Question

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

9.$\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]\mathbf{,}\left[\begin{array}{c}1\\ 9\\ -5\\ 3\end{array}\right]$

Short answer

The orthonormal vectors of the sequence is$\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right],\left[\begin{array}{c}1\\ 9\\ -5\\ 3\end{array}\right]\mathrm{is}\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right],\left[\begin{array}{c}-1/10\\ 7/10\\ -7/10\\ 1/10\end{array}\right]$.

Step-by-step solution

Determine the Gram-Schmidt process

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

Then localid="1659439775013" ${\overrightarrow{\mathbf{u}}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\parallel {\overrightarrow{v}}_1\parallel }{\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathbf{u}}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\parallel {\overrightarrow{v}}_2^{\perp }\parallel }{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}^{\mathbf{\perp }}\mathbf{,}\mathbf{\dots }\mathbf{.}\mathbf{,}{\overrightarrow{\mathbf{u}}}_{\mathbf{j}}\mathbf{=}\frac{\mathbf{1}}{\parallel {\overrightarrow{v}}_j^{\perp }\parallel }{\overrightarrow{\mathbf{v}}}_{\mathbf{j}}^{\mathbf{\perp }}\mathbf{,}\mathbf{\dots }\mathbf{.}\mathbf{,}{\overrightarrow{\mathbf{u}}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{1}}{\parallel {\overrightarrow{v}}_m^{\perp }\parallel }{\overrightarrow{\mathbf{v}}}_{\mathbf{m}}^{\mathbf{\perp }}$

Apply the Gram-Schmidt process

The given vectors are${\overrightarrow{v}}_1=\left[\begin{array}{l}1\\ 1\\ 1\\ 1\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{l}1\\ 9\\ -5\\ 3\end{array}\right]$.

Obtain the values of $u_1,u_2$according toGram-Schmidt process.

localid="1659955449914" $\begin{array}{r}{\overrightarrow{u}}_1=\frac{{\overrightarrow{v}}_1}{\parallel \overrightarrow{v}\parallel }\dots \left(1\right)\\ {\overrightarrow{u}}_2=\frac{{\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1\cdot {\overrightarrow{v}}_2\right){\overrightarrow{u}}_1}{∥{\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1\cdot {\overrightarrow{v}}_2\right){\overrightarrow{u}}_1∥}\dots \left(2\right)\end{array}$

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

localid="1659955512481" $$\begin{gathered} {\overrightarrow{u}}_1=\frac{1}{\sqrt{1^2+1^2+1^2+1^2}}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right] \\ =\frac{1}{2}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right] \end{gathered}$$

Findu→2

Now, here is need to find out the values of ${\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1.{\overrightarrow{u}}_2\right)\overrightarrow{u_1}$and$||{\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1.{\overrightarrow{u}}_2\right)\overrightarrow{u_1}||$to obtain the value of ${\overrightarrow{u}}_2$.

Consider the equations below.

$\begin{array}{c}{\overrightarrow{u}}_1\cdot {\overrightarrow{v}}_2=4\\ \left({\overrightarrow{u}}_1\cdot {\overrightarrow{v}}_2\right){\overrightarrow{u}}_1=\left[\begin{array}{l}2\\ 2\\ 2\\ 2\end{array}\right]\end{array}$

And,

$\begin{array}{r}{\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1\cdot {\overrightarrow{v}}_2\right){\overrightarrow{u}}_1=\left[\begin{array}{c}1\\ 9\\ -5\\ 3\end{array}\right]-\left[\begin{array}{c}2\\ 2\\ 2\\ 2\end{array}\right]\\ =\left[\begin{array}{c}-1\\ 7\\ -7\\ 1\end{array}\right]\end{array}$

Then,

localid="1659955715369" $$\begin{gathered} ∥{\overrightarrow{v}}_2-\left({\overrightarrow{u}}_1\cdot {\overrightarrow{v}}_2\right){\overrightarrow{u}}_1∥=\sqrt{1+49+49+1} \\ =\sqrt{100} \\ =10 \end{gathered}$$

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

${\overrightarrow{u}}_2=\frac{1}{10}\left[\begin{array}{c}-1\\ 7\\ -7\\ 1\end{array}\right]$

Thus, the values of ${\overrightarrow{u}}_1$,${\overrightarrow{u}}_2$ are $\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right],\left[\begin{array}{c}-1/10\\ 7/10\\ -7/10\\ 1/10\end{array}\right]$.

Hence, the orthonormal vectors are $\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right],\left[\begin{array}{c}-1/10\\ 7/10\\ -7/10\\ 1/10\end{array}\right]$.