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## 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.

## 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.