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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.

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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 $\ker A^T$. –  copper.hat Nov 4 '12 at 17:37
    
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? –  copper.hat Nov 4 '12 at 17:42
    
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. –  AlexW.H.B. Nov 4 '12 at 17:50
    
Sorry, I haven't had time to look at this. Note that the three parametric vectors above are linearly dependent. –  copper.hat Nov 4 '12 at 20:05
    
Finally got a chance to fill an answer in... –  copper.hat Nov 5 '12 at 5:07

1 Answer 1

up vote 1 down vote accepted

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 ${\cal R} (A)^\bot$.

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Thank you so much for your help. your answer is so thorough. –  AlexW.H.B. Nov 5 '12 at 6:42
    
You are welcome, glad to be of help... –  copper.hat Nov 5 '12 at 6:58
    
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. –  AlexW.H.B. Nov 5 '12 at 10:13
    
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. –  copper.hat Nov 5 '12 at 16:02

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