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

MATH221 April 2009

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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.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

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

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

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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 $w_{2}$.

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