Question

Consider the inner product

$\left(\stackrel{r}{v},\stackrel{r}{w}\right)\mathbf{=}\stackrel{\mathbf{r}}{{\mathbf{v}}^{\mathbf{T}}}\left[\begin{array}{cc}1& 2\\ 2& 8\end{array}\right]\stackrel{\mathbf{r}}{\mathbf{w}}$in ${\mathit{R}}^{\mathbf{2}}$. See Exercise 19.

a.Find all vectors in${\mathit{R}}^{\mathbf{2}}$ that are perpendicular to$\left[\begin{array}{c}1\\ 0\end{array}\right]$ with respect to this inner product.

b.Find an orthonormal basis of${\mathit{R}}^{\mathbf{2}}$ with respect to this inner product.

Short answer

(a) The perpendicular vector is $\left[\begin{array}{c}-2\\ 1\end{array}\right]$.

(b) The set will be span $\left[\begin{array}{c}-2\\ 1\end{array}\right]$.

Step-by-step solution

Consider for part (a).

(a)

To find the all vectors which are perpendicular to$\left[\begin{array}{c}1\\ 0\end{array}\right]$ consider the solution

The set is found as:

$$\begin{gathered} A=\left\{\left[\begin{array}{c}x\\ y\end{array}\right]:\left[\begin{array}{c}x\\ y\end{array}\right]\perp \left[\begin{array}{c}1\\ 0\end{array}\right]\right\} \\ =\left\{\left[\begin{array}{c}x\\ y\end{array}\right]:\left(\left[\begin{array}{c}x\\ y\end{array}\right]\perp \left[\begin{array}{c}1\\ 0\end{array}\right]\right)=0\right\} \\ =\left\{\left[\begin{array}{c}x\\ y\end{array}\right]:\left[xY\right]\left[\begin{array}{cc}1& 2\\ 4& 8\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\right\} \\ =\left\{\left[\begin{array}{c}-2y\\ y\end{array}\right]:y\in R\right\} \\ =\left\{\left[\begin{array}{c}-2\\ 1\end{array}\right]\right\} \end{gathered}$$

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

Consider for part (b).

(b)

From the part (a) the vectors$\left[\begin{array}{c}1\\ 0\end{array}\right]$ and$\left[\begin{array}{c}-2\\ 1\end{array}\right]$ are orthogonal, it only needs to make them unit length. Since,

$$\begin{gathered} length=||\left[\begin{array}{c}-2\\ 1\end{array}\right]|| \\ =\sqrt{4} \\ =2 \end{gathered}$$

The orthonormal basis is $\left\{\left[\begin{array}{c}-2\\ 1\end{array}\right],\left[\begin{array}{c}1\\ 0\end{array}\right]\right\}$.

Hence, the set will be span $\left[\begin{array}{c}-2\\ 1\end{array}\right]$.