}1\\0\\0\end{array}\right], \left[\begin{array}{c}0\\\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\end{array}\right]\right\}\approx \left\{\left[\begin{array}{c}1\\0\\0\end{array}\right], \left[\begin{array}{c}0\\0.707106781186548\\0.707106781186548\end{array}\right]\right\}.$$$A


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