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 $\displaystyle \left[\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}\right]$
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 $\displaystyle \left[\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}\right]$
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).
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?
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]
$