1
$\begingroup$

How do you solve a problem like this. I'm completely stumped. it seems like there should be an easy solution but I'm obviously over looking it. any help would be greatly appreciated.

enter image description here

  • 0
    You have a description of the form $x_0+A\mu$, where $A$ has the 3 columns above. You need to find two vectors in $\mathbb{R}^5$ that are orthogonal to the columns of $A$, that is vectors that span $\ker A^T$.2012-11-04
  • 0
    Sorry, I'm assuming by 'Cartesian' you mean a description of the form $Mx=b$ for some $2\times5$ matrix $M$ and some vector $b$. Is that correct?2012-11-04
  • 0
    first of all thank you very much for your quick response. secondly this is a homework problem for my linear algebra class, so I'm not 100% sure if my its supposed to be in the Mx=b form, but I would assume so.2012-11-04
  • 0
    Sorry, I haven't had time to look at this. Note that the three parametric vectors above are linearly dependent.2012-11-04
  • 0
    Finally got a chance to fill an answer in...2012-11-05

1 Answers 1

1

Let $x_0 = (1, 0,2,1,0)^T$, $a_0 = (-2, -1, 2, -1 , 1)^T$, $a_1=(0,2,-2,1,-1)^T$, $a_2=(-4,0,2,-1,1)^T$.

Let $\cal P$ denote the set in question, ie, ${\cal P} = \{ x_0+\sum_{k=0}^2 \lambda_k a_k \}_{\lambda \in \mathbb{R}^3}$. A quick check shows that $a_0 = \frac{1}{2} (a_2-a_1)$, so in fact ${\cal P} = \{ x_0+\sum_{k=1}^2 \lambda_k a_k \}_{\lambda \in \mathbb{R}^2}$, and $a_1, a_2$ are linearly independent. Let $A=\begin{bmatrix} a_1 & a_2 \end{bmatrix}$. Then, with a slight abuse of notation, we can write ${\cal P} = x_0+{\cal R} (A)$, where ${\cal R} (A)$ denotes the range of $A$.

The goal is to find a matrix $M$ such that ${\cal P} = \{ x | M(x-x_0) = 0\} = \{ v+x_0\}_{v \in \ker M}$. Equivalently, we want to find $M$ such that ${\cal P}-x_0 = {\cal R} (A) = \ker M$. (The point being that $x_0$ is sort of irrelevant here.)

We note that $x \in {\cal R} (A) $ iff $x \bot {\cal R} (A)^\bot$, so if we can find a basis $c_1,...,c_k$ for ${\cal R} (A)^\bot$, then letting $C = \begin{bmatrix} c_1 & \cdots & c_k \end{bmatrix}$, we have $x \in {\cal R} (A) $ iff $C^T x = 0$. (Then letting $M=C^T$ finishes the problem.) We note in passing that since $\dim {\cal R} (A) = 2$, we have $\dim {\cal R} (A)^\bot = 3$.

There are many ways to find a basis of ${\cal R} (A)^\bot$. Tedious inspection yields $$C=\begin{bmatrix} 0 & 1 & -2 \\ 0 & 2 & -4 \\ 0 & 1 & -18 \\ 1 & -1 & -14 \\ 1 & 1 & 14 \end{bmatrix}$$

From this we obtain the desired description $\cal P$ is the set of $(x_1,...,x_5) \in \mathbb{R}^5$ that satisfy: \begin{eqnarray} (x_4-1)+x_5 & = & 0 \\ (x_1-1) + 2 x_2 + (x_3-2) - (x_4-1) + x_5 & = & 0 \\ -2(x_1-1) -4 x_2 -18(x_3-2) - 14(x_4-1) + 14 x_5 & = & 0 \\ \end{eqnarray}

A more computational approach would be to compute matrix $\Pi$ of the orthogonal projection onto ${\cal R} (A)^\bot$, and select a maximal set of linearly independent columns of $\Pi$. From the least squares problem, it is straightforward to show that $\Pi = A(A^TA)^{-1} A^T - I$. Another tedious computation shows that $$46 \, \Pi =\begin{bmatrix} 6 & 12 & 8 & -4 & 4 \\ 12 & 24 & 16 & -8 & 8 \\ 8 & 16 & 26 & 10 & -10 \\ -4 & -8 & 10 & 41 & 5 \\ 4 & 8 & -10 & 5 & 41 \end{bmatrix}$$ It is straightforward to verify that columns $2,4$ of $\Pi$ are multiples of columns $1,5$, hence columns $1,3,5$ provide another basis for ${\cal R} (A)^\bot$.

  • 0
    Thank you so much for your help. your answer is so thorough.2012-11-05
  • 0
    You are welcome, glad to be of help...2012-11-05
  • 0
    quick question how did you calculate the C values? is it A=[a1 a2]= c , and just do row reduction? that is the only part i did not completely follow.2012-11-05
  • 0
    Just be inspection (very tedious). I noticed that the last two components of the $a_k$ were of the form $x,-x$ so that gave the first vector $(0,0,0,1,1)^T$. Then I looked for a vector orthogonal to the $a_k$ and the first vector of $C$. No recipe, just guesswork. That's why I gave the prescriptive version.2012-11-05