# Homework Help: Find the left nullspace of matrix

1. Apr 30, 2012

### sharks

The problem statement, all variables and given/known data

$$B= \displaystyle\left[ {\begin{array}{*{20}{c}} 1&0&-2&1 \\ 1&2&-2&3 \\ -2&1&3&0 \end{array}} \right]$$
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.
$$N(B)= \displaystyle\left[ {\begin{array}{*{20}{c}} 1 \\ -1 \\ 1 \\ 1 \end{array}} \right]$$
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:
$$B^T= \displaystyle\left[ {\begin{array}{*{20}{c}} 1&1&-2 \\ 0&2&1 \\ -2&-2&3 \\ 1&3&0 \end{array}} \right]$$
I then did the same steps as when finding the nullspace, by reducing to row echelon form, which is:
$$B^T= \displaystyle\left[ {\begin{array}{*{20}{c}} 1&1&-2 \\ 0&2&1 \\ 0&0&-1 \\ 0&0&1 \end{array}} \right]$$
To find the nullspace, as usual, solving: $A.x=0$ where matrix $x$ is a 3x1 column matrix, containing three variables: $x_1,\,x_2,\,x_3$.

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 $x_1=x_2=x_3=0$. 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: Apr 30, 2012
2. Apr 30, 2012

### HallsofIvy

While knowing how to solve problems by "row-reduction", etc. certainly simplifies calculations and eases the problem, don't lose track of the basic definitions!

The left null set of B is, by definition, the space of vectors $\begin{bmatrix}a & b & c & d\end{bmatrix}$, such that
$$\begin{bmatrix}a & b & c\end{bmatrix}\begin{bmatrix}1 & 0 & -2 & 1 \\ 1 & 2 & -2 & 3 \\ -2 & 1 & 3 & 0 \end{bmatrix}= \begin{bmatrix}a+ b- 2c \\ 2b+ c \\ -2a- 2b+ 3c \\ a+ 3b\end{bmatrix}= \begin{bmatrix}0 \\ 0\\ 0\\ 0\end{bmatrix}$$

That gives equations a+ b- 2c= 0, 2b+ c= 0, -2a+ 3b+ 3c= 0, and a+ 3b= 0. The "left null space" consists of all vectors [a, b, c] satisfying those equations.

3. Apr 30, 2012

### sharks

Hi HallsofIvy

According to the definition you've given, it should be possible to adapt it to: $B.N(B)=0$. Correct?

But, when multiplying, i get this vector: $[0, 4, 0]^T$. Does this mean that i've miscalculated the nullspace, N(B)?

4. Apr 30, 2012

### Staff: Mentor

B times any vector that belongs to its right nullspace has to come out to the zero vector.

Similarly, if x is in the left nullspace of B, then x*B has to be the zero vector.

5. Apr 30, 2012

### sharks

It seems that i've indeed miscalculated the right nullspace, N(B), as i don't get a zero vector.

6. Apr 30, 2012

### Ray Vickson

When you found the reduced echelon form U you essentially carried out row operations on B, so you would have gotten the results by multiplying B on the left by a lower-triangular matrix E; that is, U = EB. (You can find E by doing the same row operations on the 3x3 identity matrix that you did on B; alternatively, you could have used an augmented matrix A = [B|I] and do those row operations on A; the first 4 columns give you U and the last three give you E.)

Anyway, suppose you do have E (which is invertible). If y is a row vector we have yU = zB, where z = yE. So, if y is in the left-null space of U (and that is easy to find, because U is upper-triangular), then we get z from z = yE. The row vector z is in the left null-space of B. Conversly, if a row vector z is in the left null-space of B, the vector y = z*Inverse(E) is in the left null-space of U, so there is a 1-1 correspondence between these two null spaces.

RGV

7. Apr 30, 2012

### Staff: Mentor

No, your work is correct for the right nullspace. B * <1, -1, 1, 1>T = <0, 0, 0>T.

For the left nullspace, you want a vector x in R>3 such that x*B = <0, 0, 0, 0>T. You have it in post #1, except that it should be a vector in R3, not R4.

8. Apr 30, 2012

### sharks

So, i should aim for this:
$$N(B^T).B=0$$
These are the corresponding mxn sizes of the matrices: (1x3).(3x4)=(1x4).

I get these equations:
a+ b- 2c= 0 .......(1)
2b+ c= 0 ...........(2)
-2a+ 3b+ 3c= 0 ...(3)
a+ 3b= 0 ...........(4)

Now, to solve them...

I added (2) and (4) to get: a+5b+c=0 .....(5)
I then solved the system of 3 linear equations; (1), (3) and (5).
I get x=y=z=0. What am i doing wrong?

Last edited: Apr 30, 2012
9. Apr 30, 2012

### Staff: Mentor

Why do you think this is wrong? It's possible for a nullspace to consist of just a single vector - if so it's the zero vector.

10. Apr 30, 2012

### sharks

OK, so the left nullspace of B is: $\begin{bmatrix}0 & 0 & 0\end{bmatrix}$
According to what i learned, the left nullspace is a subspace of Rm, where m is the number of rows. Therefore, the left nullspace should be a vector in R1, not R3.

Here are the definitions from my book:
The nullspace and row space are subspaces of Rn.
The left nullspace and the column space are subspaces of Rm.
m and n, refer to the number of rows and columns respectively.

Last edited: Apr 30, 2012
11. May 1, 2012

### HallsofIvy

Well, the left null space is a subspace of Rm, a collection of vectors, not just a vector.

Te 0 vector is in any subspace so those are certainly true of the 0 vector.

Last edited by a moderator: May 1, 2012