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

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- Oct 20th 2009, 03:58 PMZockenOne-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 PMJose27
How about $\displaystyle (t,s,-(s+t))$ $\displaystyle s,t \in \mathbb{R}$

- Oct 20th 2009, 05:36 PMZocken
Thanks Jose, may I ask how you got that answer??

- Oct 20th 2009, 05:45 PMJose27
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 PMhjortur
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 PMZockenQuote:

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 21st 2009, 03:42 AMhjortur
- Oct 26th 2009, 04:24 PMZocken
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 AMhjortur
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.