Question

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

$\mathbf{1}\mathbf{.}\mathbf{}\left[\begin{array}{c}2\\ 1\\ -2\end{array}\right]$

Short answer

The orthonormal vectors of the sequence$\left[\begin{array}{c}2\\ 1\\ -2\end{array}\right]is\frac{1}{3}\left[\begin{array}{c}2\\ 1\\ -2\end{array}\right].$

Step-by-step solution

Determine the Gram-Schmidt process.

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

Then

${\stackrel{\rightharpoonup }{\mathrm{u}}}_1=\frac{1}{\left|\left|{\stackrel{\rightharpoonup }{\mathrm{v}}}_1\right|\right|}{\stackrel{\rightharpoonup }{\mathrm{v}}}_1,{\stackrel{\rightharpoonup }{\mathrm{u}}}_2=\frac{1}{\left|\left|{\stackrel{\rightharpoonup }{\mathrm{v}}}_2^{\perp }\right|\right|}{\stackrel{\rightharpoonup }{\mathrm{v}}}_2^{\perp },.....,{\stackrel{\rightharpoonup }{\mathrm{u}}}_{\mathrm{j}}=\frac{1}{\left|\left|{\stackrel{\rightharpoonup }{\mathrm{v}}}_{\mathrm{j}}^{\perp }\right|\right|}{\stackrel{\rightharpoonup }{\mathrm{v}}}_{\mathrm{j}}^{\perp },.....,{\stackrel{\rightharpoonup }{\mathrm{u}}}_{\mathrm{m}}=\frac{1}{\left|\left|{\stackrel{\rightharpoonup }{\mathrm{v}}}_{\mathrm{m}}^{\perp }\right|\right|}{\stackrel{\rightharpoonup }{\mathrm{v}}}_{\mathrm{m}}^{\perp }$

According to Gram-Schmidt process

The given vector is $\left[\begin{array}{c}2\\ 1\\ -2\end{array}\right]$ .

Consider the value of ${\stackrel{\rightharpoonup }{u}}_1$ according to Gram-Schmidt process.

${\stackrel{\rightharpoonup }{u}}_1=\frac{1}{\left|\left|{\stackrel{\rightharpoonup }{v}}_1\right|\right|}{\stackrel{\rightharpoonup }{v}}_1.....\left(1\right)$

Here, put the values in equation (1). We obtained

$$\begin{gathered} {\stackrel{\rightharpoonup }{u}}_1=\frac{1}{\sqrt{2^2+1^2+{\left(-2\right)}^2}}.\left[\begin{array}{c}2\\ 1\\ -2\end{array}\right] \\ =\frac{1}{3}.\left[\begin{array}{c}2\\ 1\\ -2\end{array}\right] \end{gathered}$$

Hence, the orthonormal vector is$\frac{1}{3}.\left[\begin{array}{c}2\\ 1\\ -2\end{array}\right]$ .