Science:Math Exam Resources/Courses/MATH221/April 2009/Question 04

MATH221 April 2009

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[hide] Question 04
Let $W={\text{span}}\{w_{1},w_{1}\}$ , where $w_{1}={\begin{pmatrix}3\\0\\4\end{pmatrix}}$ and $w_{2}={\begin{pmatrix}4\\5\\-3\end{pmatrix}}$ . If $T:\mathbb {R} ^{3}\to \mathbb {R} ^{3}$ is the orthogonal projection onto W, find the standard matrix of T.

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[show] Hint 1
Recall that the orthogonal projection to W will project vectors onto W itself.

[show] Hint 2
How can we write any vector in W in terms of the vectors $w_{1}$ and $w_{2}$ ?

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[show] Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Recall that the orthogonal projection onto W are all of the vectors that are orthogonal to the normal of W. Therefore, the orthogonal projection onto W will place objects onto W itself. If we can find an orthonormal basis for this plane $\{v_{1},v_{2}\}$ then the orthogonal projection onto W ( the linear map T), will be given by $T(v)=\langle v,v_{1}\rangle v_{1}+\langle v,v_{2}\rangle v_{2}.$ for $v\in \mathbb {R} ^{3}$ where $\langle \cdot \rangle$ is the standard dot product or inner product. We first note that in fact $w_{1}$ and $w_{2}$ are orthogonal (check!), and therefore $\{w_{1},w_{2}\}$ is an orthogonal basis for W. We can form the orthonormal basis by transforming $\{w_{1},w_{2}\}$ into unit vectors by setting $v_{1}={\frac {w_{1}}{\\|w_{1}\\|}}={\frac {1}{5}}{\begin{pmatrix}3\\0\\4\end{pmatrix}}$ and $v_{2}={\frac {w_{2}}{\\|w_{2}\\|}}={\frac {1}{5{\sqrt {2}}}}{\begin{pmatrix}4\\5\\-3\end{pmatrix}}$ . Now the standard matrix of T is the matrix of T relative to the standard basis $e_{1}={\begin{pmatrix}1\\0\\0\end{pmatrix}},e_{2}={\begin{pmatrix}0\\1\\0\end{pmatrix}},e_{3}={\begin{pmatrix}0\\0\\1\end{pmatrix}},$ i.e. it is the matrix generated by transforming each of the unit vectors where the first column of T is the transformation of $e_{1}$ , the second column the transformation of $e_{2}$ , and the third column the transformation of $e_{3}$ . Let us now determine these coefficients for each unit vector. For the first unit vector, $T(e_{1})=\langle e_{1},v_{1}\rangle v_{1}+\langle e_{1},v_{2}\rangle v_{2}={\frac {3}{5}}v_{1}+{\frac {4}{5{\sqrt {2}}}}v_{2}={\frac {3}{25}}{\begin{pmatrix}3\\0\\4\end{pmatrix}}+{\frac {4}{50}}{\begin{pmatrix}4\\5\\-3\end{pmatrix}}={\frac {17}{25}}e_{1}+{\frac {2}{5}}e_{2}+{\frac {6}{25}}e_{3},$ so the first column of the standard matrix of T is ${\begin{bmatrix}17/25\\2/5\\6/25\end{bmatrix}}$ . Continuing for the second unit vector, $T(e_{2})=\langle e_{2},v_{1}\rangle v_{1}+\langle e_{2},v_{2}\rangle v_{2}={\frac {1}{\sqrt {2}}}v_{2}={\frac {1}{10}}{\begin{pmatrix}4\\5\\-3\end{pmatrix}}={\frac {2}{5}}e_{1}+{\frac {1}{2}}e_{2}+(-{\frac {3}{10}})e_{3},$ so the second column is ${\begin{bmatrix}2/5\\1/2\\-3/10\end{bmatrix}}$ . Finally for the third unit vector, $T(e_{3})=\langle e_{3},v_{1}\rangle v_{1}+\langle e_{3},v_{2}\rangle v_{2}={\frac {4}{5}}v_{1}-{\frac {3}{5{\sqrt {2}}}}v_{2}={\frac {4}{25}}{\begin{pmatrix}3\\0\\4\end{pmatrix}}-{\frac {3}{50}}{\begin{pmatrix}4\\5\\-3\end{pmatrix}}={\frac {6}{25}}e_{1}+(-{\frac {3}{10}})e_{2}+{\frac {41}{50}}e_{3},$ so the third column is ${\begin{bmatrix}6/25\\-3/10\\41/50\end{bmatrix}}$ . We therefore conclude that the standard matrix of T (denoted $[T]$ ) is $[T]={\begin{pmatrix}17/25&2/5&6/25\\2/5&1/2&-3/10\\6/25&-3/10&41/50\end{pmatrix}}.$ Is there anyway we can think to check our answer? The original vectors $w_{1}$ and $w_{2}$ already belong to W and therefore if we project them onto W then nothing should change. For example take $w_{1}$ and project $T(w_{1})={\begin{pmatrix}17/25&2/5&6/25\\2/5&1/2&-3/10\\6/25&-3/10&41/50\end{pmatrix}}{\begin{pmatrix}3\\0\\4\end{pmatrix}}={\begin{pmatrix}3\\0\\4\end{pmatrix}}$ and indeed we have it projects onto itself! The same thing would happen if we projected $w_{2}$ .