Understanding Rank One Matrices and Their Application in Nullspace and Row Space

[applechu]

Hi:
I see an principle about rank one matrice in the book, and it says
if u=(1,2,3), $\nu$^t=[1 3 10], with Ax=0,
the equation $\nu$^tx=0;
The problem is I see an example like following:
s1=[-3
1
0]

s2=[-10
0
1]
The nullspace contains all combination of s1 and s2. and produces the plane
x+3y+10z=0, perpendicular to row(1,3,10). And it lead to the result
Nullspace perpendicular to row space. I didn't know what the result means and
how its imply, could anyone give me any instruct about that, thanks.

 

[Number Nine]

...go on.

 

[applechu]

sorry, rewrite the s1, s2;

s1=\begin{pmatrix}
-3\\
1\\
0
\end{pmatrix}
and s2= \begin{pmatrix}
-10\\
0\\
1
\end{pmatrix}

 

[HallsofIvy]

Hi:
I see an principle about rank one matrice in the book, and it says
if u=(1,2,3), $\nu$^t=[1 3 10], with Ax=0,

You defined u and $\nu^t$ but what is A and what is x?

the equation $\nu$^tx=0;
The problem is I see an example like following:
s1=[-3
1
0]

s2=[-10
0
1]
The nullspace contains all combination of s1 and s2. and produces the plane
x+3y+10z=0, perpendicular to row(1,3,10). And it lead to the result
Nullspace perpendicular to row space.

The "nullspace" and "row space"of what matrix?

I didn't know what the result means and
how its imply, could anyone give me any instruct about that, thanks.[/QUOTE]

 

[applechu]

Hi:
A is a matrix, A=\begin{bmatrix}1 & 3 & 10\\2 & 6 & 20\\3 & 9 & 30\end{bmatrix}
u=(1,2,3) , Thanks

 

[algebrat]

Hi:
A is a matrix, A=\begin{bmatrix}1 & 3 & 10\\2 & 6 & 20\\3 & 9 & 30\end{bmatrix}
u=(1,2,3) , Thanks

The trick is to row reduce, we get

\begin{bmatrix}1 & 3 & 10\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

So, let's think about this. To get to the reduced matrix, we can say we multiplied on the left by elementary matrices, leaving A to multiply any vector x on the other side. If x is is in the null space, in other words, if it solves Ax=0, wel then if and only if it's still in the null space after we multiplied by the elementary matrices. Therefore:

The nullspace of A is the same as that for the reduced echelon form.

For the row space, it's a little different, but the row reduced echelon form is helpful again. Since we were multiplying on the left, we changed the image, we change the column space. So the column space of A is not the same as the column space of the row reduced echelon form. But, the elementary matrices preserve linear dependence and linear independence, so the relations between columns is preserved. Blah blah, but what this means is:

The pivot columns of row reduced echelon form tell which columns of A span the column space of A.

So since the pivot columns of row reduced are just the leftmost column, the column space of A is spanned by it's left most column, that is (1,2,3).

For the null space of the row reduced echelon form, clearly it's where x+3y+10z=0. Trying z=0 and y=1 gives (-3,1,0), while y=0 and z=1 gives (-10,0,1).

Other problems may be more complicated, and ask for different forms of answers, but the two underlined sentences show how the row reduced form gets you closer to answers, and with a little cleverness, you can handle other problems.