Gram-Schmidt process for [1,0,1,0]', [2,1,1,1]', [1,0,0,1]'

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

Number of vectors:

Size of the vectors:

Vectors:

$$$\mathbf{\vec{v_{1}}}$$$	$$$\mathbf{\vec{v_{2}}}$$$	$$$\mathbf{\vec{v_{3}}}$$$

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Your Input

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

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

$$$\mathbf{\vec{e_{1}}} = \frac{\mathbf{\vec{u_{1}}}}{\mathbf{\left\lvert\vec{u_{1}}\right\rvert}} = \left[\begin{array}{c}\frac{\sqrt{2}}{2}\\0\\\frac{\sqrt{2}}{2}\\0\end{array}\right]$$$ (for steps, see unit vector calculator).

Step 2

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

$$$\mathbf{\vec{e_{2}}} = \frac{\mathbf{\vec{u_{2}}}}{\mathbf{\left\lvert\vec{u_{2}}\right\rvert}} = \left[\begin{array}{c}\frac{\sqrt{10}}{10}\\\frac{\sqrt{10}}{5}\\- \frac{\sqrt{10}}{10}\\\frac{\sqrt{10}}{5}\end{array}\right]$$$ (for steps, see unit vector calculator).

Step 3

$$$\mathbf{\vec{u_{3}}} = \mathbf{\vec{v_{3}}} - \operatorname{proj}_{\mathbf{\vec{u_{1}}}}\left(\mathbf{\vec{v_{3}}}\right) - \operatorname{proj}_{\mathbf{\vec{u_{2}}}}\left(\mathbf{\vec{v_{3}}}\right) = \left[\begin{array}{c}\frac{1}{5}\\- \frac{3}{5}\\- \frac{1}{5}\\\frac{2}{5}\end{array}\right]$$$ (for steps, see vector projection calculator and vector subtraction calculator).

$$$\mathbf{\vec{e_{3}}} = \frac{\mathbf{\vec{u_{3}}}}{\mathbf{\left\lvert\vec{u_{3}}\right\rvert}} = \left[\begin{array}{c}\frac{\sqrt{15}}{15}\\- \frac{\sqrt{15}}{5}\\- \frac{\sqrt{15}}{15}\\\frac{2 \sqrt{15}}{15}\end{array}\right]$$$ (for steps, see unit vector calculator).

Answer

The set of the orthonormal vectors is $$$\left\{\left[\begin{array}{c}\frac{\sqrt{2}}{2}\\0\\\frac{\sqrt{2}}{2}\\0\end{array}\right], \left[\begin{array}{c}\frac{\sqrt{10}}{10}\\\frac{\sqrt{10}}{5}\\- \frac{\sqrt{10}}{10}\\\frac{\sqrt{10}}{5}\end{array}\right], \left[\begin{array}{c}\frac{\sqrt{15}}{15}\\- \frac{\sqrt{15}}{5}\\- \frac{\sqrt{15}}{15}\\\frac{2 \sqrt{15}}{15}\end{array}\right]\right\}\approx \left\{\left[\begin{array}{c}0.707106781186548\\0\\0.707106781186548\\0\end{array}\right], \left[\begin{array}{c}0.316227766016838\\0.632455532033676\\-0.316227766016838\\0.632455532033676\end{array}\right], \left[\begin{array}{c}0.258198889747161\\-0.774596669241483\\-0.258198889747161\\0.516397779494322\end{array}\right]\right\}.$$$A