Question

In Exercises 37 through 42 , find a basis$\mathfrak{I}\mathbf{}\mathit{o}\mathit{f}\mathbf{}{\mathbf{ℝ}}^{\mathbf{n}}$ such that the $\mathfrak{I}\mathbf{}\mathbf{-}\mathit{m}\mathit{a}\mathit{t}\mathit{r}\mathit{i}\mathit{x}\mathbf{}$B of the given linear transformation T is diagonal.

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

Short answer

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

Step-by-step solution

Consider the vector.

The vector is,

${\overrightarrow{v}}_1=\left[\begin{array}{c}2\\ 3\end{array}\right]$

Consider the other vector ${\overrightarrow{v}}_2$ that is perpendicular to ${\overrightarrow{v}}_1$ to obtain the other matrix of the form:

$\mathit{B}\mathbf{=}\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$

Compute the matrix using linear combination.

Consider the vector

${\overrightarrow{v}}_1=\left[\begin{array}{c}2\\ 3\end{array}\right]$

The orthogonal vector will be,

localid="1660636146637" ${\overrightarrow{v}}_2=\left[\begin{array}{c}3\\ -2\end{array}\right]$

The matrix will be,

localid="1660636098121" $\therefore B=\left[\begin{array}{c}2\\ 3\end{array}\right]\left[\begin{array}{c}3\\ -2\end{array}\right]$

Final answer.

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