- #1
DryRun
Gold Member
- 838
- 4
Homework Statement
[tex]B=
\displaystyle\left[ {\begin{array}{*{20}{c}}
1&0&-2&1 \\
1&2&-2&3 \\
-2&1&3&0
\end{array}} \right][/tex]
Find:
1. The nullspace of B.
2. The left nullspace of B.
The attempt at a solution
I was able to find the nullspace of B. but i can't figure out why the left nullspace isn't working out, although I'm quite sure that I'm following the right procedure.
[tex]N(B)=
\displaystyle\left[ {\begin{array}{*{20}{c}}
1 \\
-1 \\
1 \\
1
\end{array}} \right][/tex]
I'm wondering if there's a shortcut way of finding the nullspace, without working the whole problem over again, starting with the transpose of B?
So, to find left nullspace of B:
[tex]B^T=
\displaystyle\left[ {\begin{array}{*{20}{c}}
1&1&-2 \\
0&2&1 \\
-2&-2&3 \\
1&3&0
\end{array}} \right][/tex]
I then did the same steps as when finding the nullspace, by reducing to row echelon form, which is:
[tex]B^T=
\displaystyle\left[ {\begin{array}{*{20}{c}}
1&1&-2 \\
0&2&1 \\
0&0&-1 \\
0&0&1
\end{array}} \right][/tex]
To find the nullspace, as usual, solving: [itex]A.x=0[/itex] where matrix [itex]x[/itex] is a 3x1 column matrix, containing three variables: [itex]x_1,\,x_2,\,x_3[/itex].
But then I'm stuck, as all the 3 variables are pivot variables! So, there are no free variables. I am unable to express the pivot variables in terms of non-existent free variables. But i still persevered to solve the equation and i got [itex]x_1=x_2=x_3=0[/itex]. This would mean that the left nullspace of B is a zero 4x1 column matrix, which i believe is wrong.
At this point, i don't know what to do, as I've checked some worked-out examples and they managed to do it with different matrices, but with this particular matrix, i think i need to try a different method or maybe I'm doing something wrong.
[tex]B=
\displaystyle\left[ {\begin{array}{*{20}{c}}
1&0&-2&1 \\
1&2&-2&3 \\
-2&1&3&0
\end{array}} \right][/tex]
Find:
1. The nullspace of B.
2. The left nullspace of B.
The attempt at a solution
I was able to find the nullspace of B. but i can't figure out why the left nullspace isn't working out, although I'm quite sure that I'm following the right procedure.
[tex]N(B)=
\displaystyle\left[ {\begin{array}{*{20}{c}}
1 \\
-1 \\
1 \\
1
\end{array}} \right][/tex]
I'm wondering if there's a shortcut way of finding the nullspace, without working the whole problem over again, starting with the transpose of B?
So, to find left nullspace of B:
[tex]B^T=
\displaystyle\left[ {\begin{array}{*{20}{c}}
1&1&-2 \\
0&2&1 \\
-2&-2&3 \\
1&3&0
\end{array}} \right][/tex]
I then did the same steps as when finding the nullspace, by reducing to row echelon form, which is:
[tex]B^T=
\displaystyle\left[ {\begin{array}{*{20}{c}}
1&1&-2 \\
0&2&1 \\
0&0&-1 \\
0&0&1
\end{array}} \right][/tex]
To find the nullspace, as usual, solving: [itex]A.x=0[/itex] where matrix [itex]x[/itex] is a 3x1 column matrix, containing three variables: [itex]x_1,\,x_2,\,x_3[/itex].
But then I'm stuck, as all the 3 variables are pivot variables! So, there are no free variables. I am unable to express the pivot variables in terms of non-existent free variables. But i still persevered to solve the equation and i got [itex]x_1=x_2=x_3=0[/itex]. This would mean that the left nullspace of B is a zero 4x1 column matrix, which i believe is wrong.
At this point, i don't know what to do, as I've checked some worked-out examples and they managed to do it with different matrices, but with this particular matrix, i think i need to try a different method or maybe I'm doing something wrong.
Last edited: