Rank and Nullity
Eigenvalues and Eigenvectors

Gram-Schmidt Process, Orthonormal Basis

Example 1.

 

We are given the set \{\left (\sqrt{2} ,\sqrt{2} ,\sqrt{2}\right ) ,\left ( -\frac{1}{\sqrt{3}} ,0 ,\frac{1}{\sqrt{3}}\right )\}. First we notice that

(1)   \begin{equation*}\left (\sqrt{2} ,\sqrt{2} ,\sqrt{2}\right ) \cdot \left ( -\frac{1}{\sqrt{3}} ,0 ,\frac{1}{\sqrt{3}}\right ) = -\frac{1}{\sqrt{3}}\left (\sqrt{2}\right ) +0 +\frac{1}{\sqrt{3}}\left (\sqrt{2}\right ) =0 \end{equation*}

and therefore the two vectors in this set are orthogonal with respect to each other (that is this set is orthogonal.) Next we want to normalize the set to get an orthonormal set. We find the norm of each vector

(2)   \begin{align*}\left \Vert \left (\sqrt{2} ,\sqrt{2} ,\sqrt{2}\right )\right \Vert =\sqrt{\left (\sqrt{2}\right )^{2} +\left (\sqrt{2}\right )^{2} +\left (\sqrt{2}\right )^{2}} =\sqrt{6} \\ \left \Vert \left ( -\frac{1}{\sqrt{3}} ,0 ,\frac{1}{\sqrt{3}}\right )\right \Vert =\sqrt{\left ( -\frac{1}{\sqrt{3}}\right )^{2} +0 +\left (\frac{1}{\sqrt{3}}\right )^{2}} =\sqrt{\frac{2}{3}}\end{align*}

and obtain the vectors

(3)   \begin{align*}\frac{1}{\sqrt{6}}\left (\sqrt{2} ,\sqrt{2} ,\sqrt{2}\right ) =\left (\frac{1}{\sqrt{3}} ,\frac{1}{\sqrt{3}} ,\frac{1}{\sqrt{3}}\right ) \\ \frac{1}{\sqrt{\frac{2}{3}}}\left ( -\frac{1}{\sqrt{3}} ,0 ,\frac{1}{\sqrt{3}}\right ) =\left ( -\frac{1}{\sqrt{2}} ,0 ,\frac{1}{\sqrt{2}}\right )\end{align*}

The set \{\left (\frac{1}{\sqrt{3}} ,\frac{1}{\sqrt{3}} ,\frac{1}{\sqrt{3}}\right ) ,\left ( -\frac{1}{\sqrt{2}} ,0 ,\frac{1}{\sqrt{2}}\right )\} is orthonormal.

 

Example 2.

 

We are given the set

(4)   \begin{equation*}U =\{\left (3 ,4 ,0 ,0\right ) ,\left ( -1 ,1 ,0 ,0\right ) ,\left (2 ,1 ,0 , -1\right ) ,\left (0 ,1 ,1 ,0\right )\} \end{equation*}

and would like to construct an orthogonal basis for the span of U. Write

(5)   \begin{equation*}v_{1} =\left (3 ,4 ,0 ,0\right ) ,v_{2} =\left ( -1 ,1 ,0 ,0\right ) ,v_{3} =\left (2 ,1 ,0 , -1\right ) ,v_{4} =\left (0 ,1 ,1 ,0\right ) \end{equation*}

Start the Gram-Schmidt process

(6)   \begin{align*}u_{1} =v_{1} =\left (3 ,4 ,0 ,0\right ) \\ \, \\ u_{2} =v_{2} -proj_{u_{1}}v_{2} \\ =\left ( -1 ,1 ,0 ,0\right ) -\frac{\left (3 ,4 ,0 ,0\right ) \cdot \left ( -1 ,1 ,0 ,0\right )}{\left (3 ,4 ,0 ,0\right ) \cdot \left (3 ,4 ,0 ,0\right )}\left (3 ,4 ,0 ,0\right ) \\ =\left ( -1 ,1 ,0 ,0\right ) -\frac{ -3 +4}{9 +16}\left (3 ,4 ,0 ,0\right ) \\ =\left ( -1 ,1 ,0 ,0\right ) -\left (\frac{3}{25} ,\frac{4}{25}\right ) \\ =\left ( -\frac{28}{25} ,\frac{21}{25} ,0 ,0\right ) \\ \, \\ u_{3} =v_{3} -proj_{u_{1}}v_{3} -proj_{u_{2}}v_{3} \\ =\left (2 ,1 ,0 , -1\right ) -\frac{\left (3 ,4 ,0 ,0\right ) \cdot \left (2 ,1 ,0 , -1\right )}{\left (3 ,4 ,0 ,0\right ) \cdot \left (3 ,4 ,0 ,0\right )}\left (3 ,4 ,0 ,0\right ) -\frac{\left ( -\frac{28}{25} ,\frac{21}{25} ,0 ,0\right ) \cdot \left (2 ,1 ,0 , -1\right )}{\left ( -\frac{28}{25} ,\frac{21}{25} ,0 ,0\right ) \cdot \left ( -\frac{28}{25} ,\frac{21}{25} ,0 ,0\right )}\left ( -\frac{28}{25} ,\frac{21}{25} ,0 ,0\right ) \\ =\left (2 ,1 ,0 , -1\right ) -\frac{10}{25}\left (3 ,4 ,0 ,0\right ) -\frac{ -\frac{56}{25} +\frac{21}{25}}{\genfrac{(}{)}{}{}{28}{25}^{2} +\genfrac{(}{)}{}{}{21}{25}^{2}}\left ( -\frac{28}{25} ,\frac{21}{25} ,0 ,0\right ) \\ =\left (2 ,1 ,0 , -1\right ) -\left (\frac{6}{5} ,\frac{8}{5} ,0 ,0\right ) -\left ( -\frac{196}{245} ,\frac{147}{245} ,0 ,0\right ) \\ =\left (\frac{392}{245} , -\frac{294}{245} ,0 , -1\right )\end{align*}

We have found three vectors so far that are perpendicular to one another. Next

(7)   \begin{align*}u_{4} =v_{4} -proj_{u_{1}}v_{4} -proj_{u_{2}}v_{4} -proj_{u_{3}}v_{4} \\ =\left (0 ,1 ,1 ,0\right ) -\frac{\left (3 ,4 ,0 ,0\right ) \cdot \left (0 ,1 ,1 ,0\right )}{\left (3 ,4 ,0 ,0\right ) \cdot \left (3 ,4 ,0 ,0\right )}\left (3 ,4 ,0 ,0\right ) -\frac{\left ( -\frac{28}{25} ,\frac{21}{25} ,0 ,0\right ) \cdot \left (0 ,1 ,1 ,0\right )}{\left ( -\frac{28}{25} ,\frac{21}{25} ,0 ,0\right ) \cdot \left ( -\frac{28}{25} ,\frac{21}{25} ,0 ,0\right )}\left ( -\frac{28}{25} ,\frac{21}{25} ,0 ,0\right ) \\ -\frac{\left (\frac{392}{245} , -\frac{294}{245} ,0 , -1\right ) \cdot \left (0 ,1 ,1 ,0\right )}{\left (\frac{392}{245} , -\frac{294}{245} ,0 , -1\right ) .\left (\frac{392}{245} , -\frac{294}{245} ,0 , -1\right )}\left (\frac{392}{245} , -\frac{294}{245} ,0 , -1\right )\end{align*}

Can you compute this vector?

 

If one wants to build an orthonormal basis, the normalization process can be done to find a unit vector in each direction found above.

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