Question

Using paper and pencil, perform the Gram-Schmidt process on the sequences of vectors given in Exercises 1 through 14.

6. $\left[\begin{array}{c}2\\ 0\\ 0\end{array}\right]\mathbf{,}\left[\begin{array}{c}3\\ 4\\ 0\end{array}\right]\mathbf{,}\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right]$

Short answer

The orthonormal vectors of the sequence $\left[\begin{array}{c}2\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}3\\ 4\\ 0\end{array}\right],\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right]$is $\overrightarrow{{\mu }_1}=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],{\overrightarrow{\mu }}_1=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right],\overrightarrow{{\mu }_3}=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$ .

Step-by-step solution

Determine the Gram-Schmidt process.

Consider a basis of a subspace Vof ${\mathit{R}}^{\mathbf{n}}$for $\mathit{j}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathit{m}$we resolve the vector ${\overrightarrow{\mathbf{v}}}_{\mathbf{j}}$into its components parallel and perpendicular to the span of the preceding vectors ${\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{j}}$,

Then

localid="1659954506203" ${\overrightarrow{\mathbf{u}}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{||{\overrightarrow{v}}_1||}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{||{{\overrightarrow{v}}_2}^{\perp }||}{{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}}^{\mathbf{\perp }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\overrightarrow{\mathbf{u}}}_{\mathbf{j}}\mathbf{=}\frac{\mathbf{1}}{||{{\overrightarrow{v}}_2}^{\perp }||}{{\overrightarrow{\mathbf{v}}}_{\mathbf{j}}}^{\mathbf{\perp }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\overrightarrow{\mathbf{u}}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{1}}{||{{\overrightarrow{v}}_m}^{\perp }||}{{\overrightarrow{\mathbf{v}}}_{\mathbf{m}}}^{\mathbf{\perp }}$

According to Gram-Schmidt process.

Let the given vectors are,${\overrightarrow{v}}_1=\left[\begin{array}{c}2\\ 0\\ 0\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{c}3\\ 4\\ 0\end{array}\right],{\overrightarrow{v}}_3=\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right]$.

Obtain the value of ${\overrightarrow{u}}_1$, ${\overrightarrow{u}}_2$and ${\overrightarrow{u}}_3$according to Gram-Schmidt process.

$$\begin{gathered} {\overrightarrow{u}}_1=\frac{\overrightarrow{v}}{||\overrightarrow{v}||}.....\left(1\right) \\ {\overrightarrow{u}}_2=\frac{{\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_2\right){\overrightarrow{u}}_1}{||{\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_2\right)||}.....\left(2\right) \\ {\overrightarrow{u}}_3=\frac{{\overrightarrow{v}}_3-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_3\right){\overrightarrow{u}}_1-\left({\overrightarrow{u}}_2.{\overrightarrow{v}}_3\right){\overrightarrow{u}}_2}{||{\overrightarrow{v}}_3-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_3\right){\overrightarrow{u}}_1-\left({\overrightarrow{u}}_2.{\overrightarrow{v}}_3\right){\overrightarrow{u}}_2||}.....\left(3\right) \end{gathered}$$

Find ${\overrightarrow{u}}_1$.

$$\begin{gathered} {\overrightarrow{u}}_1=\frac{\overrightarrow{v_1}}{||\overrightarrow{v_1}||} \\ =\frac{1}{\sqrt{2^2+0^2+0^2}}\left[\begin{array}{c}2\\ 0\\ 0\end{array}\right] \\ =\frac{1}{2}\left[\begin{array}{c}2\\ 0\\ 0\end{array}\right] \\ =\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right] \end{gathered}$$

Find u→2

As, the formula for${\overrightarrow{u}}_2$is, ${\overrightarrow{u}}_2=\frac{{\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_2\right){\overrightarrow{u}}_1}{||{\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_2\right)\overrightarrow{u_1}||}$.

