Question

In Exercises 37 through 42 , find a basis$\mathfrak{J}$ of ${\mathbf{ℝ}}^{\mathbf{n}}$ such that the$\mathfrak{J}\mathbf{-}\mathbf{matrix}$ of the given linear transformation T is diagonal.

Orthogonal projection T onto the line in${\mathbf{ℝ}}^{\mathbf{3}}$ spanned by$\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$.

Short answer

The matrix is, $B=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right]\left[\begin{array}{c}-3\\ 0\\ 1\end{array}\right]$

Step-by-step solution

Consider the vector.

The vector is,

$\overrightarrow{V_1}=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$

Consider the conditions to obtain the reflection about the line spanned as follows:

$$\begin{gathered} \mathbf{T}\left({\overrightarrow{v}}_1\right)\mathbf{=}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}} \\ \mathbf{T}\mathbf{\left(}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}\mathbf{\right)}\mathbf{=}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}} \\ \mathbf{T}\mathbf{\left(}{\overrightarrow{\mathbf{v}}}_{\mathbf{3}}\mathbf{\right)}\mathbf{=}{\overrightarrow{\mathbf{v}}}_{\mathbf{3}} \end{gathered}$$

Compute the matrix using linear combination.

Consider the vector:

$\overrightarrow{V_1}=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$

The orthogonal vectors will be,

$\overrightarrow{V_2}=\left[\left.\begin{array}{c}-2\\ 1\\ 0\end{array}\right]{\overrightarrow{V}}_3=\left[\begin{array}{c}-3\\ 0\\ 1\end{array}\right.\right]$

The matrix will be,

$\therefore B=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right]\left[\begin{array}{c}-3\\ 0\\ 1\end{array}\right]$

Final answer.

The matrix is, $\mathrm{B}=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right]\left[\begin{array}{c}-3\\ 0\\ 1\end{array}\right]$