# How can you convert from noraml to parameter form?

• Nov 17th 2007, 09:09 AM
TriKri
Linear Algebra: Convert sub room from normal to parameter form?
$\left\{\begin{array}{ll}x_{1}+x_{2}+x_{3}+x_{4}&=0 \\x_{1}+x_{2}-x_{3}&=0\end{array}\right.$
Linear algebra. Find a base that generates the vector room the equation system describes.

Just an example. I know one can find vectors which is common for the both equations by just looking and them and figuring out, but is there some method that will always work?

If you have j equations in a k-dimensional room, how can you get a base for the sub room you will get (the sub room written in parameter form)? (The sub room will have the dimension k-j, iff no of the equations is a linear combination of the other equations.)
• Nov 18th 2007, 04:19 PM
TriKri
I have now changed my post to give a better explaination of what I mean.
• Nov 21st 2007, 06:59 PM
TriKri
$\left\{\begin{array}{ll}x_{1}+x_{2}+x_{3}+x_{4}&=0 \\x_{1}+x_{2}-x_{3}&=0\end{array}\right.$

I think I have found a way.

The sub room described by the system goes through origo, since the equation system is homogen; hence it will have no constant when it's written in parameter form. In this case the sub room that's generated will have the dimension 2, hence we need to have two parameters, even if we are gonna find that out later. Eventually we would like to have the equation in this form:

$X:\ s\cdot\bar{u}\ +\ t\cdot\bar{v}$

where $\bar{u}$ and $\bar{v}$ are our base vectors. I generally don't draw arrows over the vector, but a line, because I'm lazy.

Or to make it more clear, we could write it in this form:

$\left(\begin{array}{c}x_{1}\\x_{2}\\x_{3}\\x_{4}\e nd{array}\right) = s\cdot\left(\begin{array}{c}u_{1}\\u_{2}\\u_{3}\\u _{4}\end{array}\right) + t\cdot\left(\begin{array}{c}v_{1}\\v_{2}\\v_{3}\\v _{4}\end{array}\right)$

$\left(\begin{array}{cccc}1&1&1&1\\1&1&-1&0\end{array}\right)\cdot X\ =\ \left(\begin{array}{c}0\\0\end{array}\right)$

We eliminate as far as we can:

$\left(\begin{array}{cccc}1&1&1&1\\0&0&-2&-1\end{array}\right)\cdot X\ =\ \left(\begin{array}{c}0\\0\end{array}\right)\ \Leftrightarrow\ \left(\begin{array}{cccc}1&1&1&1\\0&0&2&1\end{arra y}\right)\cdot X\ =\ \left(\begin{array}{c}0\\0\end{array}\right)$

Now, let's create a parameter $t$, and let it be equal to $x_{4}$ (since $x_{4}$ is the last $x$ with a non-zero coefficient in the last row with more than one non-zero coefficients, in the matrix before $X$):

$\left(\begin{array}{cccc}1&1&1&1\\0&0&2&1\\0&0&0&1 \end{array}\right)\cdot X\ =\ \left(\begin{array}{c}0\\0\\0\end{array}\right)\ +\ t\cdot \left(\begin{array}{c}0\\0\\1\end{array}\right)$

The row that's added represents the relation $x_{4}\ =\ t$; the other rows is affected, but the equations they represent is not since $t$ comes with a factor $0$.

Now, let's substitute $x_{4}$ with $t$ in the rest of the equations (and we can skip the null vector):

$\left(\begin{array}{cccc}1&1&1&0\\0&0&2&0\\0&0&0&1 \end{array}\right)\cdot X\ =\ t\cdot \left(\begin{array}{c}-1\\-1\\1\end{array}\right)\ \Leftrightarrow\ \left(\begin{array}{cccc}1&1&0&0\\0&0&2&0\\0&0&0&1 \end{array}\right)\cdot X\ =\ t\cdot \left(\begin{array}{c}-0.5\\-1\\1\end{array}\right)$

And now, since row 1 is the last row with more than one non-zero coefficient in the matrix before $X$, and $x_{2}$ is the last $x$ in that row with a non-zero coeffisient, let's create a new parameter $s$ which is equal to $x_{2}$ and substitute $x_{2}$ with $s$ in the rest of the rows:

$\left(\begin{array}{cccc}1&1&0&0\\0&1&0&0\\0&0&2&0 \\0&0&0&1\end{array}\right)\cdot X\ =\ t\cdot \left(\begin{array}{c}-0,5\\0\\-1\\1\end{array}\right)\ +\ s\cdot \left(\begin{array}{c}0\\1\\0\\0\end{array}\right) \Leftrightarrow$

$\left(\begin{array}{cccc}1&0&0&0\\0&1&0&0\\0&0&2&0 \\0&0&0&1\end{array}\right)\cdot X\ =\ t\cdot \left(\begin{array}{c}-0.5\\0\\-1\\1\end{array}\right)\ +\ s\cdot \left(\begin{array}{c}-1\\1\\0\\0\end{array}\right)$

And normalized:

$\left(\begin{array}{cccc}1&0&0&0\\0&1&0&0\\0&0&1&0 \\0&0&0&1\end{array}\right)\cdot X\ =\ t\cdot \left(\begin{array}{c}-0.5\\0\\-0.5\\1\end{array}\right)\ +\ s\cdot \left(\begin{array}{c}-1\\1\\0\\0\end{array}\right)\ \Leftrightarrow$

$\left/\left(\begin{array}{cccc}1&0&0&0\\0&1&0&0\\0&0&1&0 \\0&0&0&1\end{array}\right)\cdot X\ =\ E\cdot X\ =\ X\right/$

$\Leftrightarrow\ X\ =\ t\cdot \left(\begin{array}{c}-0.5\\0\\-0.5\\1\end{array}\right)\ +\ s\cdot \left(\begin{array}{c}-1\\1\\0\\0\end{array}\right)$

We now have

$\left(\begin{array}{c}x_{1}\\x_{2}\\x_{3}\\x_{4}\e nd{array}\right) = s\cdot\left(\begin{array}{c}-1\\1\\0\\0\end{array}\right) + t\cdot\left(\begin{array}{c}-0.5\\0\\-0.5\\1\end{array}\right)$

so our two base vectors is $(-1,\ 1,\ 0,\ 0)$ and $(-0.5,\ 0,\ -0.5,\ 1)$.

Also, if there are $x$:es which has no non-zero coefficient, we need to create parameters for those as well.

If someone knows how to make the vectors which is extracted be orthogonal to each other at the same time, I would like to know.

-Kristofer