Now, find the values of $\overrightarrow{u_1}.\overrightarrow{V_2}$, $\overrightarrow{v_2}-\left(\overrightarrow{u_1}.\overrightarrow{v_2}\right){\overrightarrow{u}}_1$and$||\overrightarrow{v_2}-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_2\right)\overrightarrow{u_1}||$as below:

Calculate $\overrightarrow{u_1}.\overrightarrow{v_2}$.

$$\begin{gathered} \overrightarrow{u_1}.\overrightarrow{v_2}=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right].\left[\begin{array}{c}3\\ 4\\ 0\end{array}\right] \\ =3+0+0 \\ =3 \end{gathered}$$

$$\begin{gathered} {\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_2\right)\overrightarrow{u_1}=\left[\begin{array}{c}3\\ 4\\ 0\end{array}\right]-3.\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right] \\ =\left[\begin{array}{c}3\\ 4\\ 0\end{array}\right]-\left[\begin{array}{c}3\\ 0\\ 0\end{array}\right] \\ =\left[\begin{array}{c}0\\ 4\\ 0\end{array}\right] \end{gathered}$$

$$\begin{gathered} ||{\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_2\right){\overrightarrow{u}}_1||=\sqrt{0^2+4^2+0^2} \\ =4 \end{gathered}$$

Now, find ${\overrightarrow{u}}_2$

$$\begin{gathered} {\overrightarrow{u}}_2=\frac{1}{4}\left[\begin{array}{c}0\\ 4\\ 0\end{array}\right] \\ =\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right] \end{gathered}$$

Find u→3

Now, to obtain the value of ${\overrightarrow{u}}_3$consider the equation is given below and then put the values obtained in the step 2.

Now find,role="math" localid="1659954011731" $\overrightarrow{v_3}.\overrightarrow{u_1}$and $\overrightarrow{v_3}.\overrightarrow{u_2}$.

$$\begin{gathered} \overrightarrow{v_3}.\overrightarrow{u_1}=\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right].\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right] \\ =5 \end{gathered}$$

$$\begin{gathered} \overrightarrow{v_3}-\overrightarrow{u_2}=\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right].\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right] \\ =6 \end{gathered}$$

Now fine, ${\overrightarrow{v}}_3-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_3\right){\overrightarrow{u}}_1-\left({\overrightarrow{u}}_2.{\overrightarrow{v}}_3\right){\overrightarrow{u}}_2$.

$$\begin{gathered} {\overrightarrow{v}}_3-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_3\right){\overrightarrow{u}}_1-\left({\overrightarrow{u}}_2.{\overrightarrow{v}}_3\right){\overrightarrow{u}}_2=\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right]-\left[\begin{array}{c}5\\ 0\\ 0\end{array}\right]-\left[\begin{array}{c}0\\ 6\\ 0\end{array}\right] \\ =\left[\begin{array}{c}0\\ 0\\ 7\end{array}\right] \end{gathered}$$

$$\begin{gathered} ||{\overrightarrow{v}}_3-\left({\overrightarrow{u}}_1.{\overrightarrow{v}}_3\right){\overrightarrow{u}}_1-\left({\overrightarrow{u}}_2.{\overrightarrow{v}}_3\right){\overrightarrow{u}}_2||=\sqrt{0^2+0^2+7^2} \\ =7 \end{gathered}$$

Finally, find $\overrightarrow{u_3}$.

$$\begin{gathered} \overrightarrow{u_3}=\frac{1}{7}\left[\begin{array}{c}0\\ 0\\ 7\end{array}\right] \\ =\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right] \end{gathered}$$

Thus, the values of ${\overrightarrow{u}}_1$,${\overrightarrow{u}}_2$and ${\overrightarrow{u}}_3$are ${\overrightarrow{u}}_1=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],{\overrightarrow{\mu }}_2=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right],{\overrightarrow{u}}_3=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$

Hence, the orthonormal vectors are ${\overrightarrow{u}}_1=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],{\overrightarrow{\mu }}_2=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right],{\overrightarrow{u}}_3=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$.