# One of the five given matrices represents an orthogonal projection onto a line and another represents a reflection about line. Identify both and briefly justify your choice.02510501111.png

Question
Matrices

One of the five given matrices represents an orthogonal projection onto a line and another represents a reflection about line. Identify both and briefly justify your choice.

2021-03-28

Consider L be a line in $$\displaystyle{\mathbb{{{R}}}}^{{3}}$$
Orthogonal projection of $$\displaystyle\vec{{{x}}}$$ onto the line L is,
$$\displaystyle{p}{r}{o}{j}_{{L}}{\left(\vec{{{x}}}\right)}={\left(\vec{{{x}}}\cdot\vec{{{u}}}\right)}\vec{{{u}}}$$
Here $$\displaystyle\vec{{{u}}}$$ is a unit vector parallel to the line L.
Let
$$\vec{u}=\left(\begin{array}{c}u_1\\u_2\\u_3\end{array}\right)$$
Now $$proj_L(\vec{x})=(xu_1+yu_2+zu_3)\left(\begin{array}{c}u_1\\u_2\\u_3\end{array}\right)$$
Consider the matrix $$B=\frac{1}{3}\left(\begin{array}{c}1&1&1\\1&1&1\\1&2&1\end{array}\right)$$
Consider the corresponding linear transformation that representled by the matrix B.
$$B\vec{x}=\frac{1}{3}\left(\begin{array}{c}1&1&1\\1&1&1\\1&1&1\end{array}\right)\vec{x}$$
$$=\frac{1}{3}\left(\begin{array}{c}1&1&1\\1&1&1\\1&1&1\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)$$
$$=\frac{1}{3}\left(\begin{array}{c}x+y+z\\x+y+z\\x+y+z\end{array}\right)$$
$$=(\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z)\left(\begin{array}{c}\frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}\end{array}\right)$$
This linear transformation is of the form (1)
From (1), unit vector is,
$$\vec{u}=\left(\begin{array}{c}\frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}\end{array}\right)$$
The line L is,
$$\displaystyle{x}={y}={z}$$
Hence, the matrix B represents an orthogonal projection onto the line $$\displaystyle{L}:{x}={y}={z}$$
Reflection of $$\displaystyle\vec{{{x}}}$$ onto the line L is,
$$\displaystyle{r}{e}{{f}_{{L}}{\left(\vec{{{x}}}\right)}}={2}{p}{r}{o}{j}_{{L}}{\left(\vec{{{x}}}\right)}-\vec{{{x}}}$$
$$\displaystyle={2}{\left(\vec{{{x}}}\cdot\vec{{{u}}}\right)}\vec{{{u}}}-\vec{{{x}}}$$
Now
$$\displaystyle{r}{e}{{f}_{{L}}{\left(\vec{{{x}}}\right)}}={2}{\left(\vec{{{x}}}\cdot\vec{{{u}}}\right)}\vec{{{u}}}-\vec{{{x}}}$$
$$=2(xu_1+yu_2+zu_3)\left(\begin{array}{c}u_1\\u_2\\u_3\end{array}\right)-\left(\begin{array}{c}x\\y\\z\end{array}\right)$$
Consider the matrix $$E=\frac{1}{3}\left(\begin{array}{c}-1&2&2\\2&-1&2\\2&2&-1\end{array}\right)$$
Consider the corresponding linear transformation that represented by the matrix E.
$$\vec{x}=\frac{1}{3}\left(\begin{array}{c}-1&2&2\\2&-1&2\\2&2&-1\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)$$
$$=\frac{1}{3}\left(\begin{array}{c}-x+2y+2z\\2x-y+2z\\2x+2y-z\end{array}\right)$$
$$=\frac{1}{3}\left(\begin{array}{c}-x+2y+2z\\2x-y+2z\\2x+2y-z\end{array}\right)+\left(\begin{array}{c}x\\y\\z\end{array}\right)-\left(\begin{array}{c}x\\y\\z\end{array}\right)$$
$$=\frac{1}{3}\left(\begin{array}{c}2x+2y+2z\\2x+2y+2z\\2x+2y+2z\end{array}\right)-\left(\begin{array}{c}x\\y\\z\end{array}\right)$$
This linear transformation is of the form
The line L is,
x=y=z
Hence, the matrix E represents reflection onto the line: x=y=z
Note that
$$\displaystyle{r}{e}{{f}_{{L}}{\left(\vec{{{x}}}\right)}}={2}{\left(\vec{{{x}}}\cdot\vec{{{u}}}\right)}\vec{{{u}}}-\vec{{{x}}}$$
$$\displaystyle={2}{p}{r}{o}{j}_{{L}}{\left(\vec{{{x}}}\right)}-\vec{{{x}}}$$
Therefore
2B-E=I

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