Science:Math Exam Resources/Courses/MATH152/April 2017/Question B 06 (b)

MATH152 April 2017

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Question B 06 (b)
The matrix A below represents a transformation that reflects vectors in $\mathbb {R} ^{3}$ in a plane $P$ that contains the origin. ${\begin{bmatrix}1/3&2/3&-2/3\\2/3&1/3&2/3\\-2/3&2/3&1/3\end{bmatrix}}$ Note: you can use geometric insight to greatly simplify the calculations below. (b) Find an equation for $P$ (equation form).

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
How does the given matrix $A$ transform the vectors on the plane?

Hint 2
(For alternative solution) draw the columns of this matrix as vectors in $\mathbb {R} ^{3}$.

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 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Since the transform reflects vectors with respect to the plane $P$, each vector on the plane is remained as it is, after applying the transformation. In other words, if a vector $v$ is in $P$, then $Av=v$. This means that the eigenspace of the eigenvalue $1$ is the plane $P$ which we are looking for. To find the eigenvectors corresponding to eigenvalue $1$, we solve the equation $(A-I)v=0\iff \left({\begin{bmatrix}1/3&2/3&-2/3\\2/3&1/3&2/3\\-2/3&2/3&1/3\end{bmatrix}}-{\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}}\right)v={\begin{bmatrix}-2/3&2/3&-2/3\\2/3&-2/3&2/3\\-2/3&2/3&-2/3\end{bmatrix}}v={\frac {2}{3}}{\begin{bmatrix}-1&1&-1\\1&-1&1\\-1&1&-1\end{bmatrix}}{\begin{bmatrix}v_{1}\\v_{2}\\v_{3}\end{bmatrix}}=0.$ This solution $v={\begin{bmatrix}v_{1}&v_{2}&v_{3}\end{bmatrix}}^{T}$ to this equation satisfies $v_{1}-v_{2}+v_{3}=0$, so that it can be written as an parameter equation form for the parameters $v_{1}$ and $v_{3}$; $v=v_{1}{\begin{bmatrix}1\\1\\0\end{bmatrix}}+v_{3}{\begin{bmatrix}0\\1\\1\end{bmatrix}}$ Therefore, two linearly independent eigenvectors for the eigenvalue $1$ are $e_{1}={\begin{bmatrix}1&1&0\end{bmatrix}}^{T}$ and $e_{2}={\begin{bmatrix}0&1&1\end{bmatrix}}^{T}$, and hence the corresponding eigenspace has its normal vector $n=e_{1}\times e_{2}={\begin{vmatrix}i&j&k\\1&1&0\\0&1&1\end{vmatrix}}={\begin{bmatrix}1&-1&1\end{bmatrix}}^{T}.$ Using a vector $e_{1}$ on the plane $P$, we have the equation for the plane $P$ as $n\cdot ({\text{x}}-e_{1})=0\iff \langle 1,-1,1\rangle \cdot \langle x-1,y-1,z\rangle =(x-1)-(y-1)+z=x-y+z=0$ Answer: $\color {blue}x-y+z=0$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. (Alternative solution) Let $A$ be the matrix in this question, and consider the standard basis vectors for $\mathbb {R} ^{3}$ $e_{1}={\begin{pmatrix}1\\0\\0\end{pmatrix}},e_{2}={\begin{pmatrix}0\\1\\0\end{pmatrix}},e_{3}={\begin{pmatrix}0\\0\\1\end{pmatrix}}.$ Their images under this transformation are the first, second, and third columns of $A$ respectively. Because this transformation is the reflection of $\mathbb {R} ^{3}$ about the plane $P$, the points ${\frac {1}{2}}(e_{1}+Ae_{1}),{\frac {1}{2}}(e_{2}+Ae_{2}),{\frac {1}{2}}(e_{3}+Ae_{3})$ all lie in $P$ as the norm of each of the columns of $A$ is equal to the norm of each of the standard basis vectors. The equation of $P$ is of the form $ax+by+cz={\begin{pmatrix}a&b&c\end{pmatrix}}{\begin{pmatrix}x\\y\\z\end{pmatrix}}=0$ where ${\begin{pmatrix}a\\b\\c\end{pmatrix}}$ is any vector orthogonal to all vectors in $P$. To find such a vector, we calculate two non-zero points in $P$: ${\frac {1}{2}}(e_{1}+Ae_{1})={\frac {1}{2}}({\begin{pmatrix}1\\0\\0\end{pmatrix}}+{\begin{pmatrix}1/3\\2/3\\-2/3\end{pmatrix}})={\begin{pmatrix}2/3\\1/3\\-1/3\end{pmatrix}}{\frac {1}{2}}(e_{2}+Ae_{2})={\frac {1}{2}}({\begin{pmatrix}0\\1\\0\end{pmatrix}}+{\begin{pmatrix}2/3\\1/3\\2/3\end{pmatrix}})={\begin{pmatrix}1/3\\2/3\\1/3\end{pmatrix}}$ Hence, ${\begin{pmatrix}2\\1\\-1\end{pmatrix}}$ and ${\begin{pmatrix}1\\2\\1\end{pmatrix}}$ are in $P$. So to find a vector orthogonal to all vectors in $P$, solve the following system of equations $2a+b-c=0$ $a+2b+c=0$ To that end, we apply Gaussian elimination to ${\begin{pmatrix}2&1&-1&|&0\\1&2&1&|&0\end{pmatrix}}$ to obtain ${\begin{pmatrix}0&1&1&|&0\\1&0&-1&|&0\end{pmatrix}}$ in other words, the set of solutions is given by $y=-z$ and $x=z$ . In particular, we can pick ${\begin{pmatrix}1\\-1\\1\end{pmatrix}}$ as a solution. Hence, the equation of the plane $P$ is $\color {blue}x-y+z={\begin{pmatrix}x&y&z\end{pmatrix}}{\begin{pmatrix}1\\-1\\1\end{pmatrix}}=0$