Question

In Exercises37 through 42 , find a basis $\mathfrak{I}$ of localid="1660372956863" ${\mathbf{ℝ}}^{\mathbf{n}}$ such that the localid="1660373301403" $\mathfrak{I}\mathbf{-}\mathbf{matrix}\mathbf{}\mathbf{}\mathit{B}\mathit{}$of the given linear transformation T is diagonal.

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

Short answer

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

Step-by-step solution

Consider the vector.

The vector is,

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

Consider the linear projection Tthat is projection on the line spanned by $\overrightarrow{v_1}$is:

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

Compute the matrix using linear combination.

From the transformation the required matrix is:

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

The matrix will be,

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