# One-to-one parameterization

• Oct 20th 2009, 03:58 PM
Zocken
One-to-one parameterization
Let A=$\displaystyle \left[\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right]$

Give a one-to-one parameterization of the ker(A)

Would it just be $\displaystyle \left[\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}\right]$
• Oct 20th 2009, 05:15 PM
Jose27
How about $\displaystyle (t,s,-(s+t))$ $\displaystyle s,t \in \mathbb{R}$
• Oct 20th 2009, 05:36 PM
Zocken
• Oct 20th 2009, 05:45 PM
Jose27
Notice that $\displaystyle x \in kerA$ iff $\displaystyle Ax=0$. If $\displaystyle x=(x_1,x_2,x_3)$ then we get three equations that are exactly the same $\displaystyle x_1+x_2+x_3=0$. So given any values for $\displaystyle x_1$ and $\displaystyle x_2$ $\displaystyle x_3$ is completely determined (i.e. $\displaystyle kerA$ has dimension 2).
• Oct 20th 2009, 06:16 PM
hjortur
If you multiply a general vector (on the right side!) v=(x,y,z)' with the given matrix,
you will end up with the vector (x+y+z,x+y+z,x+y+z)' <-- (transposed).

Now if A reprisents a linear map h:V->W. Then ker(A) corresponds to the
subspace which image under h is 0.

Now what that means is what does v=(x,y,z) have to be so that when you multiply it
with A you get 0?
• Oct 20th 2009, 06:29 PM
Zocken
Quote:

Now what that means is what does v=(x,y,z) have to be so that when you multiply it
with A you get 0?
I don't know, what method should I use? I am used to row reducing but in this case it looks like the bottom two rows will become 0 0 0.
• Oct 21st 2009, 03:42 AM
hjortur
Quote:

Originally Posted by Zocken
I don't know, what method should I use? I am used to row reducing but in this case it looks like the bottom two rows will become 0 0 0.

That just means you have 2 free variables. That is the reason you
can parametrise the solution.

Do you understand that, this subspace is the kernel of A?
• Oct 26th 2009, 04:24 PM
Zocken
One last question...Let b=Ae1. Find a one to one parameterization of the solution set Ax=b.

Would this makes sense using the above answer?

$\displaystyle e1+t\left[\begin{matrix} 1 \\ 0 \\ -1 \end{matrix}\right]+s\left[\begin{matrix} 0 \\ 1 \\ -1 \end{matrix}\right]$
• Oct 27th 2009, 03:57 AM
hjortur
Yes, that is correct.

The solution describes a plane with normal vector (1,1,1) going through the point e1.

So the function is not one-to-one , because all the points on that plane will go to b.