- #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.

Last edited: