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Chapter 4

Q. 4.6.2

Let \vec{v} = \begin{bmatrix}3\\4\end{bmatrix}.In Example 3.2.5, the standard matrix of the linear operator proj v : \R² → \R² was found to be

[proj_{\vec{v}}]_{S}  =\begin{bmatrix}9/25&12/25\\12/25&16/25\end{bmatrix}

Find the matrix of proj_{\vec{v}} with respect to a basis that shows the geometry of the transformation more clearly.

Step-by-Step

Verified Solution

For this linear transformation, it is natural to use a basis for \R² consisting of the vector \vec{v}, which is the direction vector for the projection, and a second vector orthogonal to \vec{v}, say  \vec{w} =\begin{bmatrix}-4\\     3\end{bmatrix}.  Then, with  \mathcal{B} = {\vec{v}, \vec{w}}, by geometry,

proj_{\vec{v}}  \vec{v} = proj_{\vec{v}} \begin{bmatrix}3\\4\end{bmatrix} =\begin{bmatrix}3\\4\end{bmatrix} = 1 \vec{v} +0\vec{w}

proj_{\vec{v}}  \vec{w} = proj_{\vec{v}} \begin{bmatrix}-4\\     3\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix} = 0\vec{v} +0\vec{w}

Hence, [proj_{\vec{v}}  \vec{v}]_{\mathcal{B}} =\begin{bmatrix}1\\0\end{bmatrix} and  [proj_{\vec{v}}  \vec{w}]_{\mathcal{B}} = \begin{bmatrix}0\\0\end{bmatrix}.Thus,

[proj_{\vec{v}}]_{\mathcal{B}}=\left[ [proj_{\vec{v}}  \vec{v}]_{\mathcal{B}}        [proj_{\vec{v}}  \vec{w}]_{\mathcal{B}}\right]=\begin{bmatrix}1&0\\0&0\end{bmatrix}

We now consider proj_{\vec{v}}  \vec{x}  for  any  \vec{x} ∈ \R². We have

[proj_{\vec{v}}  \vec{x}]_{\mathcal{B}}=[proj_{\vec{v}}]_{\mathcal{B}}[\vec{x}]_{\mathcal{B}}

                                                         =\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}b_{1}\\b_{2}\end{bmatrix}=\begin{bmatrix}b_{1}\\0\end{bmatrix}

In terms of \mathcal{B}-coordinates, proj_{\vec{v}} is described as the linear mapping that sends \begin{bmatrix}b_{1}\\b_{2}\end{bmatrix} to \begin{bmatrix}b_{1}\\0\end{bmatrix}.

This simple geometrical description is obtained when we use a basis \mathcal{B} that is adapted to the geometry of the transformation (see Figure 4.6.4 ). This example will be discussed further below.

4.6.4