1. ## One-to-one parameterization

Let A= $

\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 $\left[\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}\right]$

2. How about $(t,s,-(s+t))$ $s,t \in \mathbb{R}$

4. Notice that $x \in kerA$ iff $Ax=0$. If $x=(x_1,x_2,x_3)$ then we get three equations that are exactly the same $x_1+x_2+x_3=0$. So given any values for $x_1$ and $x_2$ $x_3$ is completely determined (i.e. $kerA$ has dimension 2).

5. 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?

6. 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.

7. 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?

8. 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?

$

e1+t\left[\begin{matrix} 1 \\ 0 \\ -1 \end{matrix}\right]+s\left[\begin{matrix} 0 \\ 1 \\ -1 \end{matrix}\right]
$

9. 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.