Science:Math Exam Resources/Courses/MATH221/April 2013/Question 10

{{#incat:MER QGQ flag|{{#incat:MER QGH flag|{{#incat:MER QGS flag|}}}}}}

MATH221 April 2013

  •  Q1  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q3  •  Q4  •  Q5  •  Q6  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q8  •  Q9  •  Q10  •  Q11  •  Q12 (a)  •  Q12 (b)  •  Q12 (c)  •  

   Other MATH221 Exams

  •  December 2008  •  December 2009  •  December 2007  •  April 2010  •  April 2009  •  December 2011  •  April 2013  •  

Question 10
Let $W={\text{span}}\left\{\left[{\begin{array}{c}1\\2\\3\\4\end{array}}\right],\left[{\begin{array}{c}1\\0\\1\\0\end{array}}\right],\left[{\begin{array}{c}0\\1\\1\\2\end{array}}\right]\right\}$ Find a basis for the orthogonal complement $W^{\bot }$ of W.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Are all three of the vectors linearly independent? If not, what is the dimension (and basis) of W. What will the dimension of the basis of $W^{\bot }$ be?

Checking a solution serves two purposes: helping you if, after having used the hint, 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

We first note that there is a non-trivial linear relationship between the three vectors, namely $\left[{\begin{array}{cccc}1&2&3&4\\\end{array}}\right]=\left[{\begin{array}{cccc}1&0&1&0\\\end{array}}\right]+2\left[{\begin{array}{cccc}0&1&1&2\\\end{array}}\right]$ Thus, we get that $W$ has dimension 2 and thus $W^{\bot }$ is also 2 dimensional. Let $V$ be the matrix of basis vectors to $W$, $V={\begin{bmatrix}1&0\\0&1\\1&1\\0&2\\\end{bmatrix}}.$ If $x=[a,b,c,d]^{T}$ is a basis vector to $W^{\bot }$ then orthogonality tells us that ${\begin{aligned}V^{T}x&=0\\{\begin{bmatrix}1&0&1&0\\0&1&1&2\end{bmatrix}}{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}}&=\mathbf {0} \end{aligned}}$ Computing this we get the following relationships ${\begin{aligned}a+c&=0\\b+c+2d&=0\end{aligned}}$ Since the dimension (or rank) of V is two we will have two free parameters. In particular, this means that the basis is not unique (it never is). We choose $a$ and $d$ as free parameters to get ${\begin{aligned}c&=-a\\b&=-2d-c=-2d+a\end{aligned}}$ and finally that $x={\begin{bmatrix}a\\-2d+a\\-a\\d\end{bmatrix}}=a{\begin{bmatrix}1\\1\\-1\\0\end{bmatrix}}+d{\begin{bmatrix}0\\-2\\0\\1\end{bmatrix}}$ Therefore the two orthogonal vectors $\left[{\begin{array}{cccc}1&1&-1&0\\\end{array}}\right],\left[{\begin{array}{cccc}0&-2&0&1\\\end{array}}\right]$ are a basis for $W^{\bot }$. Notice that the dimension of $W^{\bot }$ is two, which happens to be the same dimension as that of $W$.

{{#incat:MER CT flag||

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER QGT flag, MER Tag Orthogonal complement, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag

}}