# Find the left nullspace of matrix

• DryRun
In summary, the left nullspace of a matrix B is the space of vectors that when multiplied by B result in the zero vector. To find the left nullspace of B, you can use the same method as finding the right nullspace, by reducing B to row echelon form and solving the equation: Bx = 0. However, if all the variables in the resulting matrix are pivot variables, there are no free variables, and the left nullspace is a zero vector. Alternatively, you can find the left nullspace of B by finding the left nullspace of the reduced echelon form of B. This can be done by multiplying B on the left by a lower-triangular matrix E, and then using the row operations on
DryRun
Gold Member
Homework Statement

$$B= \displaystyle\left[ {\begin{array}{*{20}{c}} 1&0&-2&1 \\ 1&2&-2&3 \\ -2&1&3&0 \end{array}} \right]$$
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:
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.

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)?

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.

sharks said:
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)?

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

sharks said:
Homework Statement

$$B= \displaystyle\left[ {\begin{array}{*{20}{c}} 1&0&-2&1 \\ 1&2&-2&3 \\ -2&1&3&0 \end{array}} \right]$$
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.

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

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.

Mark44 said:
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.
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:
sharks said:
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?

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.

OK, so the left nullspace of B is: $\begin{bmatrix}0 & 0 & 0\end{bmatrix}$
Mark44 said:
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.

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

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

Last edited by a moderator:

## 1. What is the left nullspace of a matrix?

The left nullspace of a matrix is the set of all column vectors that, when multiplied by the matrix, result in a zero vector. In other words, it is the set of all solutions to the equation Ax = 0, where A is the given matrix.

## 2. How is the left nullspace of a matrix calculated?

The left nullspace of a matrix can be calculated by first finding the reduced row echelon form of the matrix. The columns of the matrix that correspond to the pivot columns in the reduced row echelon form will form the basis for the left nullspace.

## 3. What is the difference between the left and right nullspace of a matrix?

The left nullspace of a matrix contains vectors that satisfy the equation Ax = 0, while the right nullspace contains vectors that satisfy the equation xA = 0. In other words, the left nullspace is the set of solutions to the homogeneous system Ax = 0, while the right nullspace is the set of solutions to the system xA = 0.

## 4. Why is the left nullspace important in linear algebra?

The left nullspace is important in linear algebra because it represents the solutions to a system of linear equations that have no unique solution. It also provides insight into the linear dependence of the columns of a matrix, and can be used to determine the rank and nullity of a matrix.

## 5. How is the left nullspace used in applications?

The left nullspace has various applications in fields such as data analysis, image processing, and machine learning. It can be used to determine linear relationships between variables, identify redundant information in a dataset, and improve the performance of predictive models. Additionally, the left nullspace can be used to find a basis for the nullspace of a matrix, which is useful in solving inverse problems and optimization tasks.

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