Question

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

Orthogonal projection T onto the plane $\mathbf{3}{\mathit{x}}_{\mathbf{1}}\mathbf{+}{\mathit{x}}_{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathit{x}}_{\mathbf{3}}\mathbf{=}\mathbf{0}$ in${\mathbf{ℝ}}^{\mathbf{3}}$.

Short answer

The matrix is,$B=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$

Step-by-step solution

Consider the vector.

The vector is,

$\begin{array}{l}3x+x_{2+}+2x_3=0\\ {\overrightarrow{v}}_1=\left[\begin{array}{c}3\\ 1\\ 2\end{array}\right]\end{array}$

Consider the linear transformation that is projection on the plane is shown below:

$$\begin{gathered} \mathbf{T}\left(\overrightarrow{x}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}} \\ \mathbf{T}\left({\overrightarrow{v}}_1\right)\mathbf{=}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}} \\ \mathbf{T}\mathbf{\left(}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}\mathbf{\right)}\mathbf{=}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}} \\ \mathbf{T}\mathbf{\left(}{\overrightarrow{\mathbf{v}}}_{\mathbf{3}}\mathbf{\right)}\mathbf{=}\mathbf{0} \end{gathered}$$

Compute the matrix using linear combination.

From the transformation the matrix B will be of the form:

$B=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